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MAT3-ALG algebra 2006/7 (tnb) - lecture 0 1
MAT3-ALG algebra 2008-2009 — Toby Baileyhttp://student.maths.ed.ac.uk
lecture 0 preamble
problems
These are “easy” revision problems from year 2 Linear Algebra. It is essential
that you are on top of this earlier material and so you are strongly recommended
to do these exercises. Aim to do all of them by the end of week 2 at the latest.
Throughout, Pn denotes the vector space of polynomials of degree ≤ n in a
variable x and M denotes the vector space of 2× 2 real matrices.
done?problem 0.1 Which of the following are subspaces of the given vector space?
1. {x ∈ R3 | 2x1 − x2 + x3 = 1} ⊆ R3
2. {x ∈ R3 | x1 = 2x2} ⊆ R3
3. {P ∈ P3 |P(1) = 0} ⊆ P3
4. {A ∈M |AT = −A} ⊆M (the “T” denotes matrix transpose).
hint/solution 0.1 All except the first.
done?problem 0.2 For each of the examples in the previous question that is a
subspace give its dimension and write down a basis for the subspace.
hint/solution 0.2 The dimensions are 2, 3, 1 in order for the three examples
which are subspaces. Possible bases are
1. 210
,001
2.
x− 1, x2 − 1, x3 − 1
3. (0 1
−1 0
)
MAT3-ALG algebra 2006/7 (tnb) - lecture 0 2
done?problem 0.3 Calculate the coordinate matrix of x3 with respect to the basis
x3 − x2, x2 − x, x− 1, 1 of P3.
hint/solution 0.3 1
1
1
1
done?problem 0.4 Use the change of basis matrix to find the coordinate matrix of
x in the basis v1, v2 of R2 where
x =
(1
1
), v1 =
(1
−1
), v2 =
(2
1
).
hint/solution 0.4 Set P to be the matrix with v1, v2 as columns. Then the
coordinate matrix of x is P−1x. You can easily check your answer: if k, l are the
two entries in the coordinate matrix (which should be a column matrix) then
you should have x = kv1 + lv2.
done?problem 0.5 What is the span of a set S = {v1, . . . , vk} of vectors? What
does it mean for the set to be linearly independent?
done?problem 0.6 Let U = {x | x1 = 0} and V = {x | x2 = 0} be subspaces of R3.What is the sum U+ V of these subspaces? State the Dimension Theorem for
sums of subspaces and verify it in this example. Is this an example of a direct
sum?
hint/solution 0.6 The sum is all of R3 and the intersection is 1-dimensional.
The Dimension Theorem becomes 3 = 2+ 2− 1.
done?problem 0.7 Which of the following are linear maps?
1. T : M→ P2 where T(A) = the characteristic polynomial of the matrix A.
2. T : M→ R where T(A) = TraceA (Here “Trace” denotes the trace of a
matrix — the sum of the elements on the leading diagonal.)
3. T : P3 → P3 where T : p(x) 7→ p ′(x)
hint/solution 0.7 The second two are linear. The first is not because, for
example, if you multiply a matrix by two the charcteristic polynomial does not
get multiplied by two — even the identity matrix is a counterexample.
MAT3-ALG algebra 2006/7 (tnb) - lecture 1 3
done?problem 0.8 Define the kernel and image of a linear map. State the Rank
Theorem (a.k.a. “Rank-Nullity theorem”) for linear maps. For each of the
examples in the previous question that is linear, describe the kernel and image
and verify the theorem.
hint/solution 0.8 Example number two above is surjective and so has image
of dimension 1. The kernel is matrices of zero trace — a 3-dim subspace of M.
Example 3 has 2-dim image (polys of degree ≤ 2) and 1-dim kernel (constant
polys).
done?problem 0.9 Let A be an n × n matrix and let T : Rn → Rn be the linear
map T : x 7→ Ax. Which of the following conditions are equivalent to A having
an inverse?
1. detA 6= 0
2. ker T = {0}
3. im T = Rn
4. T is a bijection.
5. A is diagonalisable.
hint/solution 0.9 All except the last.
done?problem 0.10 Find the eigenvalues and eigenvectors of
A =
(1 2
1 1
).
Hence diagonalise A.
lecture 1 sets
problems
done?problem 1.1 Is it true that {} ⊂ B for all sets B?
hint/solution 1.1 It is true for all sets except B = {} for which it is false.
MAT3-ALG algebra 2006/7 (tnb) - lecture 1 4
done?problem 1.2 For the four properties of set algebra in §1.4.2 above, write down
(next to the original, perhaps) what they reduce to for a family of just two sets.
In each case, give a proof. For at least two of those not proved in lectures or
the text, write down a proof too of the version for families.
done?problem 1.3 Let S be a finite set. Write down a formula for the size of P(S).
Write down P(S) in the case of S being the empty set. Does the formula work
in this case?
hint/solution 1.3 ]P(S) = 2]S. For S empty, P({}) = {{}} so ]P({}) = 1
and the formula does hold since 20 = 1.
done?problem 1.4 Let L denote the set of (straight) lines through the origin in
R2. Let M denote the set of (straight) lines through (1, 1) in R2. How many
elements does L ∩M have?
hint/solution 1.4 One. (The line y = x.)
done?problem 1.5 + Hand-in for tutorial Suppose S = {x, y, z}. True or
False:
1. S ∈ P(S);
2. x ∈ P(S);
3. {x, y} ∈ S;
4. {x, y} ∈ P(S);
5. {{x, y}, {}} ∈ P(S);
6. {{x, y}, {}} ⊆ P(S);
7. {{x, y}, {}} ∈ P(P(S));
8. {{}} ∈ P(P(S)).
What is the size of P(P(S))?
hint/solution 1.5 T,F,F,T,F,T,T,T. P(P(S)) = 2]P(S) = 223
= 256
MAT3-ALG algebra 2006/7 (tnb) - lecture 1 5
done?problem 1.6 + Hand-in part 1 only for tutorial Use set algebra to
show that
1. A \ (B ∩ C) = (A \ B) ∪ (A \ C)
2. A \ (B \ C) = (A \ B) ∪ (A \ C ′)
hint/solution 1.6
1.
A \ (B ∩ C) = A ∩ (B ∩ C) ′
= A ∩ (B ′ ∪ C ′)= (A ∩ B ′) ∪ (A ∩ C ′)= (A \ B) ∪ (A \ C)
2.
A \ (B \ C) = A \ (B ∩ C ′)= A ∩ (B ∩ C ′) ′
= A ∩ (B ′ ∪ (C ′) ′)
= A ∩ (B ′ ∪ C)
= (A ∩ B ′) ∪ (A ∩ C)
= (A ∩ B ′) ∪ (A ∩ (C ′) ′)
(A \ B) ∪ (A \ C ′)
3.
A \ (B \A) = A \ (B ∩A ′)= A ∩ (B ∩A ′) ′
= A ∩ (B ′ ∪ (A ′) ′)
= A ∩ (B ′ ∪A)
= (A ∩ B ′) ∪ (AcapA)
= (A ∩ B ′) ∪A
MAT3-ALG algebra 2006/7 (tnb) - lecture 2 6
done?problem 1.7 Show that
A ∪ (A ∩ B) = A.
You will need to argue by showing that the LHS is a subset of the RHS and
that the RHS is a subset of the LHS. Use set algebra to deduce that also
A ∩ (A ∪ B) = A.
These two results (which can not be deduced from other identities of set algebra
that we have previously stated) are called the ”axioms of absorption”. Use them
together with set algebra as before to show that A \ (B \A) = A.
hint/solution 1.7 Let x ∈ A ∩ (A ∪ B). Then x ∈ A and x ∈ A ∪ B. So
x ∈ A and so LHS ⊆ RHS. Now let x ∈ A. Then x ∈ A ∪ B and so x ∈ LHS.
Now,
A \ (B \A) = A \ (B ∩A ′)= A ∩ (B ∩A ′) ′
= A ∩ (B ′ ∪ (A ′) ′)
= A ∩ (B ′ ∪A)
= Ausing the second form of the axiom of absorption.
lecture 2 cartesian products and functions
problems
done?problem 2.1 Let A,B be finite sets. What is the size of A× B?
hint/solution 2.1 ](A× B) = ]A ]B.
done?problem 2.2 Show that A × (B ∩ C) = (A × B) ∩ (A × C). (Hint: to do
this carefully, show that the LHS is a subset of the RHS and that the RHS is a
subset of the LHS.)
hint/solution 2.2 Let x ∈ A × (B ∩ C). Then x = (a, u) with a ∈ A and
u ∈ B∩C. Hence u ∈ B and u ∈ C. Hence (a, u) ∈ A×B and (a, u) ∈ A×C.
Thus x = (a, u) ∈ (A× B) ∩ (A× C). (We have shown that LHS ⊆ RHS.)
Now let x ∈ (A × B) ∩ (A × C). Then x ∈ A × B and so x = (a, b) with
a ∈ A and b ∈ B. But also x = (a, b) ∈ A×C and so b ∈ C. Hence b ∈ B×Cand so x ∈ A× (B ∩ C). (So also RHS ⊆ LHS and the proof is complete.)
MAT3-ALG algebra 2006/7 (tnb) - lecture 2 7
done?problem 2.3 Decide what relationship holds between the following pairs of
sets (one side is a subset of the other, or they are equal, or there is no relation).
Give a proof.
1. A× (B ∪ C) and (A× B) ∪ (A× C);
2. (A× B) ∩ (C×D) and (A ∩ C)× (B ∩D);
3. (A× B) ∪ (C×D) and (A ∪ C)× (B ∪D).
hint/solution 2.3
1. LHS = RHS
2. LHS = RHS
3. LHS ⊆ RHS but not generally equal. (You should find an example where
they are not equal.)
done?problem 2.4 Consider the function f : R→ R2 given by f : t 7→ (cos t, sin t).
The graph of f is a subset of R1 × R2 = R3. Sketch it.
hint/solution 2.4 The graph is {(t, cost, sin t) | t ∈ R} — a curve that screws
round the first axis direction. (It is a “helix”.)
done?problem 2.5 Consider the map f : R2 → R given by f : (x, y) 7→ √x2 + y2.
Describe each of the following using some combination of words, equations or
pictures (proofs not required):
1. f−1(2)
2. f−1([1, 2]) (here [1, 2] is the closed interval)
3. f−1(−1)
4. f(Z) where Z = {(x, y) | (x− 2)2 + (y− 2)2 = 2}
5. f(V) where V = {(x, y) |y > 0}
6. f(U) where U = {(x, y) | x2 − y2 = 1}
7. f(R2)
MAT3-ALG algebra 2006/7 (tnb) - lecture 2 8
hint/solution 2.5
1. The circle of radius 2 centred at the origin.
2. The annulus bounded by the circles of radius 1 and 2 centred at the origin
(including the boundaries).
3. {}
4. [1, 3]
5. (0,∞)
6. [1,∞)
7. [0,∞)
done?problem 2.6 + Hand-in for tutorial Let f : X→ Y be a function and
let A,B be subsets of X. Show that
f(A ∩ B) ⊆ f(A) ∩ f(B).
(Hint: You proof should begin: “Let y ∈ f(A∩B)” and should finish with “and
hence y ∈ f(A)∩f(B)”.) Give an example to show that “⊆” can not be replace
with equality.
hint/solution 2.6 Let y ∈ f(A ∩ B). Then there exists x ∈ A ∩ B such
that y = f(x). So x ∈ A and x ∈ V = B. Hence y = f(x) ∈ f(A) and
y = f(x) ∈ f(B). Hence y ∈ f(A) ∩ f(B).
Consider the squaring function f : R→ R with f(x) = x2. Let A = [−2,−1]
and B = [1, 2]. Then A ∩ B = {} and so f(A ∩ B) = {}. On the other hand,
f(A) = f(B) = [1, 4] and so f(A) ∩ f(B) = [1, 4].
done?problem 2.7 let X and Y be finite non-empty sets. Write down a formula for
the number of different functions from X to Y. Now consider the case where
one of X and Y is empty. How many functions are there in that case? Are the
results consistent with your previous formula. (Hint: For the second part use
the definition of function in terms of a subset of the cartesian product.)
MAT3-ALG algebra 2006/7 (tnb) - lecture 3 9
hint/solution 2.7 The number of functions is ] Y]X. If X is empty there is
just one function X → Y which has graph {} ⊆ X × Y = {}. Note that this
does (vacuously) satisfy the conditions for a subset of the product to define a
function. This agrees with the formula except where Y is also empty because
00 is generally taken to be undefined.
If Y is empty but X is not there are no functions X → Y since the only
subset of X× Y = {} is empty and that does not satisfy the condition to define
a function. This does agree with the formula.
lecture 3 more on functions
problems
done?problem 3.1 Give conditions on the size of the subsets f−1(y), y ∈ Y that
are characterize f being (a) injective; (b) surjective; (c) bijective.
hint/solution 3.1 The condition is that each set f−1(y) has: (a) no more
than one element; (b) at least one element; (c) exactly one element.
done?problem 3.2 + Hand-in for tutorial
1. Let g ◦ f be injective. Show that f is injective. (Hint: You may find it
best to show that if f is not injective then g ◦ f is not injective.
2. If g ◦ f is injective, does g have to be injective? Give a proof or a
counterexample.
3. What exactly can be deduced if we know that g ◦ f is surjective?
hint/solution 3.2 Suppose f : X→ Y and g : Y → Z.
1. Suppose f is not injective. Then there exist u 6= v in X such that f(u) =
f(v). Then
(g ◦ f)(u) = g(f(u)) = g(f(v)) = (g ◦ f)(v)
and so g ◦ f is not injective.
2. No. Consider f : [0, 1] → [−1, 1] where f : x 7→ x. Let g : [−1, 1] →[−1, 1] where g : x 7→ x2.
3. If g ◦ f is surjective then g is surjective. Proof: Let z ∈ Z. Since g ◦ f is
surjective there exists x ∈ X such that (g ◦ f)(x) = z. Now, f(x) ∈ Y and
g(f(x)) = z and so g is surjective.
MAT3-ALG algebra 2006/7 (tnb) - lecture 3 10
done?problem 3.3 Let f : A→ A be a map and suppose that f◦ f = f. What extra
condition on f allows us to deduce that f is the identity map?
hint/solution 3.3 f ◦ f = f means that for all a ∈ A we have f(f(a)) = f(a)
and so we can deduce precisely that f acts as the identity on the image of f.
Thus the extra condition needed is that f is surjective.
done?problem 3.4
1. Let f : A → B be a map. Suppose there exists a map g : B → A such
that g ◦ f = IA : A→ A. Show that f is injective.
2. Suppose f : A→ B is injective. Deduce that there exists a map g : B→ A
such that g ◦ f = IA : A→ A.
3. Under what circumstances is the map g in the previous part unique?
hint/solution 3.4
1. Let f(u) = f(v). Then g(f(u)) = g(f(v)) and so (since g ◦ f = IA) we
have u = v and so f is injective.
2. Let f : A→ B be injective. Fix a ∈ A. Define a map g : B→ A by
g(b) =
{a if b 6∈ im f,
u if b = f(u)..
(We note that each b ∈ im f is of the form f(u) for one and only one
u ∈ A since f is injective.) Then g ◦ f(u) = g(f(u)) = u for all u ∈ Aand so g ◦ f = IA.
3. The choice of a makes g non-unique unless f is surjective and hence a
bijection, in which case g has to be the usual inverse.
done?problem 3.5 State and prove results analogous to the previous exercise that
involve a map h : B→ A such that f ◦ h = IB?
done?problem 3.6 Show that there exists an injection A → B if and only if there
exists a surjection B→ A.
MAT3-ALG algebra 2006/7 (tnb) - lecture 4 11
done?problem 3.7 (Harder!) Let S be a set. Show that there can not exist a
surjection f : S → P(S). You might proceed as follows. Suppose there is such
a surjection f. Now consider the subset A ⊆ S defined by
A = {x ∈ S | x 6∈ f(x)}.
Deduce that A itself is not in the image of f for a contradiction.
done?problem 3.8 (Optional!) The Schroder-Bernstein Theorem states that if there
are injections A → B and B → A then there exists a bijection A → B. It is
not entirely trivial. Find a proof (Halmos’s “Naive Set Theory” or perhaps the
web) and understand it!
lecture 4 relations and quotients
problems
done?problem 4.1 + Hand-in for tutorial Show that u ∼ v if and only if
u−v ∈ Z defines an equivalence relation on R. Describe [x] for this equivalence
relation and give a set of representatives.
hint/solution 4.1
1. x− x = 0 ∈ Z and so x ∼ x.
2. Let x ∼ y. Then x− y = k ∈ Z. Then y− x = −k ∈ Z and so y ∼ x.
3. Let x ∼ y and y ∼ z. Then there exist k, l ∈ Z such that x − y = k and
y− z = l. Then x− z = (x− y) + (y− z) = k+ l ∈ Z and so x ∼ z.
So ∼ is an equivalence relation. The equivalence class of x is
[x] = {x+ k |k ∈ Z}.
A set of representatives is [0, 1).
done?problem 4.2 Does a ∼ b ⇐⇒ a + 2b = 3k where k ∈ Z define an
equivalence relation on Z? (Check carefully and investigate if you are not sure
— don’t just guess!)
MAT3-ALG algebra 2006/7 (tnb) - lecture 4 12
hint/solution 4.2 Yes it does. To check the three axioms:
1. a+ 2a = 3a and so a ∼ a.
2. Let a ∼ b. Then there exists k such that a + 2b = 3k and so b + 2a =
3(a+ b) − (a+ 2b) = 3(a+ b+ k) and so b ∼ a.
3. Let a ∼ b and b ∼ c. Then there exists k, l such that a + 2b = 3k and
b+ 2c = 3l. So a+ 2c = (a+ 2b) + (b+ 2c) − 3b = 3(k+ l− b) and
so a ∼ c.
Alternatively, notice that a+ 2b is a multiple of 3 iff a−b is and so this is just
a familiar example in disguise.
done?problem 4.3 + Hand-in for tutorial Let a be the vector (1, 1) ∈ R2.Show that
x ∼ y ⇐⇒ x− y = λa for some λ ∈ R
defines an equivalence relation on R2. Sketch the equivalence classes and show
that R = {(x, y) | x+ y = 0} is a set of representatives.
hint/solution 4.3
• x− x = 0 = 0a and so x ∼ x
• Let x ∼ y. Then x − y = λa for some λ ∈ R. Then y − x = (−λ)a and
so y ∼ x.
• Let x ∼ y and y ∼ z. Then x−y = λa and y−z = µa for some λ, ν ∈ R.
Then x− z = (x− y) + (y− z) = (λ+ µ)a and so x ∼ z.
Thus ∼ is an equivalence relation. The equivalence classes are all lines parallel
to x = y. The line R intersects every such line in a single point and so is a set
of representatives.
MAT3-ALG algebra 2006/7 (tnb) - lecture 4 13
done?problem 4.4
(a) Show that if x ∈ R2 is non-zero then there exists an invertible 2× 2 matrix
A such that Ae1 = x where e1 is the first standard basis vector in R2.
(b) Use the above to show that given two non-zero vectors x, y ∈ R2 there
exists an invertible 2 × 2 matrix P such that y = Px. (Hint: take x to e1and then e1 to y.)
(c) Let x ∼ y iff there exists an invertible 2 × 2 matrix A such that y = Ax.
Show that this defines an equivalence relation on R2.
(d) What are the equivalence classes for this equivalence relation? Give a set
of representatives. How many elements does R2/ ∼ have?
hint/solution 4.4
(a) Let z be any vector such that x, z are linearly independent. Then let A be
the matrix with x, z as columns. Then Ae1 = x and A is invertible.
(b) Let A be as above and let Be1 = y with B invertible. Then y = Be1 =
BA−1x and BA−1 is invertible with inverse AB−1.
(c) (1) x = Ix and so x ∼ x
(2) Let x ∼ y then there exists invertible A with y = Ax. Then x = A−1y
and so x ∼ y.
(3) Let x ∼ y and y ∼ z then there exists invertible A,B with y = Ax and
z = By. Then z = (BA)x and of course BA is invertible (with inverse
A−1B−1). Thus x ∼ z.
(d) There are thus two equivalence classes: all the non-zero vectors in one and
just the zero vector in the other. A set of representatives could be {e1, 0}.
The quotient has two elements.
done?problem 4.5 Consider the set X = {(a, b) |a, b ∈ Z and b 6= 0} (so an
element of X is a pair of integers with the second one non-zero). Show from
the definition that
(a, b) ∼ (k, l) ⇐⇒ al = bk
defines an equivalence relation on X.
MAT3-ALG algebra 2006/7 (tnb) - lecture 5 14
hint/solution 4.5
• (a, b) ∼ (a, b) since ab = ab.
• The relation is obviously symmetric under exchange of (a, b) and (c, d).
• Suppose (a, b) ∼ (c, d) and (c, d) ∼ (e, f). Then ad = bc and cf = de.
Multiply the first through by ef to get
afde = cfbe.
Now, assuming cf = de 6= 0 we can cancel to get af = be and so
(a, b) ∼ (e, f). If on the other hand cf = de = 0 then a, c, e are all zero
(remembering that b, d, f are all nonzero) and the result follows in this
case too.
lecture 5 the first isomorphism theorem (FIT) for sets
problems
done?problem 5.1 Prove that the operation of multiplication is well-defined in Zn.
hint/solution 5.1 Taking the equivalence relation ∼ on Z so that Zn = Z/ ∼,
let a ∼ a ′ and b ∼ b ′. Then there exist k, l ∈ Z such that a ′ − a = kn and
b ′ − b = ln. Then a ′b ′ = (a+ kn)(b+ ln) = ab+n(bk+ la+ kln) and so
ab ∼ a ′b ′.
done?problem 5.2 + Hand-in for tutorial Consider the equivalence relation
x ∼ y ⇐⇒ |x| = |y| on Z. Use the absolute value function (i.e. the modulus
function) and FIT for sets to deduce that the quotient Z/ ∼ can be identified
with N ∪ {0}.
hint/solution 5.2 Consider the surjection f : Z→ N ∪ {0} where f : n 7→ |n|.
Then x ∼ y iff f(x) = f(y). FIT for sets implies there is a canonical bijection
Z/ ∼→ N ∪ {0}.
done?problem 5.3 Consider the equivalence relation ∼ on the set X = {(a, b) |a, b ∈Z and b 6= 0} as in the problem for lecture 4. Show that setting
[(a, b)] + [(c, d)] = [(ad+ bc, bd)], [(a, b)] ∗ [(c, d)] = [(ac, bd)]
is a well-defined “addition” and “multiplication” on X/ ∼.
MAT3-ALG algebra 2006/7 (tnb) - lecture 6 15
hint/solution 5.3 Suppose that [(a, b)] = [(a ′, b ′)] and [(c, d)] = [(c ′, d ′)].
Then ab ′ = a ′b and cd ′ = c ′d. Then
(ad+ bc)b ′d ′ = adb ′d ′ + bcb ′d ′ = a ′bdd ′ + bc ′b ′d = bd(a ′d ′ + b ′c ′)
and so [(ad + bc, bd)] = [(ad + bc, bd)] and so addition is well defined. The
case of multiplication is similar (but easier).
done?problem 5.4 This continues from the previous problem. Define a map f :
X → Q (where Q is the rational numbers) by f : (a, b) 7→ a/b. Use FIT
for sets to deduce that X/ ∼ can be identified with Q. Note by the way that
the previous exercise can be taken to be a definition of Q and its arithmetic
operations which does not use the idea of fractions or real numbers.
hint/solution 5.4 Consider the map f : X→ Q defined by f : (a, b) 7→ a/b.
Then
f(a, b) = f(k, l) ⇐⇒ a/b = k/l ⇐⇒ (a, b) ∼ (k, l).
The map is clearly surjective too and so by FIT for sets we can identify the set
of equivalence classes with Q.
done?problem 5.5 Let S1 denote the unit circle, thought of as the unit-modulus
complex numbers. Consider the map f : R → S1 defined by f : x 7→ e2πxi.
Show that the equivalence relation on R defined by u ∼ v ⇐⇒ f(u) = f(v) is
that u and v are equivalent if and only if they differ by an integer. Use FIT for
sets to show that R/ ∼ can be identified with S1.
lecture 6 fields and n-dimensional space
problems
done?problem 6.1 + Hand-in for tutorial Find the multiplicative inverses
of the non-zero elements in Z7. (Just experimenting is probably easier than
using the Euclidean algorithm.)
hint/solution 6.1 By experiment, 2.4 = 1, 3.5 = 1 6.6 = 1 and so 2, 4 are
mutually inverse as are 3, 5. The element 6 is its own inverse.
done?problem 6.2 Show that if L ⊆ K is a subfield then 1, 1+ 1, 1+ 1+ 1, . . . are
all elements of L. (It is tempting to call these 1, 2, 3, . . . but note that (e.g. in
Zp) that they are not necessarily all distinct.) Deduce that Zp does not have
any subfields (other than itself).
What do you think the smallest subfield of R is?
MAT3-ALG algebra 2006/7 (tnb) - lecture 6 16
hint/solution 6.2 We know 1 ∈ L and so all the given elements are as they
are all obtained by repeated addition. If K = Zp then this exhausts all the
elements of Zp so every subfield contains (and therefore is) the whole field.
By the above argument, every subfield of R contains N and hence (closure
under additive inverses) Z. Then consider closure under division — we deduce
every subfield contains all of Q. Thus Q is the smallest subfield of R.
done?problem 6.3 + Hand-in for tutorial Do the following equations have
solutions in the fields C,R,Q,Z3,Z2?
x2 + 1 = 0, x2 − x− 1 = 0
Note: what this means in each case is this: is there an element of the given
field such that if you substitute it in to this equation and do all the arithmetic
in that field then you get zero? The answers for the first three fields should be
easy from elementary background knowledge. The last two fields have very few
elements and so you can just experiment.
hint/solution 6.3 The first equation has solutions in C but not R or Q. In
Z3 there are no solutions but in Z2 then x = 1 is a solution since 12+1 = 2 = 0
in Z2.The second equation has solutions in C and R but not Q. It has no solutions
in the two finite fields.
done?problem 6.4 Show that Q[√2] (definition in notes) is a subfield of R.
hint/solution 6.4 Q[√2] := {a+ b
√2 |a, b ∈ Q} ⊆ R. Clearly 0, 1 ∈ Q[
√2]
and if u, v ∈ Q[√2] then so are −u, u + v, uv (the latter by multiplying out).
Also, if u = a+b√2 then so is 1/(a+b
√2) as can be seen by multiplying top
and bottom by a− b√2.
done?problem 6.5 Find all the vectors in Z23 that are scalar multiples of
a =
(1
2
).
hint/solution 6.5
0a =
(0
0
), 1a = a =
(1
2
), 2a =
(2
1
),
done?problem 6.6 Find the vectors in the subspaces in §6.4.3.
MAT3-ALG algebra 2006/7 (tnb) - lecture 7 17
hint/solution 6.6
• The given 1-dim subspace of Z23 contains just the vectors(0
0
),
(1
1
),
(2
2
).
• The given 2-dim subspace of Z32 contains just the vectors
0a+0b =
000
, 1a+0b =
110
, 0a+1b =
111
, 1a+1b =
001
.lecture 7 vector spaces — revision
7.1 setting
Vector spaces over a general field F.
problems
done?problem 7.1 + Hand-in for tutorial In Z32 find all the vectors in
Span(x, y) where
x =
110
, y =
011
hint/solution 7.1 Since there are only two scalars, there are just four vectors:
0 = 0x+ 0y, x = 1x+ 0y, y = 0x+ 1y, x+ y = 1x = 1y, the last is101
.done?problem 7.2 Find all the vectors in the subspace V ⊆ Z32 given by V = {x ∈
Z32 | x1 + x2 + x3 = 0}.
hint/solution 7.2 You should find four vectors - it’s the zero vector together
with every vector containing two 1’s and one 0.
MAT3-ALG algebra 2006/7 (tnb) - lecture 7 18
done?problem 7.3 Show that in Z35 the vectors113
,202
,430
are linearly dependent.
hint/solution 7.3 Various methods: you could calculate the determinant of
the matrix with these columns, either by row expansion or Gauusian elimination,
and show that it is zero. Alternatively, hunt around for a linear dependency: in
fact the last column is the first plus twice the second.
done?problem 7.4 Give a basis of the subspace of Z35 defined by the equation
x1 + x2 + x3 = 0. What is the dimension of this subspace? How many vectors
are there in this subspace? Find the coordinate matrix of the vector
v =
433
in you chosen basis.
hint/solution 7.4 In this subspace, the general element is x1
x2
−(x1 + x2)
= x1
104
+ x2
014
and so the two vectors on the right-hand side are a basis. The dimension is
therefore 2. The coordinate matrix of v in this basis is(4
3
).
The subspace has 25 vectors: since it is two-dimensional an element is specified
by choosing two scalars.
done?problem 7.5 + Hand-in for tutorial Complete the following sentence
without using any terms from linear algebra. “If R were finite-dimensional as a
vector space over Q it would mean that there exist a finite number r1, . . . , rnof . . . such that every . . . could be written as . . . ”
MAT3-ALG algebra 2006/7 (tnb) - lecture 8 19
hint/solution 7.5 If R were finite-dimensional as a vector space over Q it
would mean that there exist a finite number r1, . . . , rn of real numbers such
that every real number r could be written as
r = q1r1 + · · ·+ qnrn
with the qk all rational.
done?problem 7.6 + Hand-in for tutorial There are seven different non-
zero vectors in Z32 and hence (since the only scalars are {0, 1}) there are seven
different 1-dimensional subspaces. There are also seven different linear equations
of the form
λ1x1 + λ2x2 + λ3x3 = 0, λj ∈ Z2
apart from the trivial one with all the λj being zero. Each
of these defines a different 2-dimensional subspace of Z32.In the picture there are seven blobs, and seven lines
(six straight together with a circle). Label each blob
with a different 1-dimensional subspace and each line
with a 2-dimensional subspace in such a way that a
blob is on a line iff the 1-dimensional subspace lies in-
side the 2-dimensional one (i.e. iff the vector satisfies
the equation). (Note by the way that this config-
uration has the property that through every pair of
points there is a unique line and every pair of lines
meet in precisely one point. It is an example of a
”finite projective plane”.)
hint/solution 7.6 One solution is to set the vertices of the triangle equal to
the standard basis vectors and the edges can then be given by the equations
x = 0, y = 0, z = 0. The centre of the triangle can be (111)T and the other
three non-zero vectors can be put in the middle of the sides. The circle has
equation x+ y+ z = 0.
done?problem 7.7 u Challenge How many 2-dimensional subspaces does Z42have? You might want to think along the lines of defining such a subspace
by choosing a non-zero vector and then choosing another vector that is not a
multiple of it — their span determines a subspace. Count how many ways there
are of doing this and then work out how many times each subspace has been
counted.
MAT3-ALG algebra 2006/7 (tnb) - lecture 8 20
lecture 8 linear maps — revision
8.1 setting
Linear maps T : U → V where U,V are finite-dimensional vector spaces over
the same field F.
problems
done?problem 8.1 + Hand-in for tutorial Consider the linear map T :
Z32 → Z32 with matrix 1 1 0
1 0 1
0 1 1
.Find all the vectors in ker T and im T .
hint/solution 8.1 The determinant is zero. There are two linearly indepen-
dent colums and so the rank is 2. The image is therefore all the vectors that
are linear combinations (with Z2 coefficients) of (say) the first two columns. In
fact, the image is precisely the three vectors that are the columns.
By the rank theorem, the kernel is 1-dimensional. A vector in the kernel is111
.Thus the kernel is this vector and the zero vector.
done?problem 8.2 How many linear maps T : Z22 → Z22 are there? How many of
them are invertible? (Hint: equivalently, how many 2 × 2 matrices are there
with entries in Z2 and how many have inverses? Remember that a 2× 2 matrix
has an inverse if its first column is non-zero and the second column is not a
multiple of the first.)
Let A be such an invertible matrix and let
e1 =
(1
0
), e2 =
(0
1
), e3 =
(1
1
).
Show that Aej 6= Aek unless j = k and that Aej 6= 0. Deduce that multi-
plication by A permutes the three non-zero vectors in Z22. Find explicitly the
permutation given by each of the six invertible matrices.
MAT3-ALG algebra 2006/7 (tnb) - lecture 8 21
hint/solution 8.2 There are two choices for each entry and so there are
24 = 16 linear maps. To count the invertible ones, there are 3 choices for a
non-zero first column. Then for each such choice there are 2 non-zero vectors
that are not multiples of the first for the second column. Hence there are
2× 3 = 6 such matrices.
The linear map given by an invertible A is injective and so the given con-
ditions hold and thus A permutes the three vectors. The images of e1, e2 are
the columns of the matrix and so we see easily that permutations are (in cycle
notation)(1 0
0 1
)! identity perm,
(0 1
1 1
)! (123),
(1 1
1 0
)! (132)(
0 1
1 0
)! (12),
(1 0
1 1
)! (13),
(1 1
0 1
)! (23)
done?problem 8.3 Let Pj[Z3] denote the vector space of polynomials of degree ≤ jwith coefficients in Z3.
1. State the dimension of Pj[Z3].
2. Show that 1+ x, x+ x2, x2 is a basis for P2[Z3].
3. Show that T : P2[Z3] → P3[Z3] where T : p(x) 7→ (x + 2)p(x) is a linear
map.
4. Calculate the matrix of T with respect to the basis above for P2[Z3] and
the basis 1, x, x2, x3 for P2[Z3].
MAT3-ALG algebra 2006/7 (tnb) - lecture 9 22
hint/solution 8.3
1. The dimension id j+ 1.
2. Suppose
λ1(1+ x) + λ2(x+ x2) + λ3x2 = 0.
The constant term implies λ1 = 0, then the linear term implies λ2 = 0
and the quadratic term then implies λ3 = 0. So the vectors are linearly
independent and since the space is 3-dimensional they form a basis. (Al-
ternatively, show that the vectors span and then appeal to dimension, or
show that the span and are linearly independent (and omit the argument
by dimension.)
3.
T(p(x)+q(x)) = (x+2)(p(x)+q(x)) = (x+2)p(x)+(x+2)q(x) = T(p(x))+T(q(x))
and also
T(λp(x)) = (x+ 2)λp(x) = λ(x+ 2)p(x) = λT(p(x)).
4. T(1+x) = (x+2)(x+1) = x2+2 and T(x+x2) = (x+2)(x2+x) = x3+2x
and T(x2) = (x+ 2)x2 = x3 + 2x2. So taking the coordinates of each in
the given basis we get that the matrix is2 0 0
0 2 0
1 0 2
0 1 1
.lecture 9 invariant subspaces and block matrices
9.1 setting
Linear maps V → V where V is a (normally finite-dimensional) vector space
over a field F.
problems
done?problem 9.1 Let T : V → V be linear. Prove ker T is an invariant subspace.
hint/solution 9.1 Let v ∈ ker T . Then Tv = 0 ∈ ker T . So ker T is invariant.
MAT3-ALG algebra 2006/7 (tnb) - lecture 9 23
done?problem 9.2 Consider rotations (about the origin) and reflections (in lines
through the origin) in the plane. Describe all 1-dimensional invariant subspaces.
hint/solution 9.2 For rotations there are none unless θ = 0, Pi in both of
which cases every 1-dimensional subspace is invariant.
done?problem 9.3 Show that every 1-dimensional invariant subspace is the span of
an eigenvector.
hint/solution 9.3 Let U ⊆ V be invariant under T and 1-dimensional. Then
U = Span(u) for some u. Then Tu ∈ U and so Tu = λu for some λ and so u
is an eigenvector.
done?problem 9.4 Consider matrix multiplication of block upper-triangular matri-
ces. Using the notation of §9.3.1, show that M is invertible if and only if the
blocks A and D are invertible.
hint/solution 9.4 The block upper-triangular case is
MM ′ =
(A B
0 D
)(A ′ B ′
0 D ′
)=
(AA ′ AB ′ + BD ′
0 CB ′ +DD ′
).
So for MM ′ = I we need AA ′ = I and DD ′ = I and so A and D must
be invertible. Conversely, suppose A,D are invertible. Then if we set B ′ =
−A−1BD ′ then we will have MM ′ = I and so M is invertible.
done?problem 9.5 + Hand-in for tutorial Show that T : V → V has an
invariant subspace of dimension l if and only if V has a basis with respect to
which the matrix of T is block lower-triangular.
Deduce that if the matrix M is n×n block upper-triangular and P is n×nsuch that Pij = 1 when i+ j = n+ 1 and zero otherwise, then P−1MP is block
lower-triangular.
hint/solution 9.5 If a linear map has block lower-triangular matrix with re-
spect to a basis, and the second diagonal block is l× l then the span of the last
l basis vectors is invariant. The proof is the same as the block upper-triangular
case.
The matrix P is the change of basis matrix from a basis to the same basis
in reverse order. Thus changing basis with it will convert block upper-triangular
to block lower-triangular and vice versa.
MAT3-ALG algebra 2006/7 (tnb) - lecture 10 24
done?problem 9.6 Let T : V → V be a linear map and suppose there is a flag {Vk}
in V such that T(Vk) ⊆ Vk−1 for k = 1, . . . , n. Show that Tn = 0.
Show that there exists such a flag for T if and only if there exists a basis
for V such that the matrix of T is strictly upper-triangular (meaning i ≥ j =⇒Tij = 0).
hint/solution 9.6 Suppose there exists a basis such that the matrix of T is
strictly upper triangular. Then setting Vk to be the span of the first k basis
vectors we have such a flag. Conversely, given such a flag, construct a basis
such that Vk is the span of the first k basis vectors. Then the matrix is strictly
upper-triangular.
lecture 10 quotients and the 1st isomorphism theo-rem
10.1 setting
Vector spaces over a field F, which may be assumed to be finite-dimensional
(and needs to be when we make statements about dimension).
problems
done?problem 10.1 Let V ⊆ X be a subspace. Check that x ∼ y ⇐⇒ x− y ∈ Vdoes define an equivalence relation on X
hint/solution 10.1 Just check the three axioms.
done?problem 10.2 Check that the addition and scalar multiplication defined in
§10.3.3 is well-defined.
hint/solution 10.2 For addition, let [x] = [x ′] and [y] = [y ′]. Then there
exist u, v ∈ V such that x− x ′ = u and y− y ′ = v. Then
(x ′ + y ′) − (x+ y) = u+ v ∈ V
and so [x + y] = [x ′ + y ′] and the addition is well-defined. The argument for
scalar multiplication is similar.
done?problem 10.3 Check that the distributive law (λ(x + y) = λx + λy for all
vectors x, y and a scalar λ) holds in X/V.
MAT3-ALG algebra 2006/7 (tnb) - lecture 10 25
hint/solution 10.3
λ([x] + [y]) = λ([x+ y])
= [λ(x+ y)]
= [λx+ λy]
= [λx] + [λy]
= λ[x] + λ[y]
done?problem 10.4 Write out a careful proof of the fact that P : X→ X/V where
P : x 7→ [x] is linear, surjective and has kernel V.
hint/solution 10.4
• It is linear because
P(λx+ µy) = [λx+ µy] = λ[x] + µ[y] = λPx+ µPy.
• It is surjective because if [x] ∈ X/V then P(x) = [x].
•Px = 0 = P0 ⇐⇒ [x] = [0] ⇐⇒ x ∼ 0 ⇐⇒ x− 0 = x ∈ V.
So ker P = V.
done?problem 10.5 + Hand-in for tutorial Suppose T : X→ Y is a linear
map and that V ⊆ X is a subspace such that V ⊆ ker T . Define a linear map
T : X/V → Y (checking that the map you have defined is linear) such that the
following diagram commutes. (The vertical map is the usual one.) Find the
dimension of the kernel of T in terms of the dimensions of V and ker T . (Hint:
apply the rank theorem to T and T .)
XT
−→ Y↓ ↗ eTX/V
MAT3-ALG algebra 2006/7 (tnb) - lecture 10 26
hint/solution 10.5 Define a map T : X/V → Y by T : [x] 7→ Tx. This is
well-defined since if [x] = [x ′] then x−x ′ = v ∈ V and then T(x−x ′) = 0 since
v ∈ ker T . The map is linear since
T(λ[x] + µ[y]) = T([λx+ µy]) = T(λx+ µy) = λTx+ µTy = λT [x] + µT [y].
(N.B. If you’re reading this solution you need to understand why each one of
that chain of equalities is true.)
The triangle commutes since if x ∈ X then the vertical map takes you to [x]
and T([x]) = T(x) by definition of T .
To find the dimension, the rank theorem for T gives
dimX− dim ker T = dim im T
and that for T gives
dimX/V − dim ker T = dim im T .
Now clearly im T = im T and so subtracting and using dimX/V = dimX−dimV
we get dim ker T = dim ker T − dimV.
done?problem 10.6 (Harder!) Let U ⊆ V ⊆ X be subspaces of X.
(a) Show that there is a canonical linear map S : V/U→ X/U.
(b) Show that S is injective.
(c) Show that there is a canonical linear map T : X/U→ X/V. Show that T is
surjective.
(d) Show that ker T = imS.
So, if we identify V/U with its image in X/U (reasonable, since S is injective)
then we can deduce that there is an isomorphism
X/U
V/U→ X/V.
(Notation suggestion: write x ∼U y if x − y ∈ U and write [x]U for the
equivalence class under this relation. Similarly for V.)
MAT3-ALG algebra 2006/7 (tnb) - lecture 10 27
hint/solution 10.6
(a) Since if v ∈ V then v ∈ X we can define S by S : [v]U 7→ [v]U.
(b) Let S([v]U) = S([v ′]U). Then v− v ′ ∈ U and so [v]U = [v ′]U.
(c) Define T by T : [x]U 7→ [x]V . This is well-defined because if [x]U = [x ′]Uthen x − x ′ = u ∈ U ⊆ V. So [x]V = [x] ′V . Clearly every [x]V is obtained
in this way so T is surjective.
(d) T [x]V = 0 iff x ∈ V and so ker T = imS.
MAT3-ALG algebra 2006/7 (tnb) - lecture 11 28
lecture 11 quotients and linear maps
11.1 setting
Finite-dimensional vector spaces over a field F. For the result that for every
linear map T : V → V there exists a basis with respect to which the matrix of
T is upper-triangular, the field is assumed to be C.
problems
done?problem 11.1 + Hand-in for tutorial The 3× 3 real matrix
A =
1 0 −1
0 −2 0
1 0 0
has a single real eigenvalue. Find a real invertible matrix P such that P−1AP is
block upper-triangular.
hint/solution 11.1 The eigenvalue is −2 with eigenvector the second stan-
dard basis vector. Thus P can be any invertible 3 × 3 matrix with that as its
first column. For example,
P =
0 1 0
1 0 0
0 0 1
.done?problem 11.2 Consider the differentiation map D : P3 → P3 where as usual
Pn is the vector space of polynomials of degree ≤ n in a variable x. Show
that D gives rise to a linear map D : P3/V → P3/V where V is the subspace
of constant polynomials. What is the matrix of D with respect to the basis
[x], [x2], [x3] of P3/V?
hint/solution 11.2 The map D is such that D(V) = {0} ⊆ V and so there is
an induced linear map D : P3/V → P3/V . The matrix of D with respect to the
basis 1, x, x2, x3 is 0 1 0 0
0 0 2 0
0 0 0 3
0 0 0 0
and so the bottom-right 3 × 3 square is the matrix of D with respect to the
given basis.
MAT3-ALG algebra 2006/7 (tnb) - lecture 12 29
done?problem 11.3 + Hand-in for tutorial Let J : R2 → R2 be rotation
anticlockwise by a rightangle. Does there exist a basis for R2 with respect to
which the matrix of J is upper-triangular? If not, explain where the proof we
gave for complex vector spaces breaks down.
hint/solution 11.3 No. The reason is that there are no real eigenvectors,
which in turn is because not all real polynomials have real roots.
done?problem 11.4 The aim of this question is to prove that if T : V → V is
nilpotent (so that Tk = 0 for some k), then there exits a basis for V such that
the matrix of T is strictly upper-triangular. (The proof in the notes applies only
if V is complex.) Let k be the least k such that Tk = 0.
1. Show that if T is nilpotent then T has zero as an eigenvalue.
2. Let U ⊆ V be an invariant subspace for nilpotent T . Show that T :
V/U→ V/U is also nilpotent.
3. Now argue analogously to the proof of the main theorem in the notes for
this lecture.
hint/solution 11.4
1. Since Tk−1 6= 0 there must be a vector v ∈ V such that u := Tk−1v 6= 0.
Then Tu = 0 and this is an eigenvector with eigenvalue 0.
2. Suppose T : V → V and Tk = 0. Let [v] ∈ V/U. Then Tk[v] = [Tkv] =
[0] = 0. So Tk = 0.
3. The proof now is completely analogous to that in the notes.
lecture 12 linear maps V → V — eigenspaces
12.1 setting
Linear maps T : V → V where V is finite-dimensional over a field F.
problems
done?problem 12.1 Write out the details of the proof of §12.4.3.
MAT3-ALG algebra 2006/7 (tnb) - lecture 12 30
hint/solution 12.1 By definition of the sum, if v ∈ V then we can write
v = u1 + · · ·+ uk, uj ∈ Uj.
Suppose that also
v = u ′1 + · · ·+ u ′k, u ′j ∈ Uj.
Then subtracting we get
0 = (u1 − u ′1) + · · ·+ (uk − u ′k)
and so by the definition of “direct sum” uj − u′j = 0 for all j.
done?problem 12.2 + Hand-in for tutorial Let x1, . . . , xk be non-zero
vectors and let Uj = Span(xj). Show that the sum of the subspaces Uj is direct
if and only if the vectors xj are linearly independent.
hint/solution 12.2 Note that Span(xj) = {αxj |α ∈ F}. Now suppose that
the xj are linearly independent. Let
u1 + · · ·+ uk = 0, uj ∈ Uj.
Then for each j we can write uj = αjxj where αj ∈ F. So
α1x1 + · · ·+ αkxk = 0
and so by linear independence all the αj are zero and so all the uj are zero.
Conversely, suppose the sum is direct and let
α1x1 + · · ·+ αkxk = 0.
Then setting uj = αjxj we have
u1 + · · ·+ uk = 0, uj ∈ Uj
and so by the sum being direct we have uj = 0 for all j. Hence λj = 0 (given
that the xj are nonzero) for all j and so the xj are linearly independent.
done?problem 12.3 Let V =⊕j=1,...,kUj be a direct sum of subspaces. Suppose
that we are given a basis for each subspace Uj. Show that the totality of all
these basis vectors forms a basis for V.
MAT3-ALG algebra 2006/7 (tnb) - lecture 12 31
hint/solution 12.3 Clearly they span. Now suppose that vj,i, i = 1, . . . , dj is
a basis for the dj-dimensional subspace Uj. Let
k∑j=1
dj∑i=1
αj,ivj,i = 0.
Then since the sum is direct and
uj :=
dj∑i=1
αj,ivj,i ∈ Uj
we deduce that uj = 0 for all j. But vj,i, i = 1, . . . , dj is a basis for Uj and so
linearly independent. So all the αj,i are zero.
done?problem 12.4 Write out the details of the proof of §12.4.4. In particular,
provide more detail on the final sentence.
In the following problems, the idea is to calculate and spot the pattern. Don’t
get too hung-up on proofs!
In the following problems we write Jn(α) for the Jordan matrix which is the
n × n matrix whose i, j-th entry is: α if i = j; 1 if j = i + 1 and 0 otherwise.
Thus for example
J3(−5) =
−5 1 0
0 −5 1
0 0 −5
.done?problem 12.5 + Hand-in for tutorial What is the characteristic
polynomial of J3(α)? Find its eigenvalues, and their algebraic and geometric
multiplicity. How does this generalize to Jn(α)?
hint/solution 12.5 The characteristic polynomial is (λ−α)3. The only eigen-
value is α and the eigenspace is the span of the first basis vector. It has algebraic
multiplicity 3 and geometric multiplicity 1.
The general result is obtained by replacing “3” with “n” in the previous
paragraph.
done?problem 12.6 Consider D : P3 → P3 (where D denotes differentiation and
Pn is the vector space of real polynomials in “x” of degree n or less). Find a
basis with respect to which D has matrix J4(0).
hint/solution 12.6 The easiest example is 1, x, x2/2!, x3/3!.
MAT3-ALG algebra 2006/7 (tnb) - lecture 13 32
done?problem 12.7 For k ≥ 1, the k-th generalized eigenspace of T : V → V with
eigenvalue λ is
Eλ,k := {v ∈ V | (T − λI)kv = 0}.
So, for k = 1 the generalized eigenspace is just the eigenspace in the usual
sense.
1. Show that if k ≤ l then Eλ,k ⊆ Eλ,l.
2. Let A = J3(α). Show that Eα,k = V for k ≥ 3. Describe Eα,2 and give
its dimension.
3. In general, what is dimEα,k for the Jordan matrix Jn(α)?
hint/solution 12.7
1. Clearly if (T − λI)kv = 0 then (T − λI)lv = 0 for all l ≥ k and so
Eλ,k ⊆ Eλ,l.
2. Computing,
(J3(α) − αI)2 =
0 0 1
0 0 0
0 0 0
The generalized eigenspace is thus 2-dimensional and it is the span of the
first 2 standard basis vectors.
3. A bit more experiment shows that generally the k-th generalized
eigenspace is k-dimensional and is the span of the first k standard ba-
sis vectors.
lecture 13 the Cayley-Hamilton theorem and the min-imal polynomial
13.1 setting
T : V → V is a linear map and V is a finite-dimensional vector space over a field
F. Its characteristic polynomial is cT (x) and its minimal polynomial is mT (x).
(We will usually write the characteristic polynomial using x as the variable in
place of the more familiar λ.)
We will write λ1, . . . , λk for the distinct eigenvalues of T or A but µ1, . . . , µlfor the eigenvalues listed with multiplicity.
MAT3-ALG algebra 2006/7 (tnb) - lecture 13 33
problems
In the problems for this lecture the idea is to calculate and spot the pattern.
Don’t get hung-up on proofs!
In the following problems we write Jn(α) for the Jordan matrix which is the
n × n matrix whose i, j-th entry is: α if i = j; 1 if j = i + 1 and 0 otherwise.
Thus for example
J3(−5) =
−5 1 0
0 −5 1
0 0 −5
.done?problem 13.1 What is the minimal polynomial of J3(α)? How does this
generalize to Jn(α)?
hint/solution 13.1 The minimal polynomial is (λ− α)3 (by experiment - no
lower power of (λ− α) annihilates J3(α)).
The general result is obtained by replacing “3” with “n”.
MAT3-ALG algebra 2006/7 (tnb) - lecture 13 34
done?problem 13.2 + Hand-in for tutorial Consider ODEs of the form
x ′ = Ax where x(t) =
(u(t)
v(t)
)and A is a constant 2 × 2 real matrix. Re-
call from CVD that one can solve such systems when A is real-diagonalizable
(“nodes”,“saddles” and “stars”) and complex-diagonalizable (“foci” and “cen-
tres”).
1. Let A be a real 2× 2 matrix that is not diagonalizable by real or complex
P. Show that the minimal polynomial of A is equal to the characteristic
polynomial which is of the form (x− λ)2, where λ is the eigenvalue.
2. Deduce (Cayley-Hamilton) that if f2 is not an eigenvector then f1 :=
(A− λI)f2 is.
3. Show that the matrix of x 7→ Ax in a basis f1, f2 (as in the previous part)
is J2(λ) and hence that A = PJ2(λ)P−1 where P has f1, f2 as columns.
4. Show that
exp(tJ2(λ)) =
(etλ tetλ
0 etλ
).
5. Show that exp(tA) = P exp(tJ2(λ))P−1 where P is the matrix with
columns f1, f2.
6. Solve
u ′ = −3u− v, v ′ = 4u+ v.
(Recall from CVD that the general solution of x ′ = Ax is x(t) =
exp(tA)
(C1
C2
).)
It is purely optional for this course, but you might like to figure out what the
phase portraits for these systems look like.
Remark: this approach to the solution obscures the relationship of the so-
lutions to the basis. Alternatively we can observe that
eλtf1, eλt(tf1 + f2)
are two independent solutions.
MAT3-ALG algebra 2006/7 (tnb) - lecture 13 35
hint/solution 13.2
1. For A not to be diagonalizable it must not have two distinct eigenvalues.
So its characteristic polynomial is (x− λ)2. Its minimal polynomial can’t
be (x − λ) otherwise A = λI which is diagonalizable. Hence its minimal
polynomial is (x− λ)2.
2. Let v2 be such that v1 := (A − λI)v2 6= 0. Then (A − λ)v1 = 0 by
Cayley-Hamilton and so v1 is an eigenvector.
3. Follows from Tf1 = λf1 and Tf2 = f1 + λf2 and the change of basis
formula.
4. Multiplying a couple of times one finds that
(tJ2(λ))k =
(tkλk ktkλk−1
0 tkλk
)and summing, the result follows. (Alternatively, use the clever trick of
writing J2(λ) as λI plus a matrix with an entry only in the top corner and
use the fact that for commuting matrices it is true that exp(A + B) =
exp(A) exp(B).
5. We know A = PJ2(λ)P−1 and so Ak = P(J2(λ))
kP−1 and sum.
6. The matrix of the system is (−3 −1
4 1
).
The characteristic equation has the single root λ = −1. Set f2 =
(0
1
)which is not an eigenvector. Then f1 = (A− λI)f2 =
(−1
2
).
So
exp(tA) = P exp(tJ2(−1))P−1
=
(−1 0
2 1
)(etλ tetλ
0 etλ
)(−1 0
2 1
)= e−t
(1− 2t −t
4t 2t+ 1
)So the general solution is
x(t) = exp(tA)
(C1
C2
)= e−t
(C1(1− 2t) − tC24tC1 + (2t+ 1)C2
)
MAT3-ALG algebra 2006/7 (tnb) - lecture 13 36
done?problem 13.3 A matrix is said to be in Jordan form if it is block-diagonal
with each diagonal block being a Jordan matrix. (There may be more than one
block with a given parameter value α.) So for example the 7× 7 matrixJ3(5) 0 0
0 J1(5) 0
0 0 J3(−2)
is in Jordan form. (Note that J1(α) is the 1 × 1 matrix (a.k.a. “number”) α
and so a diagonal matrix is an example of Jordan form where all the blocks are
of size 1.)
1. Consider a matrix A in Jordan form with just two Jordan blocks
Jp(α), Jq(β) where α 6= β. Find the characteristic polynomial, eigenval-
ues, minimal polynomial, and dimensions of the generalized eigenspaces
of A (definition is in a problem for lecture 12).
2. The same, only now assume that α = β.
3. Conjecture how this generalizes to a general matrix in Jordan form.
hint/solution 13.3
1. The characteristic and minimal polynomials are the products of those for
the blocks separately. The eigenspaces are of the same dimension as for
the blocks separately.
2. The characteristic polynomial is still the product. The minimal polynomial
is (λ−α)max(p,q). The eigenspace is 2-dimensional (corresponding to the
top entry in each block). The generalized eigenspaces are just the direct
sum of those for the blocks separately.
3. For each eigenvalue λ the characteristic polynomial has a factor (x− λ)p
where p is the sum of the size of all blocks with that λ. The minimal
polynomial has a factor (x−λ)q where q is the maximum size of all blocks
with that λ. The generalized eigenspaces just add, as before.
done?problem 13.4 Find two matrices A,B in Jordan form which have the same
minimal polynomial, characteristic polynomial and dimensions of the (ordinary,
not generalized) eigenspaces and such that A 6= B (and neither do A,B differ
only by a change in the order of the blocks down the diagonal).
MAT3-ALG algebra 2006/7 (tnb) - lecture 14 37
hint/solution 13.4 I think the smallest example is the two 7 × 7 matrices,
each made up of three Jordan blocks with the same value of α. In one the sizes
are 3, 3, 1 and in the other 3, 2, 2.
13.1.1 something worth knowing In fact, every complex square matrix is sim-
ilar to one in Jordan form. Some thought shows that the Jordan form is de-
termined (up to choosing the order of the blocks) by the dimensions of all
the generalized eigenspaces and so this provides a solution to the classification
problem for complex square matrices: two complex matrices are similar iff they
have the same eigenvalues and the same dimension for all the corresponding
generalized eigenspaces.
Knowing the characteristic and minimal polynomials is not enough, as ex-
ercise 4 demonstrates.
lecture 14 a diagonalizability theorem
14.1 setting
T : V → V is a linear map and V is a finite-dimensional vector space over a field
F. Its characteristic polynomial is cT (x) and its minimal polynomial is mT (x).
(We usually write the characteristic polynomial using x as the variable in place
of the more familiar λ.)
We will write λ1, . . . , λk for the distinct eigenvalues of T or A but µ1, . . . , µlfor the eigenvalues listed with multiplicity.
problems
done?problem 14.1 Use the process described in §14.3.1 to find quadratic polyno-
mials p1, p2, p3 such that for every quadratic polynomial q we have
q(x) = q(0)p1(x) + q(1)p2(x) + q(2)p3(x).
hint/solution 14.1 Following the formulae in §14.3.1 the polynomials are
p1(x) =1
2(x− 1)(x− 2), p2(x) = −x(x− 2), p3(x) =
1
2x(x− 1).
MAT3-ALG algebra 2006/7 (tnb) - lecture 14 38
done?problem 14.2 For what values of k is the matrix
M :=
1 1 2
0 −2 k
0 0 1
diagonalizable?
hint/solution 14.2 Clearly, cT (x) = −(x − 1)2(x + 2) and so by the main
theorem of this lecture, M is diagonalizable if and only ifmT (x) = (x−1)(x+2).
Computing, (M− I)(M+ 2I) has all entries zero except for one which is k+ 6.
So M is diagonalizable iff k = −6.
done?problem 14.3 A is a 2× 2 matrix and λ is an eigenvalue of A. Also
A− λI =
(2 −3
−4 6
).
Find a basis for R2 consisting of eigenvectors of A.
hint/solution 14.3 Solve (A − λI)v = 0 to get an eigenvector v1 and using
the previous part we know that the columns of (A−λI) are eigenvectors for the
other eigenvalue. Thus
v1 =
(3
2
), v2 =
(2
−4
)is a basis of eigenvectors.
done?problem 14.4 Let
Rθ =
(cos θ − sin θ
sin θ cos θ
), θ ∈ [0, 2π].
If you calculate you will find that the different Rθ have different (complex)
eigenvalues but the same (complex) eigenvectors. How does this relate to the
theory in this lecture?
hint/solution 14.4 Since the Rθ are rotation matrices we have RθRφ = RφRθ.
Thus, (thought of as complex matrices) if they are diagonalizable (and they all
in fact are) then they are simultaneously diagonalizable - and for one dimensional
eigenspaces that just means they have the same eigenvectors.
MAT3-ALG algebra 2006/7 (tnb) - lecture 14 39
done?problem 14.5 + Hand-in for tutorial Use the main theorem to check
that
A =
−2 −4 2
3 6 −1
−6 −4 6
is diagonalizable. Find a basis v1, v2, v3 with respect to which the matrix of
u 7→ Au is diagonal and hence write down a matrix P such that P−1AP is
diagonal. You are given that cA(x) = (x − 2)(x − 4)2. (You can (and should)
do all this without explicitly solving (A− λI)v = 0 for eigenvectors.)
hint/solution 14.5 Just check that (A− 2I)(A− 4I) = 0, which indeed it is.
Now,
A− 4I =
−6 −4 2
3 2 −1
−6 −4 2
, A− 2I =
−4 −4 2
3 4 −1
−6 −4 4
and so we can take one column from the first matrix to span the λ = 2
eigenspace (the right-most, say) and two from the second (one quarter of the
second and the last, say) to span the λ = 4 eigenspace. So set
P =
2 −1 2
−1 1 −1
2 −1 4
done?problem 14.6 Continuing the previous exercise, let
B =
−3 −4 3
5 6 −3
−5 −4 5
Check that AB = BA. Find a basis that diagonalizes both A and B. A
suggested strategy is as follows. Use your change of basis matrix P from the
previous exercise (that diagonalizes A) to find the matrix of x 7→ Bx in the
basis v1, v2, v3 (Maple, perhaps). Now you should discover that B is block-
diagonal and you just need to change basis again within the 2-dimensional
eigenspace. (There are other ways: you could think about the intersection of
the 2-dimensional eigenspaces of the two matrices, for example.)
MAT3-ALG algebra 2006/7 (tnb) - lecture 15 40
hint/solution 14.6 Using Maple to change basis for A and B (the former just
as a check) we get
P−1AP =
2 0 0
0 4 0
0 0 4
, P−1BP =
2 0 0
0 4 −6
0 0 2
.So we need to diagonalise the submatrix
S :=
(4 −6
0 2
).
This has eigenvalues 4, 2 and assembling eigenvectors as usual we discover that
Q =
(1 3
0 1
)diagonalizes S. Combining the two changes of basis we get that
R :=
2 −1 2
−1 1 −1
2 −1 4
1 0 0
0 1 3
0 0 1
=
2 −1 −1
−1 1 2
2 −1 1
diagonalizes both. It’s columns are the required basis.
There are other ways of approaching this. One way is to observe that the
basis has to consist of a vector in A’s 1-dimensional eigenspace, a vector in B’s
1-dimensional eigenspace and a vector in the intersection of the two matrices
2-dimensional eigenspaces.
lecture 15 bilinear and quadratic forms on R-vectorspaces
15.1 setting
Finite-dimensional vector spaces over R. If b is a symmetric bilinear form (SBF)
we write β for the associated quadratic form and B for the matrix of b with
respect to some basis.
problems
done?problem 15.1 Prove that an SBF is linear in the second entry (§15.3.3).
done?problem 15.2 Check the claim in §15.4.1 that if B is a symmetric n×n matrix
then b(x, y) = xTBy defines an SBF on Rn, were x, y are column vectors as
usual. (Hint: for the symmetric part, take the transpose of the 1 × 1 matrix
xTBY.)
MAT3-ALG algebra 2006/7 (tnb) - lecture 15 41
hint/solution 15.2 Firstly,
b(λx+ µy, z) = (λx+ µy)TBz = λxtBz+ µyTBz = λb(x, z) + µb(y, z).
Secondly
b(y, x) = yTBx = (yTBx)T = xTBT (yT )T = xTBy = b(x, y).
Note the trick at the start where we take the transpose of a 1× 1 matrix (a.k.a
number).
done?problem 15.3 Consider the SBF b on R2 with matrix
B =
(2 3
3 1
).
Find a vector v 6= 0 in R2 such that b(v, v) = 0. Give a sketch showing the
regions in the plane where b(x, x) is positive, negative and zero.
hint/solution 15.3 b(v, v) = 2v21 + 6v1v2 + v22. Set t = v2/v1 so that
b(v, v) = 0 iff t2 + 6t + 2 = 0. This has solns t = −3 ±√7. So e.g.
(1 − 3+√7)t is such a vector.
The sketch should show the two lines v2 = (−3 ±√7)v1 with the sign of
b(v, v) alternating in sign as you go round the four regions in such a way that
it is positive on the coordinate axes. b(v, v) = 0 on the two lines.
done?problem 15.4 Consider the SBF on R3 with matrix
B =
1 0 0
0 1 0
0 0 −1
.Sketch the regions in R3 where b(x, x) is positive, negative and zero.
hint/solution 15.4 The sketch should show a double cone with b(x, x) = 0
on the cone and negative in the two components of its interior.
MAT3-ALG algebra 2006/7 (tnb) - lecture 16 42
done?problem 15.5 + Hand-in for tutorial Let V be the vector space of
2 × 2 matrices with real entries and trace zero. Consider the SBF (known, by
the way, as the “trace form”) b(X, Y) = Trace(XY) on V. Find the matrix of b
with respect to the basis(0 1
0 0
),
(0 0
1 0
),
(1 0
0 −1
)of V.
hint/solution 15.5 The associated quadratic form is q(X) = Trace(X2). La-
beling the basis vectors f1, f2, f3 in the order given and calculating, we get
b(f1, f1) = b(f2, f2) = b(f1, f3) = b(f2, f3) = 0, b(f1, f2) = 1, b(f3, f3) = 2
and so the matrix is 0 1 0
1 0 0
0 0 2
.lecture 16 diagonalizing SBF’s
16.1 setting
Finite-dimensional vector spaces over R. If b is a symmetric bilinear form (SBF)
we write β for the associated quadratic form and B for the matrix of b with
respect to some basis.
problems
done?problem 16.1 Show that Nb := {x ∈ V |b(x, v) = 0 for all v ∈ V is a
subspace of V. (i.e. give the proof of §16.3.2.)
hint/solution 16.1 Standard subspace exercise.
done?problem 16.2 Suppose that the matrix of b in a basis is the standard form
as in §16.4.2. Identify a p-dimensional subspace on which b is positive definite.
Identify also an (n − p)-dimensional subspace on which b is “negative semi-
definite” (meaning that b(v, v) ≤ 0 for all v in the subspace). By considering
intersections, deduce that there is no subspace of dimension larger than p on
which b is positive definite.
MAT3-ALG algebra 2006/7 (tnb) - lecture 16 43
hint/solution 16.2 It is positive definite on U := span of the first p basis
vectors. It is negative semi-definite on the span W of the remaining n−p basis
vectors. Suppose there is a subspace Y of dimension > p on which b is positive
definite. Then by the dimension theorem W ∩ Y has dimension > 0 which is a
contradiction.
done?problem 16.3 + Hand-in for tutorial
1. Find a 2× 2 orthogonal matrix P such that PtSP is diagonal where
S =
(7 −6
−6 −2
).
2. Find also a matrix P such that PtSP is diagonal with diagonal entries ±1or 0.
3. What is the type of the SBF on R2 given by the matrix S? What is its
rank and what is its signature?
hint/solution 16.3
1. The matrix S =
(7 −6
−6 −2
)has eigenvalues 10 and -5. Forming the
matrix P with the corresponding unit eigenvectors as columns we get
P =
(2/√5 1/
√5
−1/√5 2/
√5
)(Note that either column could be multiplied by -1. Also the columns
would be swapped if you decided to take the eigenvalues in the reverse
order.) Then PtSP = diag(10,−5) as one can check.
2. Dividing the columns by√
|λ| we get
P =
( √2/5 1/5
−√2/10 2/5
)
3. Rank is 2, signature is zero.
MAT3-ALG algebra 2006/7 (tnb) - lecture 16 44
done?problem 16.4 + Hand-in for tutorial Let P2 denote the real vector
space of polynomials of degree ≤ 2 in a variable x.
1. Show that
b(p(x), q(x)) :=
∫10p ′(x)q ′(x)dx
defines an SBF on P2 (the “dashes” indicate derivatives).
2. Find a nonzero element of Nb and hence deduce that b is degenerate.
3. Find the matrix of b with respect to the basis {x2, x, 1} of P2 and hence
find the rank of b.
hint/solution 16.4
1. Clearly b(p(x), q(x)) = b(q(x), p(x)). Also
b(λp(x) + µq(x), r(x)) =
∫(λp(x) + µq(x)) ′r ′(x)dx
= λ
∫p ′(x)r ′(x)dx+ µ
∫q ′(x)r ′(x)dx
= λb(p(x), r(x)) + µb(q(x), r(x))
2. 1 ∈ Nb and so dimNb > 0 and so rankb < dimP2 = 3.
3. The matrix is 4/3 1 0
1 1 0
0 0 0
.This clearly has two linearly independent columns and so the rank is 2.
done?problem 16.5 Quick questions
1. True or false: An SBF is non-degenerate iff its matrix does not have zero
as an eigenvalue.
2. What are the possible types of an SBF on R3 if there exists a 2-dimensional
subspace on which it is negative-definite?
3. An SBF on Rn has type (p, q). What is the largest possible dimension
for a subspace V such that b(v, v) < 0 for all non-zero v ∈ V?
4. Same as above but now b(v, v) ≤ 0 for all non-zero v ∈ V.
MAT3-ALG algebra 2006/7 (tnb) - lecture 16 45
hint/solution 16.5
1. True
2. It could be any of the types (0, 3), (1, 2), (0, 2).
3. q
4. Slightly harder. Let V have dimension n. Then in standard form, b clearly
satisfies the condition on the (n − p)-dimensional subspace spanned by
the last n − p basis vectors. This is the largest possible dimension since
any subspace of larger dimension would have non-trivial intersection with
the subspace spanned by the first p basis vectors (on which b is positive-
definite).
done?problem 16.6 + Hand-in for tutorial Let V denote the vector space
of n× n real matrices.
1. What is the dimension of V and of the subspace of symmetric matrices
and of the subspace of antisymmetric matrices?
2. Show that b(X, Y) = Trace(XY) defines an SBF on V.
3. Show that
b(X, Y) =
n∑j=1
n∑k=1
XjkYkj.
4. Show that b is positive-definite on the subspace of symmetric matrices
and negative-definite on the subspace of antisymmetric matrices.
5. Find the type, rank and signature of b.
MAT3-ALG algebra 2006/7 (tnb) - lecture 16 46
hint/solution 16.6
1. n2, n(n+ 1)/2, n(n− 1)/2.
2. Firstly,
b(λX+µY, Z) = Trace((λX+µY)Z) = λTrace(XZ)+µTrace(YZ) = λb(X,Z)+µb(Y, Z).
Secondly
b(Y, X) = TraceXY = TraceYX = b(y, x)
because in general for matrices TraceAB = TraceBA. (Alternatively, the
properties follow from the explicit formula of the next part.)
3. The matrix product is given by
(XY)jl =
n∑k=1
XjkYkl
and so taking the trace
TraceXY =
n∑j=1
(XY)jj =
n∑j=1
n∑k=1
XjkYkj.
4. On symmetric matrices
b(X,X) =
n∑j=1
n∑k=1
XjkXkj =
n∑j=1
n∑k=1
XjkXjk
which is the sum of the squares of the entries and hence non-negative and
zero only for the zero matrix. Similarly for the antisymmetric, except that
there is a minus sign.
5. So (since V is the direct sum of the two subspaces) the type is (n(n +
1)/2, n(n− 1)/2), the rank is n2 and the signature n.
MAT3-ALG algebra 2006/7 (tnb) - lecture 16 47
done?problem 16.7 Let b be the SBF on R4 given by the matrix B given in block
form as
B =
(I2 0
0 −I2
).
Let A be a fixed 2× 2 matrix. Define
U :=
{x ∈ R4
∣∣∣∣ x =
(Av
v
), v ∈ R2
}.
(We are using “block form” notation above.) Show that U is a subspace of R4
and state its dimension.
Show that b is identically zero on U if and only if A is an orthogonal matrix
(i.e. iff AtA = I).
hint/solution 16.7 Let x, y ∈ U so that x =
(Au
u
)and y =
(Av
v
)for
some u, v ∈ R2. Then
λx+ µy =
(A(λu+ µv)
λu+ µv
)∈ U
and so U is a subspace. It has dimension 2 since a basis is obtained by taking
v to be each standard basis vector inR2.Now consider b on U. We have (taking x, y as in the proof above)
b(x, y) = xtBy =(utAt ut
)(I2 0
0 −I2
)(Av v
)= utAtAv−Utv = ut(AtA−I2)v.
This is identically zero iff AtA = I2, that is, iff A is orthogonal.
done?problem 16.8 Suppose b is a non-degenerate SBF on V. Can there exist
a subspace U of V such that b restricted to U is degenerate? What if b is
assumed to be positive-definite?
hint/solution 16.8 Yes — the SBF on R2 with matrix
(0 1
1 0
)is non-
degenerate, but it is degenerate (in fact, zero) when restricted to the span
of the first basis vector.
If however b is positive-definite then it is positive-definite on every subspace
and hence non-degenerate on every subspace.
done?problem 16.9 True or False: If an SBF is positive definite on subspaces
U,U ′ ⊆ V then it is positive definite on their sum U+U ′. Explain your answer.
MAT3-ALG algebra 2006/7 (tnb) - lecture 17 48
hint/solution 16.9 It is false. It is not hard to find a counterexample.
lecture 17 determining type — applications
17.1 setting
Finite-dimensional vector spaces over R. If b is a symmetric bilinear form (SBF)
we write β for the associated quadratic form and B for the matrix of b with
respect to some basis.
problems
done?problem 17.1 Consider the quadratic form
β = 2x2 + 3y2 + z2 − 4xy+ 2xz+ 2yz.
1. Write down the matrix B of this quadratic form.
2. By evaluating determinants only determine the type of this form.
3. Find the eigenvalues and eigenvectors of B and check that the signs of
the eigenvalues are consistent with derivation of the type in the previ-
ous part. (You might want to use Maple to find the eigenvalues — use
“evalf(LinearAlgebra[Eigenvalues](B))”.)
hint/solution 17.1
1.
B =
2 −2 1
−2 3 1
1 1 1
2. Here detB1 = 2 (top-left 1 × 1) and detB2 = 2 (top-left 2 × 2). Here
detB = −7 (row-expansion or row operations). So β is positive-definite
on a 2-dim subspace (the (x, y)-plane) but has rank three (since detB 6=0). Hence it has type (2, 1).
3. “evalf( Eigenvalues(A) )” in Maple gives eigenvalues 4.6,−0.71, 2.14
(with a tiny imaginary part that is numerical error). Two positive and
one negative, as expected.
MAT3-ALG algebra 2006/7 (tnb) - lecture 18 49
done?problem 17.2 + Hand-in for tutorial Show that the origin is a critical
point of
f(x, y, z) = 2x2 + y siny+ z2 + 2(y+ z) sin x− 2ky sin z
(where k is a constant). What can you say about the nature of the critical point
for different values of k?
hint/solution 17.2 The derivatives are
fx = 4x+2(y+z) cos x, fy = y cosy+siny+2 sin x, fz = 2z+2 sin x−2ky cos z.
These all vanish when x = y = z = 0 and so the origin is a critical point.
Taking the 2nd derivatives at zero the Hessian matrix is4 2 2
2 2 −2k
2 −2k 2
.The determinants of the top-left square submatrices are 4, 4,−16k(1+ k) (the
last after some calculation). This last number is positive for k ∈ (−1, 0) and
negative outside of [−1, 0]. Thus the critical point is a local min for k ∈ (−1, 0)
and a ”generalized saddle” for k < −1 and k > 0. One would need to look at
higher derivatives to determine what happens at k = 0 and k = −1.
done?problem 17.3 What is the type of the SBF with matrix−3 12 −7
12 4 2
−7 2 2
You do not need to evaluate a 3×3 determinant (or compute eigenvalues (don’t
even think of it) ).
hint/solution 17.3 This is positive-definite on the span of the last two basis
vectors and so has type (2, 0), (3, 0) or (2, 1). But it is negative definite on the
span of the first basis vector, which rules out the first two possibilities.
done?problem 17.4 What is the type of the SBF in problem 5 of lecture 15?
lecture 18 SBF’s on inner-product spaces
MAT3-ALG algebra 2006/7 (tnb) - lecture 18 50
18.1 setting
Finite-dimensional vector spaces over R. If b is a symmetric bilinear form (SBF)
we write B for the matrix of b with respect to some basis.
problems
done?problem 18.1 Classify the following quadrics.
1. x2 + 2y2 + 3z2 + 2xy+ 2xz = 1
2. 2xy+ 2xz+ 2yz = 1
3. x2 + 3y2 + 2xz+ 2yz− 6z2 = 1
You should be able to do all of these using “determinants” analysis if you think
carefully. You can always check your answer by asking Maple for the eigenvalues.
hint/solution 18.1
1. The matrix is
S =
1 1 1
1 2 0
1 0 3
.Calculating, detS = 1 and the top-left 1 × 1 and 2 × 2 submatrices are
also positive so S has type (3, 0) and this is an ellipsoid.
2. The matrix is
S =
0 1 1
1 0 1
1 1 0
.Here detS = 1 and so S has type (3, 0) or (1, 2). But it cannot be the
former since the first standard basis vector satisfies et1Se1 = 0. Thus this
is a hyperboloid of two sheets.
3. The matrix is
S =
1 0 1
0 3 1
1 1 −6
.Here, detS < 0 and so S has type (2, 1) and this is a hyperboloid of one
sheet.
MAT3-ALG algebra 2006/7 (tnb) - lecture 19 51
done?problem 18.2 Let β(x) be a quadratic form on Rn given by a symmetric
matrix S. How are the maximum and minimum values of β(x) on the unit
sphere xtx = 1 related to the eigenvalues of S. (Hint: orthogonal change of
coordinates to standard form.)
hint/solution 18.2 Let λ1, . . . , λn be the eigenvalues with multiplicity in non-
increasing order. Then there exists an orthogonal change of coordinates y =
P−1x such that
β = λ1y21 + · · ·+ λny2n.
Since we have changed coordinates orthogonally, the sphere will still be given
as
y21 + · · ·+ y2n = 1.
I claim that the max value of β on the sphere is λ1 and it occurs at (1, 0, . . . , 0).
Clearly that value is obtained there and clearly also for all x on the sphere
λ1y21 + λ2y
22 + · · ·+ λny2n ≤ λ1y21 + λ1y
22 + · · ·+ λ1y2n = λ1.
Similarly, the min value is λn.
done?problem 18.3 u Challenge Continuing the previous exercise, use La-
grange multipliers to find the critical points of xtSx subject to the constraint
xtx = 1.
lecture 19 simultaneous diagonalization
19.1 setting
Finite-dimensional vector spaces over R. If b is a symmetric bilinear form (SBF)
we write B for the matrix of b with respect to some basis.
problems
done?problem 19.1 + Hand-in for tutorial Consider the SBF’s on R2
given (with respect to the standard basis) by the matrices
B =
(1 3
3 3
), A =
(2 1
1 1
).
Show that one of these is positive definite and hence find a basis for R2 with
respect to which the matrices of both the SBF’s are diagonal (with the positive-
definite one having the identity as its matrix). Write down the change of basis
matrix that diagonalises both forms.
MAT3-ALG algebra 2006/7 (tnb) - lecture 19 52
hint/solution 19.1 The 2nd matrix is positive-definite since its top-left entry
is positive as is its determinant. The relative eigenvalues are the solutions of
det(B− λA) =
∣∣∣∣1− 2λ 3− λ
3− λ 3− λ
∣∣∣∣ = (3− λ)(−2− λ) = 0.
Thus the relative eigenvalues are 3 and −2.
Solving (B− λA)v = 0 in each case we get relative eigenvectors(0
1
),
(1
−1
)respectively. Finally we must scale these to be of unit length w.r.t. the positive-
definite form. Checking, in fact these already satisfy this condition. Thus they
form the required basis and the change of basis matrix is the matrix with these
as columns.
done?problem 19.2
1. Let
A =
(1 3
3 2
).
Check explicitly that 〈Ax, y〉 = 〈x,Ay〉 for all x, y ∈ R2 where the inner
product is the standard one on R2.
2. Let
A =
(1 3
−1 2
).
Find vectors x, y ∈ R2 such that 〈Ax, y〉 6= 〈x,Ay〉 where the inner
product is the standard one on R2.
hint/solution 19.2 Just calculate along the lines for the first of observing
that
Ax =
(x1 + 3x23x1 + 2x2
)and so
〈Ax, y〉 = (x1 + 3x2)y1 + (3x1 + 2x2)y2
etc.
MAT3-ALG algebra 2006/7 (tnb) - lecture 19 53
done?problem 19.3 + Hand-in for tutorial Let n ∈ N and consider
Tn =
{a0 +
n∑k=1
ak cos kx+ bk sinkx |ak, bk ∈ R
}
with the inner product
〈p(x), q(x)〉 :=
∫2π0p(x)q(x)dx.
Consider the linear map −D2 : Tn → Tn where −D2 : p(x) 7→ −p ′′(x).
1. Show that −D2 is self-adjoint. (Hint: integration by parts.)
2. What are the eigenvalues of −D2 : Tn → Tn and what is the multiplicity
of each eigenvalue? (Think ODE’s — no clever theory required.)
3. Show that the SBF associated to −D2 is b(p, q) =∫2π0 p ′(x)q ′(x)dx.
4. What is the type of the SBF just found? Relate that to the eigenvalues
of −D2.
hint/solution 19.3 Note that all the integrands appearing below are periodic
and so when we integrate by parts the boundary term will always be zero. All
integrals are from 0 to 2π.
1. Integrating by parts twice
〈p,−D2q〉 = −
∫p(x)q ′′(x)dx = −
∫p ′′(x)q(x)dx = 〈−D2p, q〉.
2. We are looking for solutions to −p ′′(x) = λp(x). So the eigenvalues are
λ = 0 with a 1-dim eigenspace consisting of the constants and λ = k2
where k > 0 with multiplicity 2 — the 2-dim eigenspace being spanned
by cos(kx) and sin(kx).
3. Integrating by parts
〈p,−D2q〉 = −
∫p(x)q ′′(x)dx =
∫p ′(x)q ′(x)dx.
4. This SBF is positive-definite on the 2n-dimensional subspace spanned by
cos(kx), sin(kx). The SBF is zero on the 1-dim subspace of constant
functions. Thus it has type (2n, 0). This agrees with −D2 having 2n
positive eigenvalues (counted with multiplicity) and one zero one.