MA322Eakin Fall 2016

Daily Quiz Items and Exam Review ProblemsThis document was last updated November 17, 2016.

1 Quiz Problems By Date

The problems for the 10-minute quizz for a given date will be straightforward adaptations or simplifi-cations of the items in this document that are listed under that date. This document may be updatedat any time but the source problems for the quiz on a particular date will not change within 40 hrs ofclass time on that date.

1.1 August 26

1. Complete the matrix below to an augmented matrix for the following linear system:{x− 4y + 1 = y, 2x+ 3 = 11y + 4}∣∣∣∣∣∣x y

1

∣∣∣∣∣∣ ans

∣∣∣∣∣∣x y RHS1 −5 −12 −11 1

∣∣∣∣∣∣• Modify your augmented matrix by one elementary row operation which results in an equiv-

alent matrix with 0 in the lower left corner . Be sure to indicate the operation by one of Pij,kRj, or Ri + cRj with appropriate values for the i, j, or c for your operation.

ans R2 − 2R1,

∣∣∣∣∣∣x y RHS1 −5 −10 −1 3

∣∣∣∣∣∣2. Write down an augmented matrix for the following linear system:{x− 4y + 1 = y, 2x+ 3 = 11y + 4}

Ans:

∣∣∣∣∣∣x y RHS1 −5 −12 −11 1

∣∣∣∣∣∣1.2 August 29

1. Definition: A Linear Equation of (or in) the variables x1, · · · , xn is an equation which can bewritten in the form a1x1 + · · ·+ anxn = b where b and {a1, · · · , an} are real or complex numbers.

2. Definition: A Linear Equation is homogenous if it can be written in the form a1x1 + · · · +anxn = 0.

3. Definition: A solution to a linear equation a1x1 + · · · + anxn = b is an ordered list of scalars(s1, s2, · · · , sn) such that the equation becomes a true statement when si is substituted for xi foreach i = 1, · · · , n. An ordered list (s1, s2, · · · , sn) is a solution to the linear system S if it is asolution to each linear equation in S.

4. Definition: Two linear systems are equivalent if they have exactly the same solutions.

1

5. Explain why x− 3y = 2 + 5y + z is a linear equation?Ans Because it can be written as x− 8y − z = 2.

6. Why is x+ 3xy = 1 not a linear equation?Ans: Because it cannot be written in the form ax+ by = c.The solutions to the first equation are {(x, 1−x

3x)|x ∈ <, x 6= 0}, Thus the first has one solution of

the form (x, ∗) for every x but one. On the other hand, if b 6= 0 then ax+ by = c has (x, c−axb

) asa solution for every x or, if b = 0 it has ( c

a, y) as a solution for every y. Thus the solution a sets

are different so the two equations cannot be equal.

7. Show that (1, 2) is a solution to the linear equation 2x− y = 0 while (1, 1) is not.Ans

• From the definition, (1, 2) is a solution since 2(1) − 2 = 0 is true; (1, 1) is not a solutionsince 2(1)− 1 = 0 is not true;

8. Show that (1, 1) is a solution to the linear equation 2x+ y = 3 while (1, 2) is not.Ans

• Also checked by direct substitution

9. Show that neither (1, 2) nor (1, 1) is a solution to the linear system S = {2x− y = 0, 2x+ y = 3}but each of them is a solution to all but one of the equations in S.Ans

• (34, 32) is a solution to the system S because it is a solution to every equation in S.

2(3

4)− 3

2= 0, and 2(

3

4) +

3

2= 3

.

• As seen in the previous two problems, each is a solution to one of the two equations in Sbut not the other.

10. This problem was edited on August 27, 2016 to remove an extraneous part (b).

Show that the linear system S = {x+ y+ z = 0, y− z = 0} is not equivalent to the linear systemT = {x+ y + z = 0, x− z = 0}Ans:We must show that one of these has a solution which is not a solution of the other. For instance(x, y, z) = (−2, 1, 1) is a solution to S but not a solution to T since it is not a solution to x−z = 0(since −2− 1 = 0 is not true).

11. Definition: The linear system S is consistent if it has at least one solution.

1.3 August 31

1. Definition: If A is an m× n matrix with rows [R1, R2, · · · , Rm] then the Pivot Column List(PC List) of A is the ordered list [c1, c2, · · · , cm] where ci is the pivot location of row i.

2. Definition: The matrix A is in Row Echelon Form (REF) if the pivot location of any rowis less than that of any row below it. (by convention ∞ < ∞). Equivalently, A is in REF if itsPC List is strictly increasing or strict.

2

3. Definition: If A is a matrix in REF then the rank of A is the number of non-zero rows in A.

4. For the following system of linear equations in x, y, write down the augmented matrix and carryout the necessary elementary row operation to solve the system.Be sure to identify the operations in correct notation introduced in class.

x+ 2 y = 5, 2x+ 5 y = 4, 3x+ 7 y = 9

(a) The augmented matrix is A =

Ans: x y RHS1 2 5

2 5 4

3 7 9

(b) The row reduced form (REF) for A is:

Ans:First R2 − 2R1 and R3 − 3R1 give:

1 2 5

0 1 −6

0 1 −6

Then R3 −R2 gives:

x y RHS1 2 5

0 1 −6

0 0 0

(c) What is the equivalent system of linear equations given by the REF of A?

Ans: x+ 2y = 5 and y = −6.

(d) The solution to the linear system is:Ans:Using backsubstitution:From the second equation we have y = −6.Substituting for y in the first equation: x + 2(−6) = 5 gives x = 17 so the solution isx = 17, y = −6.

(e) The pc list for A is :Ans:1, 1, 1

(f) The pc list for the REF of A is:Ans: 1, 2,∞.

3

(g) The rank of the matrix A is:Ans:2, since the REF has two pivots.

1.4 September 2

1. Definition: <n is the set of all column vectors of length n with entries in <.

2. Definition: The elementary row operations on a matrix are:

(a) exchange two rows

(b) multiply any row by a non-zero scalar

(c) add a multiple of any row to a different row.

The linear system {−x + 3y + 5z = 5, 2x + y + 7z = 7t, 3x + 4y + 12z = −9} has the followingaugmented matrix and associated REF

(A|B) =

x y z RHS−1 3 5 52 1 7 7t3 4 12 −9

R =

x y z RHS−1 3 5 50 −5 −3 7t− 100 0 0 −14− 7t

where t is an unknown num-

ber.

(a) Depending on the value of t, what are the ranks of A and (A|B). Explain.

Ans Regardless of the value of t the matrix A has rank 2 because there are 2 pivot columnsin its REF. If t = −2 then (A|B) has two pivot columns and therefore has rank 2. Otherwise,(A|B) has three pivot columns and therefore has rank 3.

(b) For which values of t is the system consistent?

Ans The rank of the coeficient matrix is equal to the rank of the augmented matrix if andonly if t = −2. Therefore the system is consistent if and only if t = −2.

(c) For values of t for which the system is consistent, which are the pivot variables and whichthe free variables?Ans The pivot variables are x and y; the free variable is z.

4

1.5 September 4 (Labor Day)

1.6 September 7

1. Definition: If (A|B) is an augmented matrix in REF form which has a “title row” of variablesthen the variables above pivot columns are called the pivot variables and the non-pivot variablesare called the free variables.

2. Definition: A set V ⊂ <n is a vector subspace of <n if it is closed under vector addition andscalar multiplication.

3. Definition: Let A be an m×n matrix with columns A1, A2, · · · , An. If X =

x1x2

...xn

∈ <n then

AX = x1A1 + x2A1 + · · ·+ xnAn

4. Definition: (Multiplication of a matrix times a vector) If M is an m× n matrix, written

as a row of its columns ( M = [C1, · · · , Cn] ) and X =

∣∣∣∣∣∣∣x1...xn

∣∣∣∣∣∣∣ a vector of length n, then

MX = x1C1 + · · ·+ xmCn

5. (A|B) =

∣∣∣∣∣∣∣∣x y RHS1 2 8−1 4 102 −3 −5

∣∣∣∣∣∣∣∣ and M =

∣∣∣∣∣∣∣∣=x y RHS1 2 80 6 180 0 0

∣∣∣∣∣∣∣∣ are the augmented matrix and REF of

a linear system S.

1) Express the linear system S as an equivalent vector equation x ~C1 + y ~V2 = ~B.

2) Use M to solve the system S.

3) Use your result from (2) to express your ~B in (1) as a linear combination of your vectors ~C1

and ~C2 in (1).

Ans:

1) x

∣∣∣∣∣∣1−12

∣∣∣∣∣∣+ y

∣∣∣∣∣∣24−3

∣∣∣∣∣∣ =

∣∣∣∣∣∣810−5

∣∣∣∣∣∣.2) {x+ 2y = 8, 6y = 18}, {y = 3, x = 8− 2(3)}, {x = 2, y = 3}

3) 2

∣∣∣∣∣∣1−12

∣∣∣∣∣∣+ 3

∣∣∣∣∣∣24−3

∣∣∣∣∣∣ =

∣∣∣∣∣∣810−5

∣∣∣∣∣∣.

5

1.7 September 9

Problem 1 was edited on Sept. 5 to correct a typo in < A|B >.

1. (A|B) =

x1 x2 x3 RHS1 3 −1 −12 6 −3 0−1 −3 1 1

and R =

x1 x2 x3 RHS1 3 −1 −10 0 −1 20 0 0 0

are the augmented matrix

and its REF of a linear system S

(a) Identify the free variables.Ans x2

(b) Find the general solution, written as a vector equation.Ans{x1 − x3 = −1− 3x2, x3 = −2},{x1 = −3x2 + x3 − 1, x2 = x2, x3 = −2}∣∣∣∣∣∣x1x2x3

∣∣∣∣∣∣ =

∣∣∣∣∣∣−3x2 − 3

x2−2

∣∣∣∣∣∣,∣∣∣∣∣∣x1x2x3

∣∣∣∣∣∣ =

∣∣∣∣∣∣−30−2

∣∣∣∣∣∣+ x2

∣∣∣∣∣∣−310

∣∣∣∣∣∣(c) Find the general solution to the associated homogenous sytem, also written as a vector

equation.

Ans

∣∣∣∣∣∣x1x2x3

∣∣∣∣∣∣ = x2

∣∣∣∣∣∣−310

∣∣∣∣∣∣2. Find the general solutions of the system whose augmented matrix is given below. You must

reduce the system to REF using the standard algorithm. The final answer may be given by backsubstitution or producing RREF.

Be sure to identify the operations in correct notation introduced in class, namely kRi, Ri + cRj

or Pij.

(∗∗)

x1 x2 x3 x4 x5 RHS1 2 −1 0 1 12 4 −2 1 3 2−1 1 1 1 −1 2

The REF and RREF are respectively:

1 2 −1 0 1 1

0 3 0 1 0 3

0 0 0 1 1 0

,

1 0 −1 0 5/3 −1

0 1 0 0 −1/3 1

0 0 0 1 1 0

.(a) Reduce the system to REF using the standard algorithm.

Ans: 1 2 −1 0 1 12 4 −2 1 3 2−1 1 1 1 −1 2

R2 → R2 − 2R1

1 2 −1 0 1 10 0 0 1 1 0−1 1 1 1 −1 2

R3 → R3 +R1

6

1 2 −1 0 1 10 0 0 1 1 00 3 0 1 0 3

R2 ↔ R3

1 2 −1 0 1 10 3 0 1 0 30 0 0 1 1 0

(b) Find the general solution, written as a system of linear equations. Identify the free variables.

Ans:From the given RREF the solution is:

x1 = −1 + x3 −5

3x5, x2 = 1 +

1

3x5, x4 = −x5 with x3, x5 free.

(c) The system (∗∗) can be written as AX = B. What are A and B?Ans:

A =

1 2 −1 0 1

2 4 −2 1 3

−1 1 1 1 −1

, B =

122

.

(d) The solution X =

x1x2x3x4x5

can be written as X = Xp +Xh. What are Xp, Xh?

Ans:

Xp =

−1

1000

, Xh = x3

10100

+ x5

−5

313

0−1

1

.(e) Write down a vector equation equivalent to the given system of equations in the above

questions.

x1

12−1

+ x2

241

+ x3

−1−2

1

+ x4

011

+ x5

13−1

=

122

.1.8 September 12

1. Definition: The set of vectors {v1, v2, · · · , vn} ⊂ <n is linearly dependent if there are scalarsa1, a2, · · · , an which are not all zero such that a1v1 + a2v2 + · · ·+ anvn = O , the zero vector.

2. Definition: The set of vectors {v1, v2, · · · , vn} ⊂ <n is independent if it is not dependent.

An arbitrary (possibly infinite) subset of <n is linearly independent if every finite subset isindependent.

7

3. Consider the vectors V1 =

∣∣∣∣∣∣123

∣∣∣∣∣∣, V2 =

∣∣∣∣∣∣0−26

∣∣∣∣∣∣, V3 =

∣∣∣∣∣∣−2−60

∣∣∣∣∣∣.Show that {V1, V2, V3} is a linearly dependent subset of <3.

Ans: A matrix whose columns are the given vectors isA =

1 0 −22 −2 −63 6 0

. U =

1 0 −20 −2 −20 0 0

is an REF and has two pivots. Therefore there is a free variable which guarantees that the ho-mogenous equation AX = O has a non-trivial solution.

4. Consider the vectors V1 =

∣∣∣∣∣∣123

∣∣∣∣∣∣, V2 =

∣∣∣∣∣∣0−26

∣∣∣∣∣∣, V3 =

∣∣∣∣∣∣−2−60

∣∣∣∣∣∣.Show that {V1, V2, V3} is a linearly dependent subset of <3 by finding explicit scalars a, b, c, notall zero, such that aV1 + bV2 + cV3 = O.Ans

Let A =

1 0 −22 −2 −63 6 0

and calculate the solution to the homogenous system < A|0 >,

{x = 2z, y = −z, z = z} xyz

= z

2−11

.

Choose any non-zero value of z, say z = 1 to get a non-trivial solution

2−11

which says that

2V1 − V2 + V3 = O.

1.9 September 14

1. Definition: A function L : <n → <m is a linear transformation if it satisfies

L(α v + β w) = αLv + β Lw

for every v, w ∈ <n and every α, β ∈ <.

2. Definition: If L : <n → <m is a linear transformation then the kernel of L is {X ∈ <n | L(X) =O}.

3. Definition: If L : <n → <m is a linear transformation from <n to <m then the image of L is{L(X) | X ∈ <n}.

4. Let e1 =

100

, e2 =

010

, and e3 =

001

. Suppose T is a linear transformation from <3 to

8

<3 such that

T (e1 + e2) =

305

, T (e1 + 3e2) =

−121

, T (e1 + e2 + e3) =

000

Calculate T (e1), T (e2), T (e3)

ans: T (e1 + 3e2)− T (e1 + e2) = T ((e1 + 3e2)− (e1 + e2)) = T (2e2) = 2T (e2) −121

− 3

05

=

−42−4

= 2T (e2)

T (e2) =

−21−2

T (e1 + e2) = T (e1) + T (e2) 3

05

= T (e1) +

−21−2

T (e1) =

305

− −2

1−2

=

5−17

T (e1 + e2 + e3) = T (e1 + e2) + T (e3) 0

00

=

305

+ T (e3)

T (e3) = −

305

=

−30−5

5. Let T : <2 → <3 be defined by

T (

∣∣∣∣ xy∣∣∣∣) =

∣∣∣∣∣∣2x− 3y5x+ y−x

∣∣∣∣∣∣a) If V1 =

∣∣∣∣ x1y1∣∣∣∣, V2 =

∣∣∣∣ x2y2∣∣∣∣, show that T (V1 + V2) = T (V1) + T (V2)

b) If V =

∣∣∣∣ xy∣∣∣∣, α ∈ <, show that T (αV ) = αT (V )

c) Is T a linear transformation? Why or why not?

d) Calculate the matrix AT such that T (X) = ATX for every X ∈ <2.

Ans: (a) T (V1) + T (V2) = T (

∣∣∣∣ x1y1∣∣∣∣) + T (

∣∣∣∣ x2y2∣∣∣∣) =∣∣∣∣∣∣

2x1 − 3y15x1 + y1−x1

∣∣∣∣∣∣+

∣∣∣∣∣∣2x2 − 3y25x2 + y2−x2

∣∣∣∣∣∣ =

∣∣∣∣∣∣2(x1 + x2)− 3(y1 + y2)

5(x1 + x− 2) + (y1 + y2)−(x1 + x2)

∣∣∣∣∣∣ = T (

∣∣∣∣ x1 + x2y1 + y2

∣∣∣∣) = T (V1 + V2))

9

(b) T (αV ) = T (α

∣∣∣∣ xy∣∣∣∣) = T (

∣∣∣∣ αxαy∣∣∣∣) =

∣∣∣∣∣∣2αx− 3αy5αx+ αy−αx

∣∣∣∣∣∣ = α

∣∣∣∣∣∣2x− 3y5x+ y−x

∣∣∣∣∣∣ = αT (V )

(c) Yes, by (a) and (b)

(d) If e1 =

∣∣∣∣ 10

∣∣∣∣ , e2 =

∣∣∣∣ 01

∣∣∣∣ then MT = [T (e1), T (e2) = [

∣∣∣∣∣∣2 ∗ 1− 3 ∗ 0

5 ∗ 1 + 0−1

∣∣∣∣∣∣ ,∣∣∣∣∣∣

2 ∗ 0− 3 ∗ 15 ∗ 0 + 1−0

∣∣∣∣∣∣] =∣∣∣∣∣∣2 −35 1−1 0

∣∣∣∣∣∣1.10 September 16

1. Definition: If A is an m×n matrix then Null(A), the null space of A is {X ∈ <m | AX = O}

2. Definition: If L : <n → <m is a linear transformation from <nto<m then the kernel of L is{X ∈ <n | L(X) = O}.

3. Definition: A transformation L : <n → <m is one to one or injective if L(u1) = L(u2) impliesthat u1 = u2.

4. Definition: A transformation L : <n → <m is onto or surjective if <n = image(L).

5. Definition: A transformation L is an isomorphism if it is both injective and surjective.

6. Definition: An m× n matrix, A, is a rectangular array of numbers (scalars) with m rows andn columns.

7. Let A =

(1 −1−2 2

)and T : <2 → <2 be defined by T (V ) = AV .

(a) Explain why T is not injective.

(b) Explain why T is not surjective.

ans:

(a) (one way) Check that the rank of A is 1. This means that there is a X 6= O such thatAX = O. Since AO = O this implies that T (X) = T (O) but X 6= O. So T is not injective.(another way). The linear transformation T is injective if and only if rank(AT ) = colnum(AT )where AT is the standard matrix of T . Here AT = A which has rank 1 but A has two columns.So T is not injective.

(b) (one way)The elementary row operation R2 → R2 + R1 converts < A|(xy

)> to

U =<

(1 −10 0

)|(

xx+ y

)>.

If x+y 6= 0 then U is inconsistent which means that AX =

(xy

)has no solution if x+y 6= 0.

(another way) The linear transformation T is surjective if and only if rank(AT ) = rownum(AT )where AT is the standard matrix of T . Here AT = A which has rank 1 but A has two rows.So T is not surjective.

10

1.11 September 19

1. Definition: If A is an m× n matrix then Col(A), the column space of A is the linear span ofthe columns of A. Equivalently, Col(A) = {AX | X ∈ <n}. (Note that Col(A) ⊂ <m)

2. Rownum(A) denotes the number of rows of the matrix A.

3. Suppose that A is a matrix of rank d with with 3 rows and 5 columns. Answer using complete,meaningful sentences. All examples requested need have only 0 and 1 as entries.

(a) What is the largest possible value for d? Give an example of a 3 by 5 matrix A which hasthis largest possible rank.

Ans: The largest possible rank is 3 because rank(A) ≤ min{rownum(A), colnum(A)} =

min(3, 5) = 3. As an example, consider

∣∣∣∣∣∣1 0 0 0 00 1 0 0 00 0 1 0 0

∣∣∣∣∣∣(b) What is the smallest possible value for d? Give an example of a 3 by 5 matrix which has

this least possible rank.

Ans: The smallest possible rank is 0. There is only one example

∣∣∣∣∣∣0 0 0 0 00 0 0 0 00 0 0 0 0

∣∣∣∣∣∣This section was edited Sept 19.”

4. Let A =

1 23 d4 8

and T : <2 → <3 defined by T (X) = AX. If possible, find a choice for d in

each of the following cases. If no such value exists then explain why not.

(a) A value of d for which T is injective (one to one).

Ans: An REF for A is

1 20 d− 60 0

. T is injective if and only if rank(A) = colnum(A) = 2

(i.e. there is a pivot in each column). There is a pivot in each column if and only if d isNOT 6.

(b) A value of d for which T is surjective (onto).T is surjective if and only if rank(A) = rownum(A) = 6. However A has only 2 columns sothere are at most two pivots. So T cannot be surjective, regardless of the value of d.

1.12 September 21

1. For each of the following, provide a matrix that fits the description or explain why none exists.All examples requested need have only 0 and 1 as entries.

a) A 3 by 5 matrix A such that the columns of A span all of <5

11

Ans: It is certainly not possible because the columns of A are in <3 so their linear span isin <3 which contains no elements of <5.

b) A 3 by 5 matrix A such that the columns of A are linearly independent ?

Ans: This is not possible. The rank of A is less than its number of columns so there willbe free variables in the solution to the homogenous linear system (A|O). This implies thatthere will be non-zero vectors in the nullspace of A. If X is any one of these then AX = Ois the expression of O as a non-trivial linear combination of the columns of A.

c) A 3 by 5 matrix A such that the linear system (A|B) is consistent for all B in <3?

Ans: Yes, this is possible. Consider A =

∣∣∣∣∣∣1 0 0 0 00 1 0 0 00 0 1 0 0

∣∣∣∣∣∣. Then if B =

∣∣∣∣∣∣abc

∣∣∣∣∣∣ then

A

∣∣∣∣∣∣∣∣∣∣abcxy

∣∣∣∣∣∣∣∣∣∣=

∣∣∣∣∣∣abc

∣∣∣∣∣∣ = B for any choice of x, y.

2. T is a linear transformation from <3 to <3 such that

T (

∣∣∣∣∣∣111

∣∣∣∣∣∣) =

∣∣∣∣∣∣34−2

∣∣∣∣∣∣ , T (

∣∣∣∣∣∣110

∣∣∣∣∣∣) =

∣∣∣∣∣∣2−30

∣∣∣∣∣∣ , T (

∣∣∣∣∣∣011

∣∣∣∣∣∣) =

∣∣∣∣∣∣001

∣∣∣∣∣∣. Calculate the standard matrix for T .

Ans: The standard matrix is [T (e1), T (e2), T (e3)] where e1, e2, e3 are the columns of the 3 × 3identity matrix. We note that∣∣∣∣∣∣

111

∣∣∣∣∣∣−∣∣∣∣∣∣

011

∣∣∣∣∣∣ =

∣∣∣∣∣∣100

∣∣∣∣∣∣ so

T (

∣∣∣∣∣∣100

∣∣∣∣∣∣) = T (

∣∣∣∣∣∣111

∣∣∣∣∣∣−∣∣∣∣∣∣

011

∣∣∣∣∣∣) = T (

∣∣∣∣∣∣111

∣∣∣∣∣∣)− T (

∣∣∣∣∣∣011

∣∣∣∣∣∣) =

∣∣∣∣∣∣34−2

∣∣∣∣∣∣−∣∣∣∣∣∣

001

∣∣∣∣∣∣ =

∣∣∣∣∣∣34−3

∣∣∣∣∣∣∣∣∣∣∣∣111

∣∣∣∣∣∣−∣∣∣∣∣∣

110

∣∣∣∣∣∣ =

∣∣∣∣∣∣001

∣∣∣∣∣∣ so

T (

∣∣∣∣∣∣001

∣∣∣∣∣∣) = T (

∣∣∣∣∣∣111

∣∣∣∣∣∣)− T (

∣∣∣∣∣∣110

∣∣∣∣∣∣ =

∣∣∣∣∣∣34−2

∣∣∣∣∣∣−∣∣∣∣∣∣

2−30

∣∣∣∣∣∣ =

∣∣∣∣∣∣17−2

∣∣∣∣∣∣∣∣∣∣∣∣111

∣∣∣∣∣∣−∣∣∣∣∣∣

100

∣∣∣∣∣∣−∣∣∣∣∣∣

001

∣∣∣∣∣∣ =

∣∣∣∣∣∣010

∣∣∣∣∣∣ so

T (

∣∣∣∣∣∣010

∣∣∣∣∣∣) = T (

∣∣∣∣∣∣111

∣∣∣∣∣∣)− T (

∣∣∣∣∣∣100

∣∣∣∣∣∣)− T (

∣∣∣∣∣∣001

∣∣∣∣∣∣)12

T (

∣∣∣∣∣∣010

∣∣∣∣∣∣) =

∣∣∣∣∣∣34−2

∣∣∣∣∣∣−∣∣∣∣∣∣

34−3

∣∣∣∣∣∣−∣∣∣∣∣∣

17−2

∣∣∣∣∣∣ =

∣∣∣∣∣∣−173

∣∣∣∣∣∣so T = TA where A =

∣∣∣∣∣∣3 −1 14 −7 7−3 3 −2

∣∣∣∣∣∣3. The matrix A =

∣∣∣∣∣∣1 2 3 42 4 6 83 6 7 12

∣∣∣∣∣∣ is the standard matrix of the linear transformation T : <n → <m.

a) In this case m = and n =

b) Explain why the transformation T is or is not injective (one-to-one).

c) Explain why the transformation T is or is not surjective (onto).

Ans: (a) m = 3, n = 4,

(b) The elementary row operations R2 − 2R1, R3 − 3R1 reduce A to R =

∣∣∣∣∣∣1 2 3 40 0 0 00 0 −2 0 · · ·

∣∣∣∣∣∣ so

A has rank 2. Therefore the rank of A is less than the number of columns so T is not injective.(c) The rank of A is less than the number of rows so the transformation is not surjective.

1.13 September 23

1. You are given an augmented matrix M and a matrix W which is an REF of M .

M =

x y z w RHS2 1 2 5 3−2 0 −1 0 1

6 2 5 10 5

, W =

x y z w RHS2 1 2 5 30 1 1 5 40 0 0 0 0

The sequence of elementary row operations converting M to W are, in order:

R2 → R2 +R1, R3 → R3 − 3R1, R3 → R3 +R2

(a) Find a vector B such that the linear system (A|B) is inconsistent.

Ans: We need a B such that the REF has a pivot in the rightmost column. If B =

∣∣∣∣∣∣abc

∣∣∣∣∣∣then the given sequence of row operations would take B to

∣∣∣∣∣∣a

b+ ac+ b− 2a

∣∣∣∣∣∣ so we choose a, b, c

so that c+b−2a 6= 0 for instance if a = b = 0, c = 1 then M =

x y z w RHS2 1 2 5 0−2 0 −1 0 0

6 2 5 10 1

,

13

becomes W =

x y z w RHS2 1 2 5 00 1 1 5 00 0 0 0 1

(b) Find a matrix Q such that W = QM . (Keep in mind that, for this purpose, neither the top

row with the variable names nor the vertical bar separating the coefficient matrix from therightmost column are considered to be part of M and W .Ans: by the “Voodoo” principle apply the given sequence to the identity to get∣∣∣∣∣∣

1 0 01 1 0−2 1 1

∣∣∣∣∣∣

2. If A =

2 1 20 1 0−4 0 0

0 6 −17 0 3

and B =

−1 6 2 5 83 1 0 5 1−4 5 4 −1 1

, then the number in the fourth row and

third column of AB is and the number in the third row, second column of BA is. Provide the calculations that produced your answers.

Ans:∣∣ 0 6 −1

∣∣ ∣∣∣∣∣∣204

∣∣∣∣∣∣ = 0 ∗ 2 + 6 ∗ 0 + (−1) ∗ 4 = −4,∣∣ −4 5 4 −1 1

∣∣∣∣∣∣∣∣∣∣∣∣

11060

∣∣∣∣∣∣∣∣∣∣= −4 ∗ 1 + 5 ∗

1 + 0 ∗ 4 + 6 ∗ (−1) + 0 ∗ 1 = · · ·

3. Suppose A =

1 −1 2 12 0 1 33 −1 1 2

and R is the REF of A produced by the standard algorithm.

Find a matrix M such that MA = R.

Ans Use Voodoo to get

∣∣∣∣∣∣1 0 0

0−1 1

∣∣∣∣∣∣1.14 September 26

1. Let A =

1 0 −12 1 30 1 4

and B =

−1 1 −18 −4 5−2 1 −1

.

(a) Show that B = A−1

(b) Solve AX =

110

14

Ans: (a) Multiply AB and show that it is

1 0 00 1 00 0 1

= I3

(b) Multiply X = A−1AX =

−1 1 −18 −4 5−2 1 −1

110

=

∣∣∣∣∣∣04−1

∣∣∣∣∣∣(sum of the first and second

columns)

2. Show that

(a bc d

)−1=

(d

ad−bc − bad−bc

− cad−bc

aad−bc

)provided that ad− bc 6= 0

Ans Multiply to get I.

3. Let A =

(4 39 7

), B =

(5 66 7

)(a) Calculate A−1 and B−1

Ans: 14(7)−3(9)

(7 −3−9 4

)=

(7 −3−9 4

), 1

5(7)−6(6)

(7 −6−6 5

)=

(7 −6−6 5

),

(b) Calculate (AB)−1 without calculating AB.

Ans: (AB)−1 = B−1A−1 =

(−103 45

87 −38

)(c) Solve AX = B

(01

)Ans: X = A−1(B

(01

)) =A−1

(67

)=

(7 −3−9 4

)(67

)=

(21−26

). This is also

(A−1B)

(01

)so it is the second column of A−1B =

(17 21−21 −26

)4. Suppose A and B are matrices such that AB = BA. Explain why A is square.

Ans: Let A be m × n and B be p × q. AB is defined so n = p. AB has m rows and BA has prows but they are equal so m = p. Therefore n = p = m.

5. Suppose A is an invertible matrix. Explain why A is square.

Ans: There is a matrix B such that AB = BA so A is square by the previous problem.

6. Suppose A, B and C are 2× 2 invertible matrices such that

ACA−1 =

(1 20 3

)and ABA−1 =

(1 04 1

).

Calculate ACB−1A−1

Ans: AB−1A−1 = (ABA−1)−1 = (

(1 04 1

))−1(

1 0−4 1

)so ACB−1A−1 = ACA−1AB−1A−1 =

(1 20 3

)(1 0−4 1

)=

(−7 2−12 3

)

15

The answer to this problem was edited Sept 26 to replace “surjective” by “injective”.

7. Suppose A is an invertible 2 × 2. Let L : <2 → <2 be the linear transformation defined byL(X) = AX. Explain why L is an isomorphism.

Ans: Since A is invertible its RREF is the 2× 2 identity matrix. So A has rank 2 so rank(A) =rownum(A) = colnum(A) which says that L is surjective (onto) and injective (one-to-one).Therfore L is an isomorphism.Alternately: If L(X) = L(Y ) then AX = AY so X = A−1AX = A−1AY = Y which showsinjective. If B is any element of <2 then L(X) = B is consistent since L(A−1B) = AA−1B = B.

1.15 September 28

1. You are given that the REF of the 2× 4 matrix K is the matrix K∗.

K =

1 2 1 0 0 0

2 3 0 1 0 0

−1 8 0 0 1 0

−3 5 0 0 0 1

K∗ =

1 2 1 0 0 0

0 −1 −2 1 0 0

0 0 −19 10 1 0

0 0 0 1 −1 1

Let M be the matrix formed by the first two columns of K. Use the given calculation to find a

consistency matrix for M . Use the consistency matrix to show why the vector v =

1198

is in

Col(M).

Ans: G =

∣∣∣∣ −19 10 1 00 1 −1 1

∣∣∣∣, Check that Gv =

(00

)2. Explain why the columns of an invertible matrix must be linearly independent.

Ans: If AX = O then A−1AX = AO, X = 0 so the only solution to AX = O is the trivial one.

3. Given that (AB)t = BtAt show that if A is invertible then At is invertible and (At)−1 = (A−1)t

Ans: AA−1 = I, (AA−1)t = I t = I. (A−1)tAt = I so (A−1)t = (At)−1

1.16 September 30

1. If the determinant of A =

[3 w−1 −2

]is −1 then w = (Show your work)

Ans: 5

2. If A =

1 w 0−1 −2 00 0 1

and the determinant of A is 6 then w = (Show your work)

Ans: 8

Suppose A =

a1,1 a1,2 a1,3a2,1 a2,2 a2,3a3,1 a3,2 a3,3

and that det(A) = 7. If B =

3a1,1 − a1,2 a1,3 a1,23a2,1 − a2,2 a2,3 a2,23a3,1 − a3,2 a3,3 a3,2

,

16

(a) What is the determinant of B?Justify your answer fully, using the properties of the determinant.Ans: −21

(b) What is the determinant of A−1?

Ans: 17

(c) What is the determinant of 5A?

Ans: 53(7)

1.17 October 3

(a) Definition: The linear transformation L : <n → <m is injective (one-to-one) if L(X) =L(Y ) implies X = Y .

(b) Definition: The linear transformation L : <n → <m surjective onto) if for every B ∈ <n

there is X ∈ <m such that L(X) = B.

(c) Definition: B ∈ <n is a linear combination of {v1, v2, · · · , vm} ⊂ <n if there are scalars{α1, α2, · · · , αm} such that B = α1v1 + α2v2 + · · ·αmvm.

(d) Definition: The linear span of a set of vectors {v1, v2, · · · , vm} ⊂ <n is the set of all linearcombinations of {v1, v2, · · · , vm} ⊂ <n.

(e) Definition: The set of vectors {v1, v2, · · · , vm} ⊂ <n is linearly independent if wheneverα1v1 + α2v2 + · · ·αmvm = O then α1 = α2 = · · · = αm = 0. (Here O is the zero vector.)

(f) Definition: The set of vectors {v1, v2, · · · , vm} ⊂ <n is linearly dependent if there arescalars {α1, α2, · · · , αm} not all zero such that α1v1 + α2v2 + · · ·αmvm = O. (Here O is thezero vector.)

(g) Definition: The matix A is invertible if there is a matrix B such that AB = BA = I,the identity matrix.

(h) Definition: If V ⊂ <n then V is a subspace of <n if:

• Whenever v1, v2 ∈ V then v1 + v2 ∈ V• Whenever v ∈ V and α ∈ < then α v ∈ V

Equivalently, every linear combination of a set of vectors in V is itself in V .

(i) Definition: If A is an m × n matrix then the column space of A is {AX | X ∈ <n}.Equivalently, the column space of A is the set of all linear combinations of its columns.

(j) Definition: If A matrix then the rank of A is the number of pivots in any REF of A.

(k) Definition: The matrix A is in row echelon form (REF) if its pivot column list is strictlyincreasing (by convention ∞ <∞).

(l) Definition: The matrix A is in reduced row echelon form (RREF) if it is in REF and everypivot column is a column of the identity matrix.

17

(m) Explain why any three vectors in <2 are linearly dependent.

Ans: Let A1, A2, A3 ∈ <2 and let A = [A1, A2, A3] be the matrix with columns A1, A2, A3.The matrix A is 2 × 3 so when in REF there will be a free variable. This means that thehomogenous system AX = O has infinitely many solutions.

(n) Give three vectors in <2 which are dependent but any two are independent.

Ans: For example

(10

),

(01

),

(11

). To check directly that

(01

)and

(11

)are

independent write α

(01

)+ β

(11

)=

(00

). Then α 0 + β 1 = 0, α 1 + β 1 = 0. So

α = β = 0. The other two pairs are the same or easier.

1.18 October 5

1. Suppose A = [C1, C2, C3, C4] is a matrix with 4 rows and 4 columns, written as a row if itscolumns. If V1 and V2 are in <4 and C3 = 5V1 − 7V2 then complete the following:

(a) det(A) = det([C1, C2, C3, C4]) = det( ) + det( )

(b) The alternating property of the determinant says thatdet([C1;C2;C3;C4]) = −det( )(There is more than one correct answer.)

(c) If A is the identity matrix then the normalization property says that det(A) =

2. If C1 = C3 then use the three properties of determinants to explain why det(A) = 0.Ans: permuting C1 and C3 does not change A but multiplies det(A) by −1. So det(A) = −det(A)so det(A) = 0.

3. If C1 = 2C2 − 3C3 then use the three properties of determinants to calculate det(A).Ans: 0

4. Suppose M is a 3 by 3 matrix with non-zero determinant. Explain how one could, in principle,use Cramer’s Rule to construct the inverse of M .

Ans: Since det(M) 6= 0, Cramer’s rule gives you a way to solve any equation MX = B for anyB ∈ <3. Use this to solve MXi = ei where {e1, e2, e3} is the standard basis (i.e. the columns ofI3, the 3 by 3 identity matrix. Then if N = [X1, X2, X3] , the matrix with columns X1, X2, X3

then MN = M [X1, X2, X3] = [MX1,MX2,MX3] = [e1, e2, e3] = I3

So N is the inverse of M.

5. If A =

−3 3 2−3 0 2

0 3 1

, and B =

001

solve AX = B using Cramer’s Rule. Be sure to make it

clear that you are using Cramer’s rule and not simply taking the solution from a calculator.

Ans: If X =

abc

then a69, b = 0

9, c = 0

9

18

6. Suppose A is a 4 by 4 matrix and

A

−325−1

= B.

If A = [A1A2 A3 A4] where Ai is the ith column of A and the determinant of the matrix C =[A1A2, B,A4] is 100 then what is the determinant of A?Ans: 20.

1.19 October 7

1. Is it possible to find a value of x for which the determinant of the following matrix is 10?∣∣∣∣∣∣∣∣1 2 3 42 4 6 8x x 0 11 −x 0 1

∣∣∣∣∣∣∣∣Ans: No, the second row is a multiple of the first so no matter the value of x the determinantwill be 0.

2. Suppose A = [A1, A2, A3] is a matrix written as a row of its columns and R = [R1, R2, R3] is arow echelon form of A, also written as a row of its columns. If R1 + 2R2 + 3R3 = O explain whyA1 + 2A2 + 3A3 = O.

Sln: O = R1 + 2R2 + 3R3 = R

123

says that

123

∈ Null(R), the nullspace of R. But A

and R have the same nullspace so

123

∈ Null(A). So O = A

123

= A1 + 2A2 + 3A3

1.20 October 10

1. Suppose C = [C1, C2, C3] is a matrix with 7 rows and 3 columns and A =

3 3 −4−1 0 12 1 −2

is

an invertible matrix with inverse A−1 =

−1 2 30 2 1−1 3 3

. Let B = CA and write the matrix

B = [B1, B2, B3] as a row of its columns.

(a) Suppose V = C1 + C2 + C3. Find a, b, c such that V = aB1 + bB2 + cB3.

(b) Suppose W = B1 +B2 +B3. Find e, f, g such that W = eC1 + fC2 + gC3.

19

Solution: (a) We are given V = C

111

.

From B = CA we have BA−1 = C so

V = C

111

= (BA−1)

111

= B(

−1 2 30 2 1−1 3 3

111

= B

435

so

abc

=

435

.

(b) We are given W = B

111

. From B = CA

W = B

111

= CA

111

= C

3 3 −4−1 0 12 1 −2

111

= C

201

so

efg

=

201

2 October 12

1. Let V = {f : [0, 1]→ < | f is continuous}, the set of all continuous real -valued functions on the

unit interval. Let W = {f ∈ V |∫ 1

0f(x) dx = 0}. Given that V is a vector space with the usual

addition of functions and multiplication of functions by scalars, show that V is a subspace of W .

Ans: We must show that W is closed under addition and scalar multiplication.Suppose f, g ∈ W then

∫ 1

0(f+g)(x)dx =

∫ 1

0f(x)dx+

∫ 1

0g(x)dx = 0+0 = 0. Therefore f+g ∈ W .

If r ∈ < and f ∈ W then∫ 1

0(rf)(x)dx = r

∫ 1

0f(x) = r(0) = 0. Therefore rf ∈ W .

2. Let P3 = {a0 + a1 x + a2 x2 + a3 x

3 | ai ∈ < for each i} and Q = {f ∈ P3 | f(1) = f(2) = 0}.Given that P3 is a vector space, show that Q is a subspace of P3.

Ans: We must show that Q is closed under addition and scalar multiplication.Suppose f, g ∈ Q then f(1) = f(2) = 0 and g(1) = g(2) = 0. So (f + g)(1) = f(1) + g(1) =0 + 0 = 0 and (f + g)(2) = f(2) + g(2) = 0 + 0 = 0 so f + g ∈ Q.If r ∈ < and f ∈ Q then f(1) = f(2) = 0 so (rf)(1) = rf(1) = r(0) = 0 and (rf)(2) = rf(2) =r(0) = 0. Therefore rf ∈ Q.

3. Let V = {(xy

)∈ <2 |xy = 0}. Determine whether V is or is not a subspace of <2..

Solution: It is not because it is not closed under addition. v1 =

(10

)and v2 =

(01

)are

both in V but v1 + v2 =

(11

)is not in V .

20

3 October 14

1. Consider the matrices A =

1 3 0 1 31 −1 2 0 12 2 2 1 40 4 −2 1 2

The REF of A from the “strict” algorithm is M =

1 3 0 1 30 −4 2 −1 −20 0 0 0 00 0 0 0 0

and general solution to the homogenous linear system AX = O is

xyzwt

= t

−32−12

001

+ w

−14−14

010

+ z

−3212

100

Use the information above to answer the following.

(a) Determine a basis for Col(A). You must make it clear how you use the above informationto determine the basis.

Ans: The pivot column vectors of A are a basis for Col(A). So columns 1,2 of A are a basisfor Col(A).

(b) What is the dimension of Col(A)? Explain your response.

Ans: The dimension is 2 since there is a basis with 2 elements.

(c) Determine a basis for Null(A), the nullspace of A. You must explain why your set is a basis.

Ans:The solutions to the homogenous system AX = O is the null space of A. The abovegeneral system expresses each element of the null space as a linear combination of the vectors

Bt =

−32−12

001

, Bw =

−14−14

010

, and Bz =

−3212

100

so Bt, Bw, Bz is a spanning set. This

set of vectors is also linearly independent since the last three rows of the matrix [Bz, Bw, Bt]form a 3 by 3 identity matrix which means that if αBz +βBw + γBt = O then the last three

rows of this combination are

αβγ

=

000

.

(d) What is the dimension of Null(A)? Explain your response.

Ans: 3 since it has a basis of 3 elements.

21

(e) Use the given information to express the last column of A as a linear combination of thebasis you have given without doing any calculations.

Ans. If Ai is the ith column of A then the product ABt = O says that A5 = 32A1 + 1

2A2

4 October 17

1. Suppose V and W are subspaces of the vector space Q over <. Show that V ∩W is a subspaceof W .Ans: Suppose v1, v2 ∈ V ∩W and a, b ∈ < then v1, v2 ∈ V so av1 + bv2 ∈ V and v1, v2 ∈ W soav1 + bv2 ∈ W . Therefore av1 + bv2 ∈ V ∩W . Also, the zero vector of Q is in both V and Wsince it is in every subspace of Q; so V ∩W is not empty. Therefore V ∩W is a subspace of Q.Since it is a subset of W , it is also a subspace of W ; for the same reason, it is a subspace of V .

2. If V is a vector space of dimension 4 and C ⊂ V is independent and has 4 elements then explainwhy C must be a basis for V .

Ans: C = {c1, c2, c3, c4} is independent so all that is required for it to be a basis is that itbe a spanning set for V . Let v be any element of V then {c1, c2, c3, c4, v} is 5 elements of<4 which must be dependent. This says that there are r1, r2, r3, r4, r5 not all zero such thatr1c1 + r2c2 + r3c3 + r4c4 + r5v = 0. It is not possible for r5 to be 0 since that would mean thatC is a dependent set. Since r5 is not 0 the previous equation can be solved for v as a linearcombination of the ci so any v ∈ V is in the linear span of C which makes C a basis.

3. If V is a vector space of dimension n then explain why any subspace of V of dimension n mustbe equal to V .

Ans: Let B = {b1, · · · , bn} be a basis for W a subspace of V and let v be any element of V . Theset {b1, · · · , bn, v} is in the linear span of any basis for V . Since the basis has n elements andthis set has n + 1 it is dependent. So there are numbers ri, not all zero, such that r1b1 + r2b2 +· · · + rnbn + rn+1bs+1 = O. It is not possible for rn+1 to be 0 since this would make B linearlydependent. This means we can divide through by rn+1 and solve for v in terms of B which meansthat v ∈ W . So v ∈ W which means that W = V since v was any element of V .

4. Suppose V is a four dimensional vector space and {b1, b2, b3, b4} is a basis for V . Let W be thelinear span of {b1, b2} and Q the linear span of {b3, b4}. Show that O is the only vector commonto W and Q. That is, show that W ∩Q = {O}.

Ans: If v is common to W and Q then v = αb1 + βb2 and v = γb3 + δb4. Write v − v = O andshow that all of the coefficients are 0 so v = 0.

5 October 19

1. Let P 3 be the vector space of polynomials of degree at most 3 and let B be the (ordered) basis[x3, x2, x, 1] for P 3 and C the (ordered) basis [1, x2, x3, x]. If f = (x−1)(x2 +x+ 1), what is [f ]B,

22

the coordinate vector for f , relative to the basis B? What is [f ]C?

Ans: [f ]B =

100−1

, [f ]C =

−1010

2. What are the possible dimensions of vector spaces of <4?

(b) For each possible dimension, give a supspace of <4 of that dimension.

Ans: 0, 1, 2, 3, 4. By the previous problem no subspace of dimension greater than 4 is possible.(b) The space {O} has dimension 0. If {e1, e2, e3, e4} is the standard basis for <4 then Vi, thelinear span of {e1, · · · , ei} is a basis and has i elements so it has dimension i.

3. Suppose M =

−1 2 11 0 −1−1 4 0

, N =

−1 1 −12 0 41 −1 0

,

M−1 =

2 2 −112

12

02 1 −1

, N−1 =

2 12

22 1

21

−1 0 −1

Let V be a vector space and B = [x, y, z] a basis for V . Let g1 = z−x+2y, g2 = x−z, g3 = 4y−xand let G = [g1, g2, g3].

(a) Determine [g1]B, [g2]B,[g3]B, the coordinates of g1, g2, g3 relative to B.

Ans: [g1]B =

−121

, [g2]B =

10−1

, [g3]B =

−140

(b) From the above choices, find a matrix [G]B such that B [G]B = G.

Ans: G = B [[g1]B, [g2]B, [g3]B]] = B

−1 1 −12 0 41 −1 0

so [G]B =

−1 1 −12 0 41 −1 0

(c) From the above choices, find a matrix [B]G such that G [B]G = B.

Ans: From G = B [G]B we have G [G]−1B = B so [B]G = [G]−1B =

2 12

22 1

21

−1 0 −1

(d) What is [x]B

23

Ans:

100

(e) What is [x]G

Ans: We know [x]B, x = B [x]B.From (b) B = G [B]G sox = B [x]B = (G [B]G) [x]B = G ([B]G [x]B)

So [x]G = [B]G [x]B =

2 12

22 1

21

−1 0 −1

100

=

22−1

6 October 21

1. Calculate determinant (

∣∣∣∣∣∣1 2 −13 −1 40 1 5

∣∣∣∣∣∣) by cofactor (Laplace) expansion down the third column.

You must use cofactor expansion down the third column.

Ans: (−1)(−1)1+3det(

∣∣∣∣ 3 −10 1

∣∣∣∣) + 4(−1)2+3det(

∣∣∣∣ 1 20 1

∣∣∣∣) + 5(−1)3+3det(

∣∣∣∣ 1 23 −1

∣∣∣∣)2. Calculate the determinant of A =

1 −1 −23 −3 −1−1 2 −1

by Laplace expansion across the third row.

You must use cofactor expansion across the third row.

Ans: (−1)(−1)3+1det(

∣∣∣∣ −1 −2−3 −1

∣∣∣∣) + 2(−1)3+2det(

∣∣∣∣ 1 −23 −1

∣∣∣∣) + (−1)(−1)3+3det(

∣∣∣∣ 1 −13 −3

∣∣∣∣)3. If A is an b× n matrix with ai,j the entry in the ith row and jth column of A then the adjoint ofA, denoted adj(A), is the matix whose entry in the ith row and jth column is Aj,i.

(a) If A =

(a bc d

)then adj(A) =

Ans:

(d −b−c a

)

(b) If B =

1 −2 53 −1 2−1 4 5

then adj(B) =

Ans: Using

∣∣∣∣ a bc d

∣∣∣∣ for det(

(a bc d

)), we have

24

transpose (adj(B)) =

+

∣∣∣∣ −1 24 5

∣∣∣∣ − ∣∣∣∣ 3 2−1 5

∣∣∣∣ +

∣∣∣∣ 3 −1−1 4

∣∣∣∣−∣∣∣∣ −2 5

4 5

∣∣∣∣ +

∣∣∣∣ 1 5−1 5

∣∣∣∣ − ∣∣∣∣ 1 −2−1 4

∣∣∣∣+

∣∣∣∣ −2 5−1 2

∣∣∣∣ −∣∣∣∣ 1 5

3 2

∣∣∣∣ +

∣∣∣∣ 1 −23 −1

∣∣∣∣

=

−13 −17 1130 10 −21 13 5

so

adj(B) =

−13 30 1−17 10 1311 −2 5

4. if A =

1 0 1−2 −1 13 2 −1

then adj(A) =

2 11 −4−1 −1

Ans:

adj(A) =

−1 2 11 −4 −3−1 −2 −1

7 October 24

1. Example:Let V = P 3, the polynomials of degree at most 3 and W = P 2 the polynomials of degree atmost 2. Let B = [x3, x2, x, 1] be chosen as a basis for V and C = [1, x, x2] a basis for W . LetT : V → W be defined by T (f) = d

dxf . Calculate [T (B)]C , the matrix representing T relative to

the bases B and C.

Solution: T (x3) = 3x2 = C

003

, T (x2) = 2x = C

020

, T (x) = 1 = C

100

, T (1) = 0 =

C

000

. Therefore

[T (B)]C =

0 0 1 00 2 0 03 0 0 0

If f = ax3 + bx2 + cx+ d = [x3, x2, x, 1]

abcd

then [f ]B =

abcd

and

T (f) = C [T (B)]C [f ]B

25

T (f) = C

0 0 1 00 2 0 03 0 0 0

abcd

T (f) = [1, x, x2]

c2b3a

= c+ 2bx+ 3ax2

2. Example: Let V = P 2 the polynomials of degree at most 2 with basis B = [b1, b2, b3] = [1, x, x2]and define D : V → V by D(f(x)) = (1 + x)f ′′(x).

(1) Calculate the matrix M such that [D(f)]B = M [f ]B for every f ∈ P 2.

ans: [D(B)]B = [[D(1)]B, [D(x)]B, [D(x2)]B]] =

0 0 20 0 20 0 0

f = B [f ]B so

[D(f)]B = [D(B)]B [f ]B

[D(f)]B =

0 0 20 0 20 0 0

[f ]B

(2) Calculate the image of D.The B−coordinates of the image of D are the column space of [D(f)]B. An REF of this ma-

trix will have only

220

as a pivot column so this is a basis for the column space of [D(f)]B.

Therefore B

220

= 2 + 2x is a basis for the image of D.

(3) Calculate the kernel of D.

The nullspace of [D(B)]B is the set of coordinate vectors of elements of the kernel of D.

The homogenous system [D(B)]BX = 0 has solution

xyz

= x

100

+y

010

so the vectors

{

100

,

010

} are a basis for the null space of [D(B)]B. These are the coordinate vectors of

{1, x} so {1, x} is a basis for the kernel of D.

26

3. Example: Let V = P 2 the polynomials of degree at most 2 with basis B = [b1, b2, b3] = [1, x, x2]and W also P 2 but with basis C = [c1, c2, c3] = [x2, x, 1] define T : V → W by T (f(x)) =2f(x)− xf ′(x).

(1) Calculate the matrix [T (B)]C such that [T (f)]C = [T (B)]C [f ]B for every f ∈ V .

ans: [T (B)]C = [[T (1)]C , [T (x)]C , [T (x2)]C ]] =

[[2]C , [x]C , [0]C ]] =

0 0 00 1 02 0 0

f = B [f ]B so

[T (f)]C = [T (B)]C [f ]B

[T (f)]C =

0 0 00 1 02 0 0

[f ]B

(2) Calculate the image of T .

The C−coordinates of the image of T are the column space of [T (B)]C . The columns {

002

,

010

},are a linearly independent spanning set so this is a basis for the column space of [T (f)]C . There-

fore C

002

= 2, C

010

= x is a basis for the image of T .

(3) Calculate the kernel of T .The nullspace of [T (B)]C is the set of coordinate vectors with respect to B of elements of thekernel of T .

The homogenous system [T (B)]CX = 0 has solution

xyz

= z

001

so the polynomial

B{

001

} = x2 is a basis for the kernel of T .

8 October 26

1. Let V be the subspace of P 2 spanned by S = {x2 − 5x+ 1, x2 − 2x+ s, sx2 + x+ 1}.Find all values of s for which S is not a basis for P 2.

Ans: If B = [x2, x, 1] is chosen as a basis for P 2 then [S]B =

∣∣∣∣∣∣2 1 s−5 −2 11 s 1

∣∣∣∣∣∣.S is a basis if and only if [S]B is invertible. det([S]B) = −5s2 + 2 so S is a basis if and only if s

27

is not ±√

25

2. Suppose S = {v1, v2, v3} and T = {w1, w2, w3, w4} are both subsets of a vector space V and thatT is contained in the linear span of S. Explain why T is linearly dependent.

Ans: span(S) is a vector space of dimension at most 3 so no linearly independent subset ofspan(S) can have more than 3 elements.

3. Suppose A =

1 1 00 0 00 1 2

and T : <3 → <3 is given by T (v) = Av − tv.

(a) Calculate the standard matrix for T .

Ans:

1− t 1 00 0− t 00 1 2− t

(b) Find a value of t for which A(

001

= t

001

Ans: 2.

9 October 28

1. Determine whether each of the following is a subspace of the given vector space:

(a) A = {(xy

)∈ <2 | (x− y)(x2 + y2) = 0}

(b) B = {f(x) ∈ C0[0, 1] | f(0) = f(1) = 0} where C0[0, 1] = {f : [0, 1]→ < |f is continuous }.

(c) C = {(xy

)∈ <2 | sin(xy)2 ≤ 1}

(d) D = {(xy

)∈ <2 | xy = 0}

ans: A is because the condition is equivalent to a = b = 0. B is a subspace because the mapT : C0[0, 1] → <2 defined by T (f) = (f(0), f(1)) is a linear map and the kernel of a linear mapis a subspace. C is because the condition holds for all x, y, D is not because, for example, each

of

(10

)and

(01

)is in D but their sum is not.

2. If L : V → <3 is a linear transformation, f, g are linearly independent elements of V , L(f) = 101

, and L(g) =

202

, then f + g is a non-zero element of kernel(L) because ...

Ans: 2,−1 since L(2f − g) = 2L(f)− L(g). (why is 2f − g not O?)

28

3. If A is a 7 by 7 matrix then the determinant of (5A) is *determinant(A).

Ans: 57.

4. If L : V → W is an injective linear transformation and {f, g, h} is a linearly independent set inV , then {L(f), L(g), L(h)} is a linearly set in W .(explain).

Ans: Independent. This is because if aL(f) + bL(g) + cL(h) = O then L(a(f) + b(g) + c(h)) = O(why?) and since L is injective this means that a(f) + b(g) + c(h) = O which means thata = b = c = 0 (why?)

5. If L : V → W is a surjective linear transformation and V is finite-dimensional, explain why thedimension of W is less than or equal to the dimension of V .

Ans: The fundamental equation isdim(Image(L))) +dim(kernel(L)) = dim(V ).Since L is surjective, Image(L) = W sodim(W ) +dim(kernel(L)) = dim(V ).

6. If L : V → W is an injective linear transformation and both V and W are finite dimensional,explain why the dimension of V is less than or equal to the dimension of W .

Ans: dim(Image(L) +dim(kernel(L)) = dim(V ).Since L is injective, kernel(L)= {O}Image(L) is a subspace of W so dim(Image(L) )≤ dim(W )so dim(V ) = dim(Image(L)) ≤ dim(W )

10 October 31

1. Calculate the adjoint of A =

1 1 12 3 44 9 16

Ans:

12 −7 1−16 12 −2

6 −5 1

2. Suppose A and B are two square matrices of the same size. Explain why AB is invertible if and

only if A is invertible and B is invertible.

Ans: The square matrix M is invertible if and only if det(M) 6= 0.

det(AB) = det(A)det(B) so det(AB) = 0 if and only if det(A) = 0 or det(B) = 0.

Therefore det(AB) 6= 0 if and only if det(A) 6= 0 and det(B) 6= 0.

29

3. Let T : R3 → R3 be T = TA for some matrix A and let T 2(v) = T (T (v))

(a) Why is ker (T ) a subspace of ker T 2?

(b) Why is image(T 2) a subspace of image(T )?

(c) Show that A2 is AT 2 , the standard matrix for T 2.

Ans: (a) If T (v) = O then T (T (v)) = T (O) = O(b) If v ∈ image(T 2) then v = T (T (u)) for some u so if w = T (u) then v = T (w) ∈ image(T )

(c) AT 2 = [T 2(e1), T2(e2), T

2(e3)] =[T (Ae1), T (Ae2), T (Ae3)] =[AAe1, AAe2, AAe3)] = [A2e1, A

2e2, A2e3)] = A2

4. Let A =

−4 3 32 −3 −2−1 0 −2

. Find a number c such that A

3−21

= c

3−21

ans: −5

11 November 2

1. Let b and c be distinct numbers and define f(x) = det

1 1 1x b cx2 b2 c2

.

(a) Use cofactors to show that f(x) is a polynomial of degree at most 2. Evaluate the coefficientu of x2 to show that the degree is exactly 2.

Ans: Expanding the determinant using cofactors of the first column expresses f(x) as apolynomial whose coefficients are the coeffactors. In particular, the coefficient u of x2 is

u = det

(1 1b c

)= −(b− c).

(b) Explain why f(b) = f(c) = 0. Why is it true that f(x) = u(x− b)(x− c)?Ans: If x = b, then the first two columns of the determinant defining f(b) are equal and sof(b) = 0; similarly, f(c) = 0. Since b and c are roots of the polynomial f(x), it is divisibleby (x− b)(x− c) and since u is the coefficient of x2, we must have f(x) = u(x− b)(x− c).

(c) Let a, b, and c be numbers. The Vandermonde determinant for a, b, and c is defined to be

D = det

1 1 1a b ca2 b2 c2

. Prove that the Vandermonde determinant is nonzero if and only

if the numbers a, b, and c are pairwise distinct.

Ans: If any two of a, b, and c are equal, then the Vandermonde determinant is zero sincetwo columns are equal. Otherwise, its value is (c− b)(a− b)(a− c) which is zero if and onlyif two of the quantities are equal. Analogous results are true regardless of the number of

30

quantities a, b, c, etc. one has. You should check to see what the value of the Vandermondedeterminant is when one has four or more numbers; use care to get the signs right! Thegeneralization can then be used in extending the next problem.

2. Explain why given any three points (a, d), (b, d), and (c, d) in the plane with distinct x-coordinates,it is always possible to find one and only one parabola y = sx2+tx+u whose graph passes throughall three points. For example, when the points are (-2, 13), (1, 4), and (3,8), what is this uniqueparabola?

Ans: The parabola is the unique solution of s+at+a2u = d, s+bt+b2u = e, and s+ct+c2u = f .The solution is unique because the determinant of the coefficient matrix is nonzero when a, b,c are distinct. Solving the system of equations for the particular points gives the parabolay = 5− 2x+ x2.

3. Let a, b, and c be three numbers and define D = det

eax ebx ecx

(eax)′ (ebx)′ (ecx)′

(eax)′′ (ebx)′′ (ecx)′′

where the prime

and double prime indicate first and second derivative.

(a) Explain why D = e(a+b+c)xdet

1 1 1a b ca2 b2 c2

.

Ans: Expanding out the derivatives, one sees that the columns are multiples of eax, ebx

and ecx respectively and so these factors can be taken out of the determinant leading to theformula to be proved.

(b) Explain why the functions eax, ebx, and ecx are linearly independent if and only if a, b, andc are pairwise distinct.

Ans: Since e(a+b+c)x is never zero, the result follows from the last part of the last problem.Generalize the result to the case where one has four or more numbers a, b, c, d, etc.

12 November 4

1. If A is a square n by n matrix with entries in a field F and v ∈ F n is a vector, then v is aneigenvector of A with associated eigenvalue λ if v 6= 0 and Av = λv. Any λ ∈ F is called aneigenvalue of A if there is an eigenvector v ∈ F n for which λ is the associated eigenvalue; in thiscase we say that v is an eigenvector associated with λ or simply that v is a λ-eigenvector of A.

(a) For any λ ∈ F , the nullspace of A − λI (where I is the n by n identity matrix) is calledthe λ-eigenspace of A. Why is the λ-eigenspace of A a subspace of F n? Explain why thenonzero vectors in the λ-eigenspace of A are eigenvectors of A with associated eigenvalue λ.

Ans: The nullspace of any m by n matrix is a subspace of F n. A nonzero vector v is aneigenvector of A with associated eigenvalue λ if and only if Av = λv, i.e. v is in the nullspaceof A− λI.

(b) The polynomial f(λ) = det(A−λI) is called the characteristic polynomial of A. Explainwhy the eigenvalues of A are just the roots of f(λ) = det(A− λI) which are in F .

Ans: The number λ is an eigenvalue of A if and only if there is a nonzero vector v suchthat (A− λI)v = 0 if and only if the columns of A− λI are linearly dependent if and onlyif f(λ) = det(A− λI) = 0.

31

2. Let A =

(0 1−6 5

)∈ <4.

(a) Find the characteristic polynomial f(λ) of A.

Ans: f(λ) = det(

(−λ 1−6 5− λ

)) = −λ(5− λ)− 1(−6) = λ2 − 5λ+ 6.

(b) Find the all the eigenvalues of A in < by solving f(λ) = 0. For each of these λ find a basisof the λ-eigenspace of A.

Ans: The roots of f(λ) are 2 and 3. The nullspace of A− 2I =

(−2 1−6 3

)is span(

(12

))

and the nullspace of A− 3I =

(−3 1−6 2

)is span(

(13

)). These are the 2-eigenspace of A

and the 3-eigenspace of A respectively.

(c) Choosing v1 =

(12

)a 2-eigenvector of A and v2 =

(13

)a 3-eigenvector of A, explain

why B = (v1, v2) is a basis of <2. The map TA : <2 → <2 is defined to be the linear mapsuch that TA(X) = AX for every X ∈ <2. Calculate the matrix of TA with respect to thebasis B for both the domain and codomain. Note that it is a diagonal matrix.

Ans: Because the vj are eigenvectors, we have Av1 = 2v1 and Av2 = 3v2. Putting these to-

gether, we getB(B−1AB) = AB = A(v1, v2) = (Av1, Av2) = (2v1, 3v2) = (B

(20

), B

(03

)) =

B

(2 00 3

). But this says that

(2 00 3

)= [AB]B = [TA(B)]B which is the matrix of TA

with respect to the bases B for both the domain and the codomain. The matrix is diagonalwith the eigenvalues along the main diagonal.

13 November 7

1. Now consider the matrix A =

(0 −11 0

)∈ <4. The matrix A is the matrix of a linear transfor-

mation T which rotates the plane about the origin by 90 degrees.

(a) Explain geometrically why there are no real eigenvectors.

Ans: Every vector in the real plane is rotated by 90 degrees and so its direction is changed;so no vector can be be an eigenvector.

(b) Calculate the characteristic polynomial and find all the complex eigenvectors and their as-sociated eigenvalues.

Ans: The characteristic polynomial is f(λ) = det(A−λI) = det(

(−λ −11 λ

)) = λ2+1. Its

roots are the eigenvalues i and −i. The i-eigenspace is the nullspace of A−iI =

(−i −11 −i

)which is just span(

(1i

)); and the −i-eigenspace is the nullspace of A + iI =

(i −11 i

)which is just span(

(1−i

)).

32

(c) Find a basis B of eigenvectors and the matrix of TA respect to this basis B for both thedomain and codomain. Again, you should get a diagonal matrix.

Ans: For example, letting B =

(1 1i −i

), the fact that the columns of B are eigenvectors

with eigenvalues i and −i can be expressed as B

(i 00 −i

)= AB. So,

(i 00 −i

)= B−1AB

is the matrix of A with respect to the basis B.

2. Consider the matrix A =

(0 1−4 4

)∈ <4.

(a) Find all the eigenspaces of A for all real and complex eigenvalues. Is there a basis ofeigenvectors of A for either the real or complex plane?

Ans: The characteristic polynomial is f(λ) = det(A − λI) = (−λ)(4 − λ) − (1)(−4) =λ2 − 4λ + 4 = (λ − 2)2. So, the only eigenvalue is λ = 2. The corresponding eigenspace is

the nullspace of A − 2I =

(−2 1−4 2

)which is span(

(12

)). Since the plane is dimension

2, there cannot be a basis of eigenvectors.

(b) Explain why the matrix of TA with respect to any basis B (for both the domain andcodomain) can never be a diagonal matrix.

Ans: If there were a basis B such that the matrix of TA with respect to this basis were adiagonal matrix D, then we would have D = B−1AB and so BD = AB. But this saysthat the columns of B are eigenvectors of A with associated eigenvalues the correspondingdiagonal entries of D. This contradicts the result of part(a).

Understanding A is hard because Kernel A− 2I does not span <2. But notice that Kernel

(A−2I)2 does span <2 (because (A−2I)2 =

(−2 1−4 2

)2

= 0). Choose any v2 ∈ Kernel(A−

2I)2 which is not in Kernel A− 2I and let v1 = (A− 2I)v2. Then v1 is an eigenvector of Aand v2 is called a generalized eigenvector of A. Let B = (v1, v2). Then the matrix of TA with

respect to the basis B for both the domain and codomain if B−1AB =

(2 10 2

). Try to

understand this without calculating B−1AB and then calculate it to verify. We, of course,didn’t get a diagonal matrix; but it is still pretty easy to understand: v2 maps to v1 whichmaps to 0 where the map is TA−2I .

14 November 9

The quiz will ask you to examine specific 2 by 2, matrices to determine those that are diagonalizable.For those that are diagonalizable, you will be asked to find all possible diagonalizations using variousbases of eigenvectors. If a matrix is not diagonalizable, you will need to prove that this is the case.

For examples of how to calculate diagonalizations, see the Quiz examples for November 4 and 7.What follows is just an explanation of why insufficient eigenvectors cause lack of diagonalizability andwhy permuting the order of the eigenvectors gives all possible diagonalizations.

An n by n matrix A is said to be diagonalizable if there is an invertible n by n matrix B such thatD = B−1AB is a diagonal matrix. In this case any such D is called a diagonalization of A. WritingB = (v1, ..., vn) where vj are column vectors and letting λ1, ..., λn be the entries on the main diagonal

33

of D, the equation BD = AB is the same as the n equations λjvj = Avj for j = 1, ..., n. So, if A isdiagonalizable, then it has a basis of eigenvectors B, n.b. the vj must be nonzero since B is linearlyindependent. So any matrix A that does not have a basis of eigenvectors is NOT diagonalizable.

Let A be an n by n matrix, λ1, ..., λr be all the (distinct) eigenvalues of A, d1, ..., dr be the dimensionsof the λj-eigenspaces. Then d1 + ... + dr ≤ n, and we have equality if and only if A has a basis ofeigenvectors. In particular, any basis of eigenvectors must have exactly dj λj-eigenvectors in it. (Whyis this true?) So, if A has a diagonalization D, then λj must appear exactly dj times on the maindiagonal of D. This means that any two diagonalizations D1 and D2 then we can obtain D2 from D1

by permuting the entries on the main diagonal. In particular, if n = 2, then A can have at most twodiagonalizations if A has two distinct eigenvalues and can have only one if A has a single eigenvalue.

Suppose A is a diagonalizable two by two matrix, i.e. D = B−1AB where B = (v1, v2) and

D =

(λ1 00 λ2

). Then B2 = (v2, v1) = BE where E =

(0 11 0

). (Use the Voodoo principle to find

column swap column operation matrix!). Notice that E−1 = E is also the row swap row operationmatrix. So B−12 = E−1B−1 = EB−1. In particular, we have B−12 AB2 = EB−1ABE = EDE is D

with both columns swapped and both rows swapped, i.e. B−12 AB2 =

(λ2 00 λ1

). All of which just

says that if we swap the order of the eigenvectors, we swap the order of the diagonal entries in thediagonalization. You can do the same calculation on an n by n matrix with the same result; soall thediagonalizations are just the matrices obtained by permuting the entriesof the main diagonal of anygiven diagonalization.

Example: Let A be a diagonalizable n by n matrix. If A has only one diagonalization, show thatA is a scalar multiple of the identity matrix.

Ans: Let D be a diagonalization of A. The entries on the main diagonal of D are all the eigenvaluesof A, each repeated dimension of the eigenspace times. The other diagonalization are just like D exceptthat the the main diagonal entries are permuted. Since there are no other diagonalizations, it followsthat A has a single eigenvalue and its eigenspace has dimension n, i.e. D = λI. Since D = B−1ABwhere B = (v1, ..., vn) is a matrix of eigenvectors, we have A = BDB−1 = B(λI)B−1 = λBIB−1 = λI.

15 November 11

1. If A, S are matrices such that S−1AS = D then show that A5 = SD5S−1.

2. The matrix A =

(−18 −10

30 17

)has eigenvalues −3, 2 which respectively have eigenvectors(

−23

)and

(−1

2

). If S =

(−2 −1

3 2

)and D =

(−3 0

0 2

)then show that AS = SD

so S−1AS = D.

(a) Write A5 =

(−2 −1

3 2

)M5Q where M and Q are specific matrices. (no floating point).

(b) Give a formula for Ak for any k.Ans(a): From S−1AS = D we have A = SDS−1

so A5 = SDS−1SDS−1SDS−1SDS−1SDS−1 = SD5S−1

34

so A5 = S

(−3 0

0 2

)5

S−1 = S

((−3)5 0

0 25

)S−1

A5 =

(−2 −13 2

)((−3)5 0

0 25

)(−2 −13 2

)Ans(b):

(4(−3)k − 3(2)k 2(−3)k − 2(2k)−6(−3)k + 6(2)k −3(−3)k + 4(2k)

)

3. Let A =

−7 21 6−3 9 2−3 7 4

. This matrix satisfies det(A − λI) = −(λ − 2)3 and the reduced row

echelon form of A− 2I is R =

1 −7/3 −2/30 0 00 0 0

.

(a) Find a basis for the 2-eigenspace of A. Why is A not diagonalizable?

Ans: The 2-eigenspace is the subspace of solutions of (A− 2I)X = 0.

Since R is the RREF of A − 2I we can read off the basis

7/310

,

2/301

for the

nullspace. Since this is the only eigenspace of A, A has only two linearly independent eigen-vectors. But three are required for A to be diagonalizable.

(b) Show that (A− 2I)2 = 0 so it is all of C3.

Let v3 be any vector in Kernel (A−2I)2 which is not in Kernel (A−2I). Set v2 = (A−2I)v3,and let v1 be any vector in Kernel (A−2I) such that v1, v2 is linearly independent (so it is abasis of Kernel A− 2I). Given that v1, v2, v3 are linearly independent1, it is a basis for C3.What is the matrix of the linear transformation TA : C3 → C3 with respect to the basisv1, v2, v3 for both the domain and codomain?

Ans: The vectors satisfy: 2v1 = Av1, 2v2 = Av2, and 2v3 + v2 = Av3. These amount to the

single equation: (v1, v2, v3)

2 0 00 2 10 0 2

= A(v1, v2, v3). So, the matrix of TA with respect

to the basis v1, v2, v3 is

2 0 00 2 10 0 2

.

Note that, while not diagonal, this AT is “close” to diagonal and certainly much simplerthan A.

16 later review

1. The matrix A is similar to the matrix B if there is an invertible matrix S such that S−1AS = B.

Let A =

3 1 2 00 −1 2 10 0 7 60 0 0 9

. Explain why A is similar to a diagonal matrix and give one of the

1This is something that needs to be proven. Look for a chance to do so on a later quiz.

35

diagonal matrices similar to A. Do not do any calculating.Ans: A has 4 distinct eigenvalues {3,−1, 7, 9} so there are 4 linearly independent eigenvectorsV1, V2, V3, V4 for A. If S = [V1, V2, V3, V4] is the matrix with columns V1, V2, V3, V4. Then Sis invertible and AS = SD where D is diagonal so S−1AS = D and S is similar to D. Allpossible D have the eigenvalues of A on the diagonal so one diagonal matrix similar to A is

D =

3 0 0 00 −1 0 00 0 7 00 0 0 9

.

17 November 14

1. The matrix B is similar to the matrix A if there is an invertible matrix S such that S−1AS = B.If B is similar to A, prove that A is similar to B.

Ans: Suppose B = S−1AS. Set P = S−1. Then P−1 = (S−1)−1 = S soP−1BP = S(S−1AS)S−1 = A.

2. Suppose B is similar to A. Prove that A and B have the same characteristic polynomial.

Ans: If B = S−1AS then S−1(A − xI)S = (S−1A − S−1xI)S = S−1AS − S−1xIS = S−1AS −xS−1IS = B − xIso detB−xI = detS−1(A−xI)S = (detS−1)(detA−xI)(detS) = detA−xI where we have usedthe fact that the determinant of a product is product of the determinants and that, in particular,detS−1 = (detS)−1.

3. The matrix A =

1 4 −43 −3 13 −6 4

has eigenvalues 3,−2, 1. Calculate its characteristic polyno-

mial.

ans: A has three linearly independent eigenvectors since it has three different eigenvalues andeigenvectors for different eigenvalues are linearly independent. So S, a matrix with those eigen-vectors as columns is invertible and S−1AS = D is a diagonal matrix with 3, 2,−1 (in some order)on the diagonal. Since similar matrices have the same characteristic polynomial

det(A− xI) = det(

3− x 0 00 −2− x 00 0 1− x

) = (3− x)(−2− x)(1− x)

36

18 November 16

There are two questions below. The first will be used for the quiz on November 16; the second will beused during dead week to allow you to add points to your quiz score before the end of the semester.

1. Two column vectors v, w ∈ <n are said to be perpendicular (orthogonal) if vTw = 0, i.e. thestandard inner product of v and w is zero. Any two linearly independent column vectors u, v ∈ <n

span a subspace of dimension 2, a plane. The problem is to find a column vector w in this subspacewhich is perpendicular to u. Do this for u = (2,−1, 2)T and v = (4, 2, 5)T .

(a) Suppose w is a nonzero column vector in the plane which is perpendicular to u. Explainwhy v ∈ span(u,w)?

Ans: If w = cu with c ∈ <, then 0 = wTu = cuTu and so c = 0. So, w is independent of u,and w, u span a two dimensional subspace of the two dimensional space span(u, v). v mustbe in the span of u,w.

(b) So v = au + bw for some a, b ∈ < But then v − au = bw is perpendicular to u, i.e.(v − au)Tu = 0. Solve this linear system to find a.

Ans: The linear system is (v−au)Tu = 0. which is just: ((4, 2, 5)−a(2,−1, 2))(2,−1, 2)T =0, one linear equation in one unknown. The solution is a = 16/9.

(c) Find a vector w ∈ span(u, v) which is perpendicular to u.

Ans: We have (v−au)Tu = 0. So w = v−au = ((4, 2, 5)−16/9(2,−1, 2))T = (4/9, 34/9, 13/9)T

is such a vector.

2. Every square n by n matrix A over the complex numbers is similar to an upper triangular matrix.

To see why, we verify the result for the matrix A =

3 −1/3 −1/3−2 4/3 −5/35 −1/3 8/3

.

(a) Let T = TA. Find a basis B such that the the matrix of T with respect to B had its firstcolumn a multiple of e1. Why does any basis whose first vector is an eigenvector of A satisfythe condition? In our case, the characteristic polynomial of A is p(X) = (X − 3)(X − 2)2 sofind an eigenvector u with eigenvalue 3 and complete the basis with standard basis vectorsu2, u3.

Ans: To find the eigenvector, solve (A − 3I)u = 0. The reduced row echelon form of

A− 3I is

1 0 00 1 10 0 0

. So the 3-eigenspace is spanned by u =

0−11

. A possible basis

is B = (u, e1, e2).

(b) Calculate the matrix A2 of T with respect to the basis B1 = u, u2, u3.

Ans: The matrix A2 = B−1AB. One can calculate B−1 using the adjoint to get B−1 = 0 0 11 0 00 1 1]

, and then calculate the product. Or, one can calculate each column of A2

individually: for example, Au =

0−33

= 3u, Ae1 =

3−25

= 5u + 3e1 + 3e2, and

37

Ae2 =

−1/34/3−1/3

= −1/3u− 1/3e1 + e2. Either way, one has A2 =

3 5 −1/30 3 −1/30 3 1

.

(c) Let C be the submatrix of A2 obtained by deleting its first row and first column. Find aneigenvector v of C and a standard basis vector w such that v, w is a basis of C2.

Ans: The matrix C =

(3 −1/33 1

)has characteristic polynomial (2 − λ)2 and its 2-

eigenspace is spanned by v =

(13

)and so one can choose w =

(10

).

(d) Let S be the linear transformation whose standard matrix is A2. Find the matrix A3 of the

transformation S with respect to the basis B2 = e1,

(0w

),

(0u

).

(e) You have calculated an upper triangular matrix A3 which is similar to A and obtainedby changing the basis to B = B1B2. (If A had more than three rows, one could need tokeep repeating the process of finding eigenvectors of smaller and smaller matrices to get anuppertriangular matrix similar to the original matrix. N.B. You get an upper triangularmatrix, not a diagonal one.)

Ans: We have B2 =

1 0 00 1 10 3 0

. Its inverse computed with an adjoint is B−12 = 1 0 00 0 1/30 1 −1/3

. So A2 = B−12 A2B2 =

3 4 50 2 10 0 2

.

19 November 18

The quiz problem for 11/18 will be:

1. Two linearly independent vectors v and w determine a parallelogram whose diagonals are v + wand v − w and two of whose sides are v and w. The parallelogram law says that the sum of thesquares of the lengths of diagonals of a parallelogram is equal to the sum of lengths of its foursides.

(a) Verify the parallelogram law in the special case of a parallelogram in the plane determinedby two vectors v = (2, 5), w = (3,−4).

Ans: ||v + w||2 + ||v − w||2 = ((2 + 3)2 + (5 − 4)2) + ((2 − 3)2 + (5 + 4)2) = 108 and2(||v||2 + ||w||2) = 2(22 + 52) + (32 + (−4)2)) = 108.

(b) Prove the parallelogram law in the special case where the parallelogram is determined bytwo arbitrary vectors in R1.

Ans: Let v = (a) and w = (b). Then ||v + w||2 + ||v − w||2 = (a + b)2 + (a − b)2 =(a2 + 2ab+ b2) + (a2 − 2ab+ b2) = 2(a2 + b2) = 2||v||2 + 2||w||2.

(c) If v = (a, b) and w = (c, d), use the fact that the parallelogram law is true for the vectors(a) and (c) as well as for the vectors (c) and (d) to explain why it is also true for theparallelogram determined by v and w. Why is the parallelogram law true for all vectors inRn.

38

Ans: One has (a+ c)2 + (a− c)2 = 2(a2 + c2) and (b+ d)2 + (b− d)2 = 2(b2 + d2). Addingthe two equations gives ||v + w||2 + ||v − w||2 = 2(||v||2 + ||w||2) The result for vectors inRn is proved in the same way, you have the result for each coordinate and adding them upgives the result for the vectors.

(d) Use the properties of inner products to expand and simplify (v+w) ·(v+w)+(v−w) ·(v−w)giving another proof of the parallelogram law true for any inner product space.

Ans: Using bilinearity of the inner product (i.e. the distributive law), one has ||v + w||2 +||v−w||2 = (v+w)·(v+w)+(v−w)·(v−w) = (v·v+v·w+w·v+w·w)+(v·v−v·w−w·v+w·w) =2(v · v + 2w · w) = 2(||v||2 + ||w||2). Can you prove the result for Cn?

20 November 21

The quiz problems for 11/21 will be taken from:

1. Inner products on real vector spaces can be used to define the length of a vector as ||v|| =√v · v.

It is more surprising that this length function determines the inner product. Explain why v ·w =||v+w||2−||v−w||2

4for all vectors v, w in a real inner product space.

Ans: The easiest way to see that the identity is true is to express the right hand side as innerproducts and simplify the expression. One has ||v + w||2 − ||v − w||2 = (v + w) · (v + w) + (v −w) · (v−w) = (v · v+ v ·w+w · v+w ·w)− (v · v− v ·w−w · v+w ·w) = 2(u ·w+w · v) = 4v ·w.Dividing by 4 gives the result.

2. Show that the square of the average of n real numbers is at most equal to the average of the

squares of these numbers, i.e. (x1+···+xn

n)2 ≤ x2

1+···x2n

n. (Hint: x1 + · · ·xn = u · v where u = (1, . . . 1)

and v = (x1, . . . xn)).

Ans: The Cauchy-Schwarz Inequality gives u ·v ≤ ||u||||v||. Use ||u||2 = n to complete the proof.

3. Let u and v be vectors in a real inner product space. If ||u|| = 3, ||u+ v|| = 4, and ||u− v|| = 5,what is ||v||? (Hint: Draw a picture.)

Ans: We are dealing with a parallelogram. So the parallelogram law says ||u+ v||2 + ||u− v||2 =

2(||u||2 + ||v||)@ which amounts to 42 + 52 = 2(32 + ||v||2) or ||v|| =√

232

.

39