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8/2/2019 Ma 1506 Chap 6
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CHAPTER 6
LINEAR TRANSFORMATIONS
6.1 WHAT IS A LINEAR TRANSFORMA-
TION?
You know what a function is - its a RULE which
turns NUMBERS INTO OTHER NUMBERS: f(x) =
x2 means please turn 3 into 9, 12 into 144 and so
on.
Similarly a TRANSFORMATION is a rule whichturns VECTORS into other VECTORS. For exam-
ple, please rotate all 3-dimensional vectors through
an angle of 90 clockwise around the z-axis. A LIN-
EAR TRANSFORMATION T is one that ALSOsatisfies these rules: if c is any scalar, and
u and
v
are vectors, then
T(cu ) = cT(
u ) and T(
u +
v ) = T(
u ) + T(
v ).
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EXAMPLE: Let I be the rule Iu =
u for all
u .
You can check that I is linear! Called IDENTITY
Linear Transformation.
EXAMPLE : Let D be the rule Du = 2
u for all
u .
D(cu ) = 2(c
u ) = c(2
u ) = cD
u
D(u +
v ) = 2(
u +
v ) = 2
u + 2
v = D
u + D
v
LINEAR!
Note: Usually we write D(u ) as just D
u .
6.2. THE BASIC BOX, AND THE MATRIX
OF A LINEAR TRANSFORMATION
The usual vectors i and j define a square:
Lets call this the BASIC BOX in two dimensions.
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Similarly, i, j, and k define the BASIC BOX in 3
dimensions.
Now let T be any linear transformation. You know
that any 2-dimensional vector can be written as ai +
bj, for some numbers a and b. So for any vector, we
have
T(ai + bj) = aTi + bTj.
This formula tells us something very important: IF
I KNOW WHAT T DOES TO i and j, THEN I
KNOW EVERYTHING ABOUT T - because now I
can tell you what T does to ANY vector.
EXAMPLE: Suppose I know that T(i) = i + 14j
and T(j) = 14 i +j. Then what is T(2i + 3j)?
Answer: T(2i + 3j) = 2Ti + 3Tj = 2
i + 14j
+
314 i +
j
= 2i + 12j + 34 i + 3
j = 114 i +72
j.
Since Ti and Tj tell me everything I need to know,
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this means that I can tell you everything about T by
telling you WHAT IT DOES TO THE BASIC BOX.
EXAMPLE: Let T be the same transformation as
above, T(i) = i + 14j and T(j) = 14 i +
j.
The basic box has been squashed a bit! Pictures
of WHAT T DOES TO THE BASIC BOX tell us
everything about T!
EXAMPLE: If D is the transformation Du = 2
u ,
then the Basic Box just gets stretched:
So every LT can be pictured by
seeing what it does to the Basic Box.
There is another way!
Let Ti =
a
c
and Tj =
b
d
. Then we DEFINE
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THE MATRIX OF T RELATIVE TO i, j as
a b
c d
,
that is, the first COLUMN tells us what happened
to i, and the second column tells us what happened
to j.
EXAMPLE: Let I be the identity transformation.
Then Ii =
i =
10
, Ij =
j =
01
, so the matrix of
the identity transformation relative to ij is
1 00 1
.
EXAMPLE: If Du = 2
u , then Di =
20
and
Dj =
02
so the matrix of D relative to i, j is2 00 2
.
EXAMPLE: If Ti = i + 14j and Tj = 14 i +j, then
the matrix is
1
1
414 1
.
EXAMPLE: If Ti = j and Tj = i, the matrix is0 11 0
. Basic box is REFLECTED
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EXAMPLE: Suppose in 3 dimensions Ti = i+4j +
7k, Tj = 2i + 5j + 8k, Tk = 3i + 6j + 9k, then the
matrix is
1 2 34 5 67 8 9
, relative to ijk.
EXAMPLE: Suppose Ti = i + j + 2k and Tj =
i
3k. This is an example of a LT that eats 2-
dimensional vectors but PRODUCES 3-dimensional
vectors. But it still has a matrix,
1 11 0
2 3
. Its
just that this matrix is not a SQUARE MATRIX,
that is, it is not 2 by 2 or 3 by 3. Instead it is 3 by
2.
We shall say that a linear transformation is a 2-
dimensional L.T. if it eats 2-dimensional vectors AND
produces 2-dimensional vectors. A 2-dimensional
L.T. has a square, 2 by 2 matrix relative to i, j. Sim-
ilarly a 3-dimensional linear transformation has a 3
by 3 matrix. In Engineering applications, most lin-
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ear transformations are 2-dimensional or 3-dimensional,
so we are mainly interested in these two cases.
EXAMPLE: Suppose T is a linear transformation
that eats 3-dimensional vectors and produces
2-dimensional vectors according to the rule Ti = 2i,
Tj = i + j, Tk = i
j. What is its matrix?
Answer:
2 1 10 1 1
, a 2 by 3 matrix.
EXAMPLE: Suppose, in solid mechanics, you take
a flat square of rubber and SHEAR it, as shown.
In other words, you dontchange its volume,
you just push it like a pack
of cards. The base stays fixed but the top moves
a distance tan(). (The height remains the same, 1unit.) Clearly the shearing transformation S satisfies
Si = i, Sj = i tan + j, so the matrix of S relative
to i, j is
1 tan 0 1
.
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EXAMPLE: Suppose T i = i + j and Tj = i +
j. Matrix is
1 11 1
and basic box is SQUASHED
FLAT!
EXAMPLE: Rotations in the plane. Suppose you
ROTATE the whole plane through an angle (anti-
clockwise). Then simple trigonometry shows you
that
Ri = cos i + sin j
Rj = sin i + cos jSo the rotation matrix is
R() =
cos sin sin cos
.
Application: Suppose an object is moving on a cir-
cle at constant angular speed . What is its accel-
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eration?
Answer: Let its position vector at t = 0 b e
r0.Because the object is moving on a circle, its position
at a later time t is given by rotatingr0 by an angle
(t). So
r (t) =
cos sin sin cos
r0
Differentiate
d
rdt
=
sin cos cos
sin
r0 by the chain rule. Here
is actually , so
d
rdt
=
sin cos cos sin
r0. Differentiate again,
d2
rdt2
= cos sin sin cos
2r0
= 2
cos sin sin cos
r0
.
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Substitute the equation forr (t),
d
2
rdt2 = 2
r ,
which is formula you know from physics.
6.3. COMPOSITE TRANSFORMATIONS
AND MATRIX MULTIPLICATION.
You know what it means to take the COMPOSITE
of two functions: if f(u) = sin(u), and u(x) = x2,
then f
u means: please do u FIRST, THEN f, so
f u(x) = sin(x2). NOTE THE ORDER!!u f(x) = sin2(x), NOT the same!
Similarly ifA and B are linear transformations, then
AB means do B FIRST, then A.
NOTE: BE CAREFUL! According to our definition,
A and B both eat vectors and both produce vectors.
But then you have to take care that A can eat what
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B produces!
EXAMPLE: Suppose A eats and produces 2-dimensionalvectors, and B eats and produces 3-dimensional vec-
tors. Then AB would not make sense!
EXAMPLE: Suppose B eats 2-d vectors and pro-
duces 3-d vectors (so its matrix relative to ijk looks
like this:
b11 b12b21 b22
b31 b32
, a 3 by 2 matrix) and suppose
A eats 3-d vectors and produces 2-d vectors. Then
AB DOES make sense, because A can eat what B
produces. (In this case, BA also makes sense.).
IMPORTANT FACT: Suppose aij is the matrix
of a linear transformation A relative to ijk, and sup-
pose bij is the matrix of the Linear Transformation
B relative to ijk. Suppose that AB makes sense.
Then the matrix of AB relative to ij or ijk is just
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the matrix product of aij and bij.
EXAMPLE: What happens to the vector 12 if
we shear 45 parallel to the x axis and then rotate
90 anticlockwise? What if we do the same in the
reverse order?
Answer: Shear
1 tan 0 1
so in this case it is
1 10 1
. A rotation through
has matrix
cos sin sin cos
, so here it is
0 11 0
Hence
SHEAR, THEN ROTATE
0
1
1 0 1 1
0 1
= 0
11 1
ROTATE, THEN SHEAR
1 10 1
0 11 0
=
1 11 0
.
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So shear, then rotate
1
2
0 11 1
1
2 =
23 .
Rotate, then shear12
1 11 0
12
=
11
Very different!
EXAMPLE: Suppose B is a LT with matrix 1 00 1
1 1
and A is a LT with matrix
0 1 1
1 1 0
.
What is the matrix of AB? Of BA?
Answer:0 1 1
1 1 0 1 00 1
1 1
=
1 01 1
= AB
2 by 3 3 by 2 2 by 2 1 00 1
1 1
0 1 11 1 0
=
0 1 11 1 0
1 0 1
3 by 2 2 by 3 3 by 3
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EXAMPLE: Suppose you take a piece of rubber in
2 dimensions and shear it parallel to the x axis by
degrees, and then shear it again by degrees. What
happens?
1 tan 0 1
1 tan 0 1
= 1 tan + tan
0 1
which is also a shear, but NOT through + !
The shear angles dont add up, since tan +tan =tan( + ).
EXAMPLE: Rotate 90 around z-axis, then rotate
90 around x-axis in 3 dimensions. [Always anti-
clockwise unless otherwise stated.] Is it the same
if we reverse the order? Rotate about z axis i becomes j, j becomes i, k stays the same, so 0 1 01 0 0
0 0 1
. Rotate about x axis, i stays the same,
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j becomes k, k becomes j, so
1 0 00 0 10 1 0
, and
1 0 00 0 1
0 1 0
0 1 01 0 0
0 0 1
= 0
1 0
1 0 00 0 1
1 0 00 0 10 1 0
,
so the answer is NO!
6.4 DETERMINANTS
You probably know that the AREA of the box de-
fined by two vectors is
|
u
v |, magnitudeof the vector product.
If you dont know it, you can easily check it, since
the area of any parallelogram is given by
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AREA = HEIGHT Base= |
v
|sin |
u
|= |u | |v | sin = |u v |.
Similarly, the VOLUME of a three-dimensional par-
allelogram [called a PARALLELOPIPED!] is given
by
VOLUME = (AREA OF BASE) HEIGHT.
If you take any 3 vectors in 3 dimensions, sayu ,
v ,
w,
then they define a 3-dimensional parallelogram. The
area of the base is |u v |,height is
|w
| |sin
2 |
where is the angle between
u v and w, so VOLUMEdefined by
u ,
v ,
w is just
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|u v | |w| | sin
2
|
=
|u
v
| |w
| |cos
|=|u v w|.
[Check: Volume of Basic Box defined by ijk is
|i j k| = |k k| = 1, correct!
Now let T be any linear transformation in two di-
mensions. [This means that it acts on vectors in the
xy plane and turns them into other vectors in the xy
plane.]
We let T act on the Basic Box, as usual.
Now Ti and Tj still lie in the same plane as i and
j, s o (Ti) (Tj) must be perpendicular to thatplane. Hence it must be some multiple of k. We
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define the DETERMINANT of T to be that multi-
ple, that is, by definition, det(T) is the number given
[STRICTLY IN 2 DIMENSIONS] by
(Ti) (Tj) = det(T)k.
EXAMPLE: If I = identity, then
Ii I
j =
i
j =
k = 1
k
so det(I) = 1.
EXAMPLE: Du = 2
u
Di
Dj = 4i
j = 4k
det(D) = 4
EXAMPLE: Ti = i + 14j, Tj = 14 i +
j,
Ti Tj =
i +1
4j
1
4i + j
= i j + 1
16j i
= 1516i j = 1516 k det T = 1516 .
EXAMPLE: Ti = j, Tj = i,
Ti Tj = j i = k det T = 1
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EXAMPLE: Shear, Si = i, Sj = i tan + j,
Si Sj = k det S = 1.
EXAMPLE: Ti = i + j = Tj,
Ti Tj = 0 det T = 0.
EXAMPLE: Rotation
Ri Rj = (cos i + sin j) ( sin i + cos j)= (cos2 sin2 )k = k det(R) = 1.
The area of the Basic Box is initially |i j| = 1.After we let T act on it, the area becomes
|Ti T
j| = |det T| |
k| = | det T|.
So
Final Area of Basic Box
Initial Area of Basic Box=
| det T|1
= | det T|
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so | det T| TELLS YOU THE AMOUNT BY WHICHAREAS ARE CHANGED BY T. So det T =
1
means that the area is UNCHANGED (Shears, ro-
tations, reflections) while det T = 0 means that the
Basic Box is squashed FLAT, zero area.
Take a general 2 by 2 matrix M =
a bc d
. We
know that this means Mi = ai + cj, Mj = bi + dj.
Hence Mi Mj =
ai + cj
bi + dj
= (ad bc)k, so
det
a bc d
= ad bc.
Check: det
2 00 2
= 4, det
1 tan 0 1
= 1,
det
cos sin sin cos
= 1, det
1 11 1
= 0.
IN THREE dimensions there is a similar gadget.
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The Basic Box is defined by ijk, and we can let any
3-dimensional L.T. act on it, to get a new box de-
fined by Ti, Tj, Tk. We define
det T =
Ti
Tj
Tk
where the dot is the scalar product, as usual. Since
|Ti Tj Tk| is the volume of the 3-dimensionalparallelogram defined by Ti, Tj, Tk, we see that
| det T| = Final Volume of Basic BoxInitial Volume of Basic Box
,
that is,|
det T|
tells you how much T changes vol-
umes. If T squashes the Basic Box flat, then
det T = 0.
Just as det a b
c d = ad
bc, there is a formula
for the determinant of a 3 by 3 matrix. The usual
notation is this. We DEFINE
a b
c d
= det
a b
c d
= ad bc.
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Similarly
a11 a12 a13a21 a22 a23a31 a32 a33
is the determinant of
a11 a12 a13a21 a22 a23
a31 a32 a33
and there is a formula for it, as
follows:
a11 a12 a13a21 a22 a23a31 a32 a33
= a11
a22 a23a32 a33 a12
a21 a23a31 a33+ a13
a21 a22a31 a32 .
In other words, we can compute a three-dimensional
determinant if we know how to work out 2-dimensional
determinants.
COMMENTS:
[a] We worked along the top row. Actually, a THE-
OREM says that you can use ANY ROW OR ANYCOLUMN!
[b] How did I know that a12 had to multiply the par-
ticular 2-dimensional determinant
a21 a23a31 a33
? Easy:
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I just struck out EVERYTHING IN THE SAME
ROW AND COLUMN as a12 :
a21
a23
a31 a33
and
just kept the survivors!
This is the pattern, for example if you expand along
the second row you will get
a21
a12 a13 a32 a33
+ a22
a11 a13
a31 a33
a23
a11 a12 a31 a32
[c] What is the pattern of the + and signs? It is
an (Ang Moh ) CHESSBOARD, starting with a +
in the top left corner:+ + + + +
[d] You can do exactly the same thing in FOUR di-
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mensions, following this pattern, using
+ + + ++
+
+ +because now you know how to work out 3-dimensional
determinants. And so on!
Example:
1 1 01 1 12 0 0
= 1 1 10 0
1 1 12 0
+ 0 1 12 0
= 0 + 2 + 0 = 2
(expanding along the top row) or, if you use the sec-
ond row,
1 1 01 1 12 0 0
= 1
1 00 0+ 1
1 02 0 1
1 12 0
= 0 + 0 + 2 = 2
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or
1
1 0
1 1 12 0 0
= 1
1 10 0 1
1 00 0+ 2
1 01 1
= 0 + 0 + 2 = 2
(expanding down the first column).
Important Properties of Determinants
[a] Let S and T be two linear transformations such
that det S and det T are defined. Then
det ST = det T S = (det S) (det T).
Therefore, det[ST U] = det[U ST] = det[T U S] and
so on: det doesnt care about the order. Remember
however that this DOES NOT mean that ST U =
U ST etc etc.
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[b] If M is a square matrix, then
det MT = det M.
[c] Ifc is a number and M is an n by n matrix, then
det(cM) = cn det M.
EXAMPLE: Remember from Section 2[g] of Chap-
ter 5 that an ORTHOGONAL matrix satisfies M MT =
I. So det(M MT) = det I = 1. But det(M MT) =
det(M) det(MT) = det(M) det(M) = (det M)2,thus
det M = 1
for any orthogonal matrix.
6.5. INVERSES.
If I give you a 3-dimensional vectoru and a 3-
dimensional linear transformation T, then T sends
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u to a particular vector, it never sends
u to two
DIFFERENT VECTORS! So this picture is impos-
sible:
Tu
u
Tu
But what about this picture:u
Tu = T
v
v
Can T send TWO DIFFERENT VECTORS TO
ONE? Yes!
1 0 00 0 00 0 0
010
= 000
= 1 0 00 0 00 0 0
001
So it can happen! Notice that this transformation
destroys j (and also k). In fact if u = v and
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Tu = T
v , then T(
u v ) = 0, that is, Tw = 0
wherew IS NOT THE ZERO VECTOR. So if this
happens, T destroys everything in the w direction.
That is, T SQUASHES 3-dimensional space down
to two or even less dimensions. This means that
T LOSES INFORMATION it throws away all ofthe information stored in the w direction. Clearly T
squashes the basic box down to zero volume, so
det T = 0
and we say T is SINGULAR.
SUMMARY: A SINGULAR LINEAR TRANSFOR-
MATION
[a] Maps two different vectors to one vector
[b] Destroys all of the vectors in at least one direction
[c] Loses all information associated with those direc-
tions
[d] Satisfies det T = 0.
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Conversely, a NON-SINGULAR transformation never
maps 2 vectors to one,
u Tuv Tv
Different
Therefore if I give you Tu , THERE IS EXACTLY
ONEu . The transformation that takes you from T
u
back tou is called the INVERSE OF T. The idea is
that since a NON-SINGULAR linear transformation
does NOT destroy information, we can re-constructu if we are given T
u . Clearly T HAS AN INVERSE,
CALLED T1, if and only if det T = 0. All thisworks in other dimensions too.
EXAMPLE:
1 11 1
34
=
77
,
1 11 1
43
=
77
,
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two different vectors to one!
1 11 1
11 = 1 1
1 1 22
=
1 11 1
13.59
13.59
=
00
It destroys everything
in that direction!Finally det
1 11 1
= 0.
So it is SINGULAR and
has NO inverse.
1
1
EXAMPLE: Take
0 11 0
and suppose it acts on
and
a
b
and sends them to the same vector,
so
0 11 0
= 0 11 0
a
b
.
Then
=
ba
= a
= b
=
a
b
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so
and
a
b
are the same this transforma-
tion never maps different vectors to the same vector.
No vector is destroyed, no information is lost, noth-
ing gets squashed! And det
0 11 0
= 1, NON-
SINGULAR.
How to FIND THE INVERSE.
By definition, T1 sends Tu to
u , i.e.
T1(T(u )) =
u = T(T1(
u )).
Butu = I
u (identity) so T1 satisfies
T1T = T T1 = I.
So to find the inverse of 0 11 0 we just have to
find a matrix
a b
c d
such that
a b
c d
0 11 0
=
1 00 1
, b = 1, a = 0, d = 0, c = 1 so answer is
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0 1
1 0
. In fact its easy to show that
a bc d
1
= 1ad bc
d b
c a
.
For example, when we needed to find the matrix S
in Section 4 of Chapter 5, we needed to find a way
of solvingS
0.7 0.4
0.5 0.7
= I.
This just means that we need to inverse of
0.7 0.4
0.5 0.7
,
and the above formula does the job for us.
For bigger square matrices there are many tricks
for finding inverses. A general [BUT NOT VERY
PRACTICAL] method is as follows:
[a] Work out the matrix of COFACTORS. [A cofac-
tor is what you get when you work out the smaller
determinant obtained by striking out a row and a
column, for example the cofactor of 6 in
1 2 34 5 67 8 9
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is
1 27 8
= 6. You can do this for each elementin a given matrix, to obtain a new matrix of the
same size. For example, the matrix of cofactors of 1 0 10 1 0
0 0 1
is
1 0 00 1 0
1 0 1
.
[b] Keep or reverse the signs of every element accord-
ing to+ + + + +
(you get
1 0 00 1 0
1 0 1
above.)
[c] Take the TRANSPOSE,
1 0 10 1 0
0 0 1
.
[d] Divide by the determinant of the original ma-
trix. THE RESULT IS THE DESIRED INVERSE.
1 0 10 1 00 0 1
in this example. Check:
1 0 10 1 0
0 0 1
1 0 10 1 0
0 0 1
=
1 0 00 1 0
0 0 1
.
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INVERSE OF A PRODUCT:
(AB)1 = B1A1
Note the order! Easily checked:
(AB)1
AB = B1
(A1
A)B = B1
IB = I.
APPLICATION: SOLVING LINEAR SYSTEMS.
Suppose you want to solve
x + 2y + 3z = 14x + 5y + 6z = 27x + 8y + 9z = 4.
One way is to write it as
1 2 34 5 6
7 8 9
xy
z
=
12
4
.
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Then all you have to do is find
1 2 34 5 6
7 8 9
1
and
multiply it on both sides, so you get xy
z
=
1 2 34 5 6
7 8 9
1 12
4
= answer.
So this is a systematic way of solving such problems!
Now actually det
1 2 34 5 6
7 8 9
= 0, and you can see
why:
1 2 34 5 6
7 8 9
1
21
=
00
0
.So this transformation destroys everything in the di-
rection of
12
1
. In fact it squashes 3-dimensional
space down to a certain 2-dimensional space. [We
say that the matrix has RANK 2. If it had squashed
everything down to a 1-dimensional space, we would
say that it had RANK 1.] Now actually
12
4
. DOES
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NOT lie in that two-dimensional space. Since
1 2 34 5 67 8 9
squashes EVERYTHING into that two-dimensional
space, it is IMPOSSIBLE for
1 2 34 5 6
7 8 9
xy
z
to be
equal to
124
. Hence the system has NO solutions.
If we change
12
4
to
12
3
, this vector DOES lie
in the special 2-dimensional space, and the system
1 2 34 5 67 8 9
xyz
= 123
DOES have a solution in fact it has infinitely many!
SUMMARY:
Any system of linear equations can be written as
Mr =
a
where M is a matrix,r = the vector of variables,
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anda is a given vector. Suppose M is square.
[a] If det M= 0, there is exactly one solution,
r = M1
a .
[b] If det M = 0, there is probably no solution. But
if there is one, then there will be many.
PRACTICAL ENGINEERING PERSPECTIVE:
In the REAL world, NOTHING IS EVER EXACTLY
EQUAL TO ZERO! So if det M = 0, either [a] you
have made a mistake, OR [b] you are pretending that
your data are more accurate than they really are!
1 2 34 5 67 8 9
REALLY means
1.01 2.08 3.033.99 4.97 6.027.01 7.96 8.98
and of course the determinant of THIS is non-zero!
Actually, det = 0.597835!
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6.6 EIGENVECTORS AND EIGENVALUES.
Remember we said that a linear transformation
USUALLY changes the direction of a vector. But
there may be some special vectors which DONT
have their direction changed!
EXAMPLE:
1 22 2
clearly DOES change the
direction of i and j, since
12
is not parallel to i
and 22 is not parallel to j. BUT
1 22 2
21
=
42
= 2
21
which IS parallel to 21.
In general if a transformation T does not change the
direction of a vectoru , that is
Tu =
u
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for some (SCALAR), thenu is called an EIGEN-
VECTOR of T. The scalar is called the EIGEN-
VALUE ofu .
6.7 FINDING EIGENVALUES AND EIGEN-
VECTORS.
There is a systematic way of doing this. Take the
equation
Tu =
u
and write u = Iu , I = identity. Then
(T I)u = 0
Lets supposeu = 0 [of course, 0 is always an eigen-
vector, that is boring]. So the equation says that
T I SQUASHES everything in the u direction.Hence
det(T I) = 0.
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This is an equation which can be SOLVED to find
.
EXAMPLE: Find the eigenvalues of
1 22 2
:
det
1 22 2
1 00 1
= 0
det 1
2
2 2 = 0 (1 )(2 + ) 4 = 0 = 2 OR 3
So there are TWO answers for a 2 by 2 matrix. Sim-
ilarly, in general there are three answers for 3 by 3
matrices, etc.
What are the eigenvectors for = 2, = 3?
IMPORTANT POINT: Let
u be an eigenvectorof T. Then 2
u is also an eigenvector with the same
eigenvalue!
T(2u ) = 2T
u = 2
u = (2u ).
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Similarly 3u , 13.59
u etc are all eigenvectors! SO
YOU MUST NOT EXPECT A UNIQUE ANSWER!
OK, with that in mind, lets find an eigenvector for
= 2. Lets call an eigenvector
. Then
(T I)u = 0
1 22 2
= 0
1 2
2 4
= 0
+ 2 = 0
2 4 = 0But these equations are actually the SAME, so we
really only have ONE equation for 2 unknowns. We
arent surprised, because we did not expect a unique
answer anyway! We can just CHOOSE = 1 (or13.59 or whatever) and then solve for . Clearly
= 12 , so an eigenvector corresponding to = 2
is
112
. But if you said
21
or
10050
that is also
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correct!
What about =
3?
4 22 1
= 0
4 + 2 = 0
2 + = 0
Again we can set = 1, then = 2, so an eigen-vector corresponding to = 3 is
1
2
or
2
4
or 10
20 etc.
EXAMPLE: Find the eigenvalues, and correspond-
ing eigenvectors, of
0 11 0
.
Answer: We have det
11
= 0 2 + 1 =
0 = i, i = 1.Eigenvector for i: we set
i 11 i
1
= 0
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i = 0 = i so an eigenvector for i is1
i
. For =
i we have
i 11 i
1
= 0
i = 0 = i so an eigenvector for iis 1
i. Note that a REAL matrix can have COM-
PLEX eigenvalues and eigenvectors! This is hap-
pening simply because
0 11 0
is a ROTATION
through 90, and of course such a transformation
leaves NO [real] vectors direction unchanged (apart
from the zero vector).
6.8. DIAGONAL FORM OF A LINEAR TRANS-
FORMATION.
Remember that we defined the matrix of a linear
transformation T WITH RESPECT TO i, j by let-
ting T act on i and j and then putting the results in
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the columns. So to say that T has matrix
a b
c d
with respect to i, j means that
Ti = ai + cj
Tj = bi + dj.
Whats so special about the two vectors i and j?
Nothing, except that EVERY vector in two dimen-
sions can be written as i + j for some , .
Now actually we only really use i and j for CONVE-
NIENCE. In fact, we can do this with ANY pair of
vectors u , v in two dimensions,
PROVIDED that they are not parallel.
That is, any vectorw can be
expressed as
w = u + v
for some scalars , . You can see this from the
diagram by stretching u to u and v to v , wecan make their sum equal to
w.
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We callu ,
v a BASIS for 2-dimensional vectors. Let
u = P11i + P21j = P11
P21
v = P12i + P22j =
P12P22
Then the transformation that takes
i, j
to (u ,
v )
has matrixP11 P12
P21 P22
= P. In order for
u ,
v to be
a basis, P must not squash the volume of the Basic
Box down to zero, since otherwiseu and
v will be
parallel. So we must have
det P = 0.
The same idea works in 3 dimensions: ANY set of
3 vectors forms a basis PROVIDED that the matrix
of components satisfies det P
= 0.
EXAMPLE: The pair of vectorsu =
10
,v =
11
forms a basis, because det
1 10 1
= 1 = 0.
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Now of course the COMPONENTS of a vector will
change if you choose a different basis. For example,12
= 1i + 2j BUT
12
= 1u + 2v .
Instead,
12
= u + 2v , so the components of this
vector relative tou ,
v are
12
(
u,
v )
. Where did I
get these numbers?
As usual, set u = Pi, v = Pj where P =
1 10 1
.
We want to find , such that
12
=
u +
v .
We have, in this particular case,u = i,
v = i + j, so
12
= i + [i + j] =
+
= P
We know P is not singular, so we can take P over to
the left side by multiplying both sides of this equa-
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tion by the inverse of P. So we get
= P
1 1
2
and this is our answer: this is how we find and !
So to get and we just have to work out
P1
12
=
1 10 1
12
=1
2
,
that is, the components of this vector relative tou ,
v
are found as
12
(
u,
v )
= P1
12
(i,j)
THE COMPONENTS RELATIVE TOu ,
v ARE
OBTAINED BY MULTIPLYING P
1
INTO THECOMPONENTS RELATIVE TO i, j. Similarly for
linear transformations if a certain linear transfor-mation T has matrix
1 20 1
ij
relative to i, j it
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will have a DIFFERENT matrix relative tou ,
v .
We have 1 20
1
(i,j)
12
(i,j)
= 5
2
(i,j)
That is, the matrix ofT relative to i, j sends
12
(ij)
to
5
2(ij)
. In the same way, the matrix of T rela-
tive to (
u ,
v ), which we dont know and want to find,
sends
12
(
u,
v )
to
7
2(
u,
v )
, because these are
the components of these two vectors relative tou ,
v ,
as you can show by multiplying P1 into
12
(ij)
and
5
2(ij)
respectively.
So the unknown matrix we want satisfies
? ?? ?
(
u,
v )
12
(
u,
v )
=
7
2(
u,
v )
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But we know
1
2(
u,
v ) = P1 1
2(i,j) and
72
(
u,
v )
= P1
52
(i,j)
so
? ?? ?
(
u,
v )
P1
12
(ij)
= P1
52
(ij)
.
Multiply both sides by P and get
P ? ?
? ?(
u,
v )P1 1
2(ij)
= 52
(ij)
Compare this with
1 20
1
(i,j)
12
(i,j)
=
5
2
(i,j)
1 20 1
(i,j)
= P
? ?? ?
(
u,
v )
P1
? ?? ?
(
u,
v )
= P1
1 20 1
(i,j)
P.
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[In the last step, we multiplied both sides on the
LEFT by P1, and on the RIGHT by P.]
We conclude that THE MATRIX OF T REL-
ATIVE TOu ,
v , IS OBTAINED BY MULTIPLY-
ING P1 ON THE LEFT AND P ON THE RIGHT
INTO THE MATRIX OF T RELATIVE TO i, j.
In this example,
? ?? ?
(
u,
v )
=
1 10 1
1 20 1
1 10 1
= 11
0 1 1 3
0 1 = 1 4
0 1 .
So now we know how to work out the matrix of any
linear transformation relative to ANY basis.
Now let T be a linear transformation in 2 dimensions,
with eigenvectorse1,
e2, eigenvalues 1, 2. Now
e1
ande2 may or may not give a basis for 2-dimensional
space. But suppose they do.
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QUESTION: What is the matrix of T relative toe1,
e2?
ANSWER: As always, we see what T does to e1
ande2, and put the results into the columns!
By definition of eigenvectors and eigenvalues,
Te1 = 1
e1 = 1
e1 + 0
e2
Te2 = 2
e2 = 0
e1 + 2
e2
So the matrix is
1 00 2
(
e1,
e2)
.
We say that a matrix of the form
a 00 d
or
0 00 00 0
is DIAGONAL. So we see that THE MATRIX OF
A TRANSFORMATION RELATIVE TO ITS OWN
EIGENVECTORS (assuming that these form a ba-sis) is DIAGONAL.
EXAMPLE: We know that the eigenvectors of1 22 2
are
112
and
1
2
. So here P =
1 112 2
,
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P1 = 25 2 1
12 1
,
P1
1 22 2
P = 2
5
2 1 12 1
1 22 2
1 112 2
= 25
2 1 12 1
2 31 6
= 2
5
5 0
0 152
=
2 00 3
as expected since the
eigenvalues are 2 and -3.
EXAMPLE: The shear matrix
1 tan 0 1
.
Eigenvalues: det
1 tan
0 1
= 0 (1)2 =
0
= 1. Only one eigenvector, namely
1
0, so
the eigenvectors DO NOT give us a basis in this case
NOT possible to diagonalize this matrix!
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6.9 APPLICATION
MARKOV CHAINS.
We saw back in Section 3 of Chapter 5 that to predict
the weather 4 days from now, we needed the 4th
power of the matrix
0.6 0.30.4 0.7
.
But suppose I want the weather 30 days from now
I need M30! There is an easy way to work thisout using eigenvalues.
Suppose I can diagonalize M, that is, I can writeP1M P = D =
1 00 2
for some matrix P. Then
M = P DP1
M2 = (P DP1)(P DP1)
= P DP1P DP1 = P D2P1
M3 = M M2 = P DP1P D2P1 = P D3P1 etc
M30 = P D30P1.
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But D30 is very easy to work out it is just
301 00 302
.
Lets see how this works!
Eigenvectors and eigenvalues of
0.6 0.30.4 0.7
are
1
1
(eigenvalue 0.3) and
143
(eigenvalue 1) so
P = 1 11 43 , D =
0.3 00 1
, P
1
= 4
7 3
737
37
D30 =
(0.3)30 0
0 1
2 1016 00 1
so
M30 =
1 1
1 43
2 1016 0
0 1
47 3737
37
= 17
3 + 8 1016 3 6 10164 8 1016 4 + 6 1016
37
37
47
47
So if it is rainy today, the probability of rain tomor-
row is 60%, but the probability of rain 30 days from
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now is only 37 43%. As we go forward in time, thefact that it rained today becomes less and less im-
portant! The probability of rain in 31 days is almost
the same as the probability of rain in 30 days!
6.10 THE TRACE OF A MATRIX.
Let M be any square matrix. Then the TRACE
of M, denoted T rM, is defined as the sum of the
diagonal entries: T r
1 00 1
= 2, T r
1 2 34 5 6
7 8 9
=
15, T r
1 5 167 2 15
11 9 8
= 11, etc.
In general it is NOT true that T r(M N) = TrM TrN
BUT it is true that TrMN = TrNM.
Proof: T rM =i
Mii so
TrMN =i
j
MijNji =j
i
NjiMij = TrNM.
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Hence T r(P1AP) = T r(AP P1) = T rA so if A is
diagonalizable, T rA = T r
1 00 2
= 1 + 2.