MATH 520 LECTURE NOTES I

B. Cole
March 31, 2014
Linear Transformations
These notes cover some of the lectures from March 1721, plus a few additions.
1. Basic Theorems
Theorem 1.1. Let T: V W be a linear transformation. Let dimV = n, and let
dimW = m. Let B = {v
1
, . . . , v
n
} be a basis for V , and let C = {w
1
, . . . , w
m
} be a basis
for W. Then, there exists a unique mn matrix M so that
[Tx]
C
= M[x]
B
for all x in V . In particular, the j-th column of M is [Tv
j
]
C
for 1 j n.
We say that M is the matrix representing T with respect to B and C. We write
M = [T]
C
B
. All features of T are reected in features of M.
Proposition 1.2. Let T, B, C, M be as above. Then, y Im(T) if and only if [y]
C

Col(M). In particular, y
1
, . . . , y
l
is a basis for Im(T) if and only if [y
1
]
C
, . . . , [y
l
]
C
is a basis
for Col(M).
Proposition 1.3. Let T, B, C, M be as above. Then, x Ker(T) if and only if [x]
B

Nul(M). In particular, x
1
, . . . , x
k
is a basis for Ker(T) if and only if [x
1
]
B
, . . . , [x
k
]
B
is a
basis for Nul(M).
Denition 1.4. rank(T) = dimIm(T).
Theorem 1.5. Let T, B, C, M be as above. Then,
(i) dimCol(M) = dimIm(T),
(ii) dimNul(M) = dimKer(T),
(iii) rank(M) = rank(T).
Corollary (rank-nullity theorem). Let T: V W be a linear transformation. Then,
dimIm(T) + dimKer(T) = dimV .
Example 1.1. Let T: P
3
P
3
be dened by Tp = q where
q(t) = 3p(t) tp

(2 t) + (1 +t
2
)p

(1 +t).
(Check: Do you know what this means? Practice calculating Tp = q when p(t) = 4t +t
2
.
First calculate p

(t) = 4+2t and p

(t) = 2. We get p

(2t) = 4+2(2t) = 4+42t = 82t.

So, q(t) = 3(4t +t
2
) t(8 2t) + (1 +t
2
)2 which simplies to q(t) = 2 20t +t
2
.)
Calculate the matrix M = [T]
B
B
with respect to B, the standard basis for P
3
.
The elements of B are given by p
k
(t) = t
k1
for k = 1, . . . , 4. Set Tp
k
= q
k
for
k = 1, . . . , 4. We calculate q
1
(t) = 3, q
2
(t) = 4t, q
3
(t) = 24t+t
2
, q
4
(t) = 66t+6t
2
.
Hence,
[q
1
]
B
=

3
0
0
0

, [q
2
]
B
=

0
4
0
0

, [q
3
]
B
=

2
4
1
0

, [q
4
]
B
=

6
6
6
0

.
So, we have
M =
[
[q
1
]
B

[q
2
]
B

[q
3
]
B

[q
4
]
B
]
=

3 0 2 6
0 4 4 6
0 0 1 6
0 0 0 0

.
Example 1.2. Using the linear transformation T from example 1.1, what is rank T?
Answer: rank T = rank M = 3.
Example 1.3. Using the linear transformation T from example 1.1, nd a basis for
Ker(T).
We nd that

12
9
12
2

is a basis for Nul(M). Hence, 12 + 9t 12t

2
+ 2t
3
is a basis
for Ker(T).
Note: Ker(T) is a subspace of P
3
. Hence, the basis for Ker(T) must consist of poly-
nomials. Also, since we use the standard basis for P
3
, the calculations are particularly
simple.
2. Spaces of matrices
Let M
m,n
be the set of mn matrices.
Theorem 2.1. The vector space M
m,n
has dimension mn.
Let us calculate a basis for M
2,2
, the set of 2 2 matrices. Let
E
1
=
[
1 0
0 0
]
, E
2
=
[
0 1
0 0
]
, E
3
=
[
0 0
1 0
]
, E
4
=
[
0 0
0 1
]
.
2
It is easy to see that
c
1
E
1
+c
2
E
2
+c
3
E
3
+c
4
E
4
=
[
c
1
c
2
c
3
c
4
]
.
Hence, if c
1
E
1
+c
2
E
2
+c
3
E
3
+c
4
E
4
= 0, then we have c
1
= 0, c
2
= 0, c
3
= 0, and c
4
= 0.
So, E
1
, . . . , E
4
are linearly independent.
On the other hand, given any matrix A =
[
a b
c d
]
, then we have A = c
1
E
1
+c
2
E
2
+
c
3
E
3
+c
4
E
4
for c
1
= a, c
2
= b, c
3
= c, and c
4
= d.
This shows that E
1
, . . . , E
4
is a basis for M
2,2
. We call it the standard basis for this
space.
For M
3,2
, we take the standard basis to consist of these matrices:
E
1
=

1 0
0 0
0 0

, E
2
=

0 1
0 0
0 0

, E
3
=

0 0
1 0
0 0

, E
4
=

0 0
0 1
0 0

, E
5
=

0 0
0 0
1 0

, E
6
=

0 0
0 0
0 1

.
For M
m,n
, we dene the standard basis in a similar way. Here is a technical descrip-
tion. For 1 k mn, let E
k
be the k-th element of that basis. It consists of all zero
entries except for a single 1 appear in row j and column i where 1 + (j 1)n k jn,
i = k (j 1)n. A simple way to think about this is that as k increases, the single one
moves to the right along each row until it reaches the end, then moves to the rst entry in
the next row.
Example 2.2. Consider T: M
2,2
M
3,2
by the formula T(A) = BAwhere B =

1 2
2 4
3 6

.
Use the standard basis B for M
2,2
and the standard basis C for M
3,2
. Let B be
E
1
, . . . , E
4
, and let C be F
1
, . . . , F
6
.
We calculate M = [T]
C
B
. Note that M must be a 6 4 matrix since dimM
3,2
= 6 and
dimM
2,2
= 4 .
We calculate T(E
1
) = BE
1
=

1 0
2 0
3 0

= 1 F
1
+0 F
2
+2 F
3
+0 F
4
+3 F
5
+0 F
6
.
So, [T(E
1
)]
C
= {1, 0, 2, 0, 3, 0} which gives the rst column of M. Similarly, [T(E
2
)]
C
=
{0, 1, 0, 2, 0, 3}, [T(E
3
)]
C
= {2, 0, 4, 0, 6, 0}, [T(E
4
)]
C
= {0, 2, 0, 4, 0, 6}. So,
M =

1 0 2 0
0 1 0 2
2 0 4 0
0 2 0 4
3 0 6 0
0 3 0 6

.
Example 2.3. Using the linear transformation T from example 2.2, what is rank T?
Answer: rank T = rank M = 2.
3
Example 2.4. Using the linear transformation T from example 2.2, nd a basis for
Ker(T).
We nd that w
1
, w
2
is a basis for Nul(M) where w
1
=

2
0
1
0

and w
2
=

0
2
0
1

. So,
P
1
, P
2
is a basis for Ker(T) where [P
1
]
B
= w
1
and [P
2
]
B
= w
2
. We see that
P
1
=
[
2 0
1 0
]
, P
2
[
0 2
0 1
]
.
Note: Ker(T) is a subspace of M
2,2
. Hence, the basis for Ker(T) must consist
of 2 2 matrices. Also, since we use the standard basis for M
2,2
the calculations are
particularly simple.
Example 2.5. Using the linear transformation T from example 2.2, nd a basis for Im(T).
We nd that v
1
, v
2
, the rst 2 columns of M, are the pivot columns. Hence, they form
a basis for Col(M). So, Q
1
, Q
2
is a basis for Im(T) where [Q
1
]
C
= v
1
and [Q
2
]
C
= v
2
. We
see that
Q
1
=

1 0
2 0
3 0

, Q
2
=

0 1
0 2
0 3

.
Note: Im(T) is a subspace of M
3,2
. Hence, the basis for Im(T) must consist of 32
matrices. Also, since we use the standard basis for M
3,2
the calculations are particularly
simple.
3. Mappings from V to V
Denition 3.1. For a square matrix A, the trace of A, denoted tr(A), is the sum of the
diagonal elements of A.
Proposition 3.2. Let A be an n n matrix. Let A be diagonalizable with eigenvalues

1
, . . . ,
n
. Then, tr(A) =
1
+ +
n
and det(A) =
1

n
.
Let T: V V be a linear transformation. Let dimV = n, and let B = {v
1
, . . . , v
n
}
be a basis for V . Let M = [T]
B
B
. For simplicity we write M = [T]
B
. Note that M is an
n n matrix.
Proposition 3.3. Given two bases B
1
, B
2
for V , the matrices [T]
B
1
and [T]
B
2
are similar.
Hence, they have the same trace, determinant, characteristic polynomial, and eigenvalues.
We use M = [T]
B
to dene interesting quantities for T.
4
Denition 3.4. For T, B, M be as above,
(i) tr(T) = tr(M),
(ii) det(T) = det(M),
(iii) the characteristic polynomial of T is the characteristic polynomial of M.
Denition 3.5. Let T: V V be a linear transformation. Consider the vector x V
and the scalar . We say that x is an eigenvector with corresponding eigenvalue if
Tx = x and x = 0. We say that T is diagonalizable if there is a basis for V consisting of
eigenvectors.
Proposition 3.6. Let T, B, M be as above. Then, x is an eigenvector with eigenvalue
for T if and only if [x]
B
is an eigenvector with eigenvalue for M. Hence, T and M
have the same eigenvalues, and T is diagonalizable if and only if M is diagonalizable.
Furthermore, x
1
, . . . , x
n
is a basis of eigenvectors for T if and only if [x
1
]
B
, . . . , [x
n
]
B
is a
basis of eigenvectors for M.
Example 3.7. Use the linear transformation T from example 1.1 where T: P
3
P
3
.
Find the eigenvalues for T.
In example 1.1, we determined M = [T]
B
for a certain basis. Since M is upper
triangular with 3, 4, 1, 0 on the diagonal, those 4 numbers are the eigenvalues for M
and hence for T.
Example 3.8. Using the linear transformation T from example 3.1, is T diagonalizable?
If so, nd a basis of eigenvectors.
Since the 4 4 matrix M has 4 distinct eigenvalues, it is diagonalizable. A basis of
eigenvectors for M is given by
v
1
=

0
1
0
0

, v
2
=

1
0
0
0

, v
3

12
9
12
2

, v
4

5
8
10
0

.
corresponding to the eigenvalues 4, 3, 0, 1. Let [u
i
]
B
= v
i
for i = 1, . . . , 4. So, u
1
, . . . , u
4
is a basis of eigenvectors for T corresponding to the same list of eigenvalues. We calculate
that u
1
(t) = t, u
2
(t) = 1, u
3
(t) = 12 + 9t 12t
2
+ 2t
3
, u
4
(t) = 5 8t + 10t
2
,
Note: Since T: P
3
P
3
, the eigenvectors for T must be polynomials in P
3
. Also,
since we use the standard basis for P
3
the calculations are particularly simple.
Example 3.9. Consider T: M
2,2
M
2,2
by the formula T(A) = BA+AC where
B =
[
1 2
2 4
]
, C
[
1 1
3 3
]
.
Is T diagonalizable? If so, nd a basis of eigenvectors.
5
Use B, the standard basis for M
2,2
. Let B = E
1
, . . . , E
4
. We nd M = [T]
B
.
We calculate
T(E
1
) = BE
1
+E
1
C =
[
2 1
2 0
]
.
So, [T(E
1
)]
B
= {2, 1, 2, 0}. Similarly, [T(E
2
)]
B
= {3, 2, 0, 2}, [T(E
3
)]
B
= {2, 0, 5, 1},
[T(E
4
)]
B
= {0, 2, 3, 1}. So,
M =

2 3 2 0
1 2 0 2
2 0 5 3
0 2 1 1

.
We nd that det(M I) = 30
2
6
3
+
4
= ( + 2)( 3)( 5). So, the
eigenvalues for M are given by 2, 0, 3, 5. Since we have 4 distinct eigenvalues, M and
hence T are diagonalizable.
Corresponding to this list of eigenvalues, we nd eigenvectors {v
1
, v
2
, v
3
, v
4
} and put
them in columns of this matrix.
P = [v
1
|v
2
|v
3
|v
4
] =

2 6 1 3
2 2 1 1
1 3 2 6
1 1 2 2

.
So the basis of eigenvectors for T consist of the matrices A
i
where [A
i
]
B
= v
i
for i = 1, . . . , 4.
We obtain
A
1
=
[
2 2
1 1
]
, A
2
=
[
6 2
3 1
]
, A
3
[
1 1
2 2
]
, A
4
=
[
3 1
6 2
]
.
Check:
T(A
4
) = BA
4
+A
4
C =
[
15 5
30 10
]
= 5A
4
,
as expected.
What we have done here is to nd solutions to the equation BA + AC = A where
B and C are given and A and are unknown. It turns out that there are only 4 values
of for which a non-zero A can be found. Those 4 values are determined by the matrices
B and C. However, the solution is probably not obvious except by turning this into an
eigenvalue problem for the linear transformation T.
6