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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 dim V = n, and letdim W = m. Let B={v1, . . . , vn} be a basis forV, and let C={w1, . . . , wm} be a basisforW. Then, there exists a uniquem n matrixM so that

[T x]C =M[x]B

for allx in V. In particular, thej-th column ofM is[T vj ]C for1 j n.

We say that M is the matrix representing T with respect to B and C. We writeM= [T]C

B. All features ofTare reflected in features ofM.

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, y1, . . . , yl is a basis forIm(T)if and only if[y1]C, . . . , [yl]C is a basisforCol(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, x1, . . . , xk is a basis forKer(T) if and only if [x1]B, . . . , [xk]B is abasis forNul(M).

Definition 1.4. rank(T) = dim Im(T).

Theorem 1.5. Let T, B, C, Mbe as above. Then,

(i) dim Col(M) = dim Im(T),

(ii) dim Nul(M) = dim Ker(T),

(iii) rank(M) = rank(T).

Corollary (rank-nullity theorem). Let T: V W be a linear transformation. Then,dim Im(T) + dim Ker(T) = dim V.

Example 1.1. Let T: P3 P3 be defined by T p= qwhere

q(t) =3p(t) tp(2 t) + (1 + t2)p(1 + t).

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(Check: Do you know what this means? Practice calculating T p= qwhen p(t) = 4t + t2.First calculatep(t) = 4+2tandp(t) = 2. We getp(2t) = 4+2(2t) = 4+42t= 82t.So, q(t) =3(4t + t2) t(8 2t) + (1 + t2)2 which simplifies to q(t) = 2 20t + t2.)

Calculate the matrix M= [T]BB

with respect to B, the standard basis for P3.

The elements of B are given by pk(t) = tk1

for k = 1, . . . , 4. Set T pk = qk fork= 1, . . . , 4. We calculateq1(t) = 3,q2(t) = 4t,q3(t) = 24t+t

2,q4(t) =66t+6t2.

Hence,

[q1]B=

3000

, [q2]B=

04

00

, [q3]B =

24

10

, [q4]B =

66

60

.

So, we have

M=

[q1]B[q2]B[q3]B[q4]B=

3 0 2 6

0 4 4 60 0 1 60 0 0 0

.

Example 1.2. Using the linear transformation Tfrom 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, find a basis forKer(T).

We find that

12

9122

is a basis for Nul(M). Hence,12 + 9t 12t2 + 2t3 is a basisfor Ker(T).

Note: Ker(T) is a subspace ofP3. Hence, the basis for Ker(T)must consist of poly-nomials. Also, since we use the standard basis forP3, the calculations are particularlysimple.

2. Spaces of matrices

Let Mm,n be the set ofm n matrices.

Theorem 2.1. The vector spaceMm,n has dimension m n.

Let us calculate a basis for M2,2, the set of 2 2 matrices. Let

E1=

1 00 0

, E2=

0 10 0

, E3=

0 01 0

, E4=

0 00 1

.

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It is easy to see that

c1E1+ c2E2+ c3E3+ c4E4=

c1 c2c3 c4

.

Hence, ifc1E1+ c2E2+ c3E3+ c4E4= 0, then we havec1= 0, c2= 0, c3= 0, and c4= 0.So, E1, . . . , E 4 are linearly independent.

On the other hand, given any matrix A=

a b

c d

, then we have A= c1E1+ c2E2+

c3E3+ c4E4 for c1= a, c2= b, c3= c, and c4= d.

This shows that E1, . . . , E 4 is a basis forM2,2. We call it the standard basis for thisspace.

ForM3,2, we take the standard basis to consist of these matrices:

E1=

1 00 0

0 0

, E2=

0 10 0

0 0

, E3=

0 01 0

0 0

, E4=

0 00 1

0 0

, E5=

0 00 0

1 0

, E6=

0 00 0

0 1

.

ForMm,n, we define the standard basis in a similar way. Here is a technical descrip-tion. For 1 k mn, let Ek be the k-th element of that basis. It consists of all zeroentries 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 onemoves to the right along each row until it reaches the end, then moves to the first entry inthe next row.

Example 2.2. Consider T: M2,2 M3,2by the formula T(A) =B A where B =

1 22 4

3 6

.

Use the standard basis B for M2,2 and the standard basis C for M3,2. Let B beE1, . . . , E 4, and let Cbe F1, . . . , F 6.

We calculateM= [T]CB

. Note thatMmust be a 6 4 matrix since dim M3,2= 6 anddim M2,2= 4 .

We calculateT(E1) =BE1=

1 02 0

3 0

= 1 F1+ 0 F2+ 2 F3+ 0 F4+ 3 F5+ 0 F6.

So, [T(E1)]C = {1, 0, 2, 0, 3, 0} which gives the first column of M. Similarly, [T(E2)]C ={0, 1, 0, 2, 0, 3}, [T(E3)]C ={2, 0, 4, 0, 6, 0}, [T(E4)]C ={0, 2, 0, 4, 0, 6}. So,

M=

1 0 2 0

0 1 0 22 0 4 00 2 0 43 0 6 00 3 0 6

.

Example 2.3. Using the linear transformation Tfrom example 2.2, what is rank T?

Answer: rank T= rank M= 2.

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Example 2.4. Using the linear transformation T from example 2.2, find a basis forKer(T).

We find that w1, w2 is a basis for Nul(M) where w1=

2

0

10

and w2=

0

2

01

. So,P1, P2 is a basis for Ker(T) where [P1]B=w1 and [P2]B =w2. We see that

P1=

2 0

1 0

, P2

0 20 1

.

Note: Ker(T) is a subspace of M2,2. Hence, the basis for Ker(T) must consistof 2 2 matrices. Also, since we use the standard basis for M2,2 the calculations areparticularly simple.

Example 2.5. Using the linear transformationTfrom example 2.2, find a basis for Im(T).

We find that v1, v2, the first 2 columns ofM, are the pivot columns. Hence, they forma basis for Col(M). So, Q1, Q2 is a basis for Im(T) where [Q1]C =v1 and [Q2]C =v2. Wesee that

Q1=

1 02 0

3 0

, Q2=

0 10 2

0 3

.

Note: Im(T) is a subspace ofM3,2. Hence, the basis for Im(T)must consist of3 2matrices. Also, since we use the standard basis for M3,2the calculations are particularlysimple.

3. Mappings from V to V

Definition 3.1. For a square matrixA, the trace ofA, denotedtr(A), is the sum of thediagonal elements ofA.

Proposition 3.2. Let A be an n n matrix. Let A be diagonalizable with eigenvalues1, . . . , n. Then, tr(A) =1+ + n anddet(A) =1 n.

Let T: V V be a linear transformation. Let dim V = n, and let B= {v1, . . . , vn}be a basis for V. LetM = [T]B

B. For simplicity we write M = [T]B. Note that M is an

n nmatrix.

Proposition 3.3. Given two basesB1,B2forV, the matrices[T]B1 and[T]B2 are similar.Hence, they have the same trace, determinant, characteristic polynomial, and eigenvalues.

We use M= [T]B to define interesting quantities for T.

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Definition 3.4. ForT, B, Mbe as above,

(i) tr(T) = tr(M),

(ii) det(T) = det(M),

(iii) the characteristic polynomial ofTis the characteristic polynomial ofM.

Definition 3.5. Let T: V V be a linear transformation. Consider the vector x Vand the scalar . We say that x is an eigenvector with corresponding eigenvalue ifT x= x andx= 0. We say that T isdiagonalizableif there is a basis forV consisting ofeigenvectors.

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 Mhave the same eigenvalues, and T is diagonalizable if and only if M is diagonalizable.Furthermore, x1, . . . , xn is a basis of eigenvectors forTif and only if [x1]B, . . . , [xn]B is abasis of eigenvectors forM.

Example 3.7. Use the linear transformation T from example 1.1 where T: P3 P3.Find the eigenvalues for T.

In example 1.1, we determined M = [T]B for a certain basis. Since M is uppertriangular with 3, 4, 1, 0 on the diagonal, those 4 numbers are the eigenvalues for Mand hence for T.

Example 3.8. Using the linear transformation Tfrom example 3.1, is T diagonalizable?If so, find a basis of eigenvectors.

Since the 4 4 matrix M has 4 distinct eigenvalues, it is diagonalizable. A basis of

eigenvectors for Mis given by

v1=

0100

, v2=

1000

, v3

129

122

, v4

5810

0

.

corresponding to the eigenvalues4, 3, 0, 1. Let [ui]B =vi fori = 1, . . . , 4. So,u1, . . . , u4is a basis of eigenvectors for Tcorresponding to the same list of eigenvalues. We calculatethat u1(t) =t, u2(t) = 1, u3(t) =12 + 9t 12t

2 + 2t3, u4(t) = 5 8t + 10t2,

Note: Since T: P3 P3, the eigenvectors for T must be polynomials in P3. Also,

since we use the standard basis for P3 the calculations are particularly simple.

Example 3.9. Consider T: M2,2 M2,2 by the formula T(A) =B A + ACwhere

B=

1 22 4

, C

1 13 3

.

Is Tdiagonalizable? If so, find a basis of eigenvectors.

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UseB, the standard basis for M2,2. Let B= E1, . . . , E 4. We find M= [T]B.

We calculate

T(E1) =BE1+ E1C=

2 12 0

.

So, [T(E1)]B = {2, 1, 2, 0}. Similarly, [T(E2)]B ={3, 2, 0, 2}, [T(E3)]B ={2, 0, 5, 1},[T(E4)]B ={0, 2, 3, 1}. So,

M=

2 3 2 01 2 0 2

2 0 5 30 2 1 1

.

We find that det(M I) = 30 2 63 +4 = (+ 2)( 3)( 5). So, theeigenvalues for Mare given by 2, 0, 3, 5. Since we have 4 distinct eigenvalues,M andhence T are diagonalizable.

Corresponding to this list of eigenvalues, we find eigenvectors {v1, v2, v3, v4} and put

them in columns of this matrix.

P = [v1|v2|v3|v4] =

2 6 1 32 2 1 11 3 2 6

1 1 2 2

.

So the basis of eigenvectors for Tconsist of the matrices Aiwhere [Ai]B =vifor i= 1, . . . , 4.We obtain

A1= 2 21 1

, A2= 6 23 1

, A3 1 12 2

, A4= 3 16 2

.

Check:

T(A4) = B A4+ A4C=

15 530 10

= 5A4,

as expected.

What we have done here is to find solutions to the equation BA +AC = A whereB and C are given and A and are unknown. It turns out that there are only 4 valuesof for which a non-zero Acan be found. Those 4 values are determined by the matricesB and C. However, the solution is probably not obvious except by turning this into an

eigenvalue problem for the linear transformation T.

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