Section CB  Change of Basis

From A First Course in Linear Algebra
Version 1.08
© 2004.
Licensed under the GNU Free Documentation License.
http://linear.ups.edu/

We have seen in Section MR that a linear transformation can be represented by a matrix, once we pick bases for the domain and codomain. How does the matrix representation change if we choose different bases? Which bases lead to especially nice representations? From the infinite possibilities, what is the best possible representation? This section will begin to answer these questions. But first we need to define eigenvalues for linear transformations and the change-of-basis matrix.

Subsection EELT: Eigenvalues and Eigenvectors of Linear Transformations

We now define the notion of an eigenvalue and eigenvector of a linear transformation. It should not be too surprising, especially if you remind yourself of the close relationship between matrices and linear transformations.

Definition EELT
Eigenvalue and Eigenvector of a Linear Transformation
Suppose that T : V V is a linear transformation. Then a nonzero vector v V is an eigenvector of T for the eigenvalue λ if T v = λv.

We will see shortly the best method for computing the eigenvalues and eigenvectors of a linear transformation, but for now, here are some examples to verify that such things really do exist.

Example ELTBM
Eigenvectors of linear transformation between matrices
Consider the linear transformation T : M22M22 defined by

T ab cd = 17a + 11b + 8c 11d57a + 35b + 24c 33d 14a + 10b + 6c 10d41a + 25b + 16c 23d

and the vectors

x1 = 01 01 x2 = 11 10 x3 = 13 23 x4 = 26 14

Then compute

T x1 = T 01 01 = 02 02 = 2x1 T x2 = T 11 10 = 22 20 = 2x2 T x3 = T 13 23 = 13 23 = (1)x3 T x4 = T 26 14 = 412 2 8 = (2)x4

So x1, x2, x3, x4 are eigenvectors of T with eigenvalues (respectively) λ1 = 2, λ2 = 2, λ3 = 1, λ4 = 2.

Here’s another.

Example ELTBP
Eigenvectors of linear transformation between polynomials
Consider the linear transformation R: P2P2 defined by

R a + bx + cx2 = (15a + 8b 4c) + (12a 6b + 3c)x + (24a + 14b 7c)x2

and the vectors

w1 = 1 x + x2 w 2 = x + 2x2 w 3 = 1 + 4x2

Then compute

R w1 = R 1 x + x2 = 3 3x + 3x2 = 3w 1 R w2 = R x + 2x2 = 0 + 0x + 0x2 = 0w 2 R w3 = R 1 + 4x2 = 1 4x2 = (1)w 3

So w1, w2, w3 are eigenvectors of R with eigenvalues (respectively) λ1 = 3, λ2 = 0, λ3 = 1. Notice how the eigenvalue λ2 = 0 indicates that the eigenvector w2 is a non-trivial element of the kernel of R, and therefore R is not injective (Exercise CB.T15).

Of course, these examples are meant only to illustrate the definition of eigenvectors and eigenvalues for linear transformations, and therefore beg the question, “How would I find eigenvectors?” We’ll have an answer before we finish this section. We need one more construction first.

Subsection CBM: Change-of-Basis Matrix

Given a vector space, we know we can usually find many different bases for the vector space, some nice, some nasty. If we choose a single vector from this vector space, we can build many different representations of the vector by constructing the representations relative to different bases. How are these different representations related to each other? A change-of-basis matrix answers this question.

Definition CBM
Change-of-Basis Matrix
Suppose that V is a vector space, and IV : V V is the identity linear transformation on V . Let B = v1,v2,v3,,vn and C be two bases of V . Then the change-of-basis matrix from B to C is the matrix representation of IV relative to B and C,

CB,C = MB,CIV = ρC IV v1 ρC IV v2 ρC IV v3 ρC IV vn = ρC v1 ρC v2 ρC v3 ρC vn

Notice that this definition is primarily about a single vector space (V ) and two bases of V (B, C). The linear transformation (IV ) is necessary but not critical. As you might expect, this matrix has something to do with changing bases. Here is the theorem that gives the matrix its name (not the other way around).

Theorem CB
Change-of-Basis
Suppose that v is a vector in the vector space V and B and C are bases of V . Then

ρC v = CB,CρB v

Proof  

ρC v = ρC IV v  Definition IDLT = MB,CIV ρB v  Theorem FTMR = CB,CρB v  Definition CBM

So the change-of-basis matrix can be used with matrix multiplication to convert a vector representation of a vector (v) relative to one basis (ρB v) to a representation of the same vector relative to a second basis (ρC v).

Theorem ICBM
Inverse of Change-of-Basis Matrix
Suppose that V is a vector space, and B and C are bases of V . Then the change-of-basis matrix CB,C is nonsingular and

CB,C1 = C C,B

Proof   The linear transformation IV : V V is invertible, and its inverse is itself, IV (check this!). So by Theorem IMR, the matrix MB,CIV = C B,C is invertible. Theorem NI says an invertible matrix is nonsingular.

Then

CB,C1 = M B,CIV 1  Definition CBM = MC,BIV 1   Theorem IMR = MC,BIV  Definition IDLT = CC,B  Definition CBM

Example CBP
Change of basis with polynomials
The vector space P4 (Example VSP) has two nice bases (Example BP),

B = 1,x,x2,x3,x4C = 1, 1 + x, 1 + x + x2, 1 + x + x2 + x3, 1 + x + x2 + x3 + x4

To build the change-of-basis matrix between B and C, we must first build a vector representation of each vector in B relative to C,

ρC 1 = ρC (1) 1 = 1 0 0 0 0 ρC x = ρC (1) 1 + (1) 1 + x = 1 1 0 0 0 ρC x2 = ρ C (1) 1 + x + (1) 1 + x + x2 = 0 1 1 0 0 ρC x3 = ρ C (1) 1 + x + x2 + (1) 1 + x + x2 + x3 = 0 0 1 1 0 ρC x4 = ρ C (1) 1 + x + x2 + x3 + (1) 1 + x + x2 + x3 + x4 = 0 0 0 1 1

Then we package up these vectors as the columns of a matrix,

CB,C = 11 0 0 0 0 1 1 0 0 0 0 1 1 0 0 0 0 1 1 0 0 0 0 1

Now, to illustrate Theorem CB, consider the vector u = 5 3x + 2x2 + 8x3 3x4. We can build the representation of u relative to B easily,

ρB u = ρB 5 3x + 2x2 + 8x3 3x4 = 5 3 2 8 3

Applying Theorem CB, we obtain a second representation of u, but now relative to C,

ρC u = CB,CρB u  Theorem CB = 11 0 0 0 0 1 1 0 0 0 0 1 1 0 0 0 0 1 1 0 0 0 0 1 5 3 2 8 3 = 8 5 6 11 3  Definition MVP

We can check our work by unraveling this second representation,

u = ρC1 ρ C u  Definition IVLT = ρC1 8 5 6 11 3 = 8(1) + (5)(1 + x) + (6)(1 + x + x2) + (11)(1 + x + x2 + x3) + (3)(1 + x + x2 + x3 + x4)  Definition VR = 5 3x + 2x2 + 8x3 3x4

The change-of-basis matrix from C to B is actually easier to build. Grab each vector in the basis C and form its representation relative to B

ρB 1 = ρB (1)1 = 1 0 0 0 0 ρB 1 + x = ρB (1)1 + (1)x = 1 1 0 0 0 ρB 1 + x + x2 = ρ B (1)1 + (1)x + (1)x2 = 1 1 1 0 0 ρB 1 + x + x2 + x3 = ρ B (1)1 + (1)x + (1)x2 + (1)x3 = 1 1 1 1 0 ρB 1 + x + x2 + x3 + x4 = ρ B (1)1 + (1)x + (1)x2 + (1)x3 + (1)x4 = 1 1 1 1 1

Then we package up these vectors as the columns of a matrix,

CC,B = 11111 01111 00111 00011 00001

We formed two representations of the vector u above, so we can again provide a check on our computations by converting from the representation of u relative to C to the representation of u relative to B,

ρB u = CC,BρC u  Theorem CB = 11111 01111 00111 00011 00001 8 5 6 11 3 = 5 3 2 8 3  Definition MVP

One more computation that is either a check on our work, or an illustration of a theorem. The two change-of-basis matrices, CB,C and CC,B, should be inverses of each other, according to Theorem ICBM. Here we go,

CB,CCC,B = 11 0 0 0 0 1 1 0 0 0 0 1 1 0 0 0 0 1 1 0 0 0 0 1 11111 01111 00111 00011 00001 = 10000 01000 00100 00010 00001

The computations of the previous example are not meant to present any labor-saving devices, but instead are meant to illustrate the utility of the change-of-basis matrix. However, you might have noticed that CC,B was easier to compute than CB,C. If you needed CB,C, then you could first compute CC,B and then compute its inverse, which by Theorem ICBM, would equal CB,C.

Here’s another illustrative example. We have been concentrating on working with abstract vector spaces, but all of our theorems and techniques apply just as well to m, the vector space of column vectors. We only need to use more complicated bases than the standard unit vectors (Theorem SUVB) to make things interesting.

Example CBCV
Change of basis with column vectors
For the vector space 4 we have the two bases,

B = 1 2 1 2 , 1 3 1 1 , 2 3 3 4 , 1 3 3 0 C = 1 6 4 1 , 4 8 5 8 , 5 13 2 9 , 3 7 3 6

The change-of-basis matrix from B to C requires writing each vector of B as a linear combination the vectors in C,

ρC 1 2 1 2 = ρC (1) 1 6 4 1 + (2) 4 8 5 8 + (1) 5 13 2 9 + (1) 3 7 3 6 = 1 2 1 1 ρC 1 3 1 1 = ρC (2) 1 6 4 1 + (3) 4 8 5 8 + (3) 5 13 2 9 + (0) 3 7 3 6 = 2 3 3 0 ρC 2 3 3 4 = ρC (1) 1 6 4 1 + (3) 4 8 5 8 + (1) 5 13 2 9 + (2) 3 7 3 6 = 1 3 1 2 ρC 1 3 3 0 = ρC (2) 1 6 4 1 + (2) 4 8 5 8 + (4) 5 13 2 9 + (3) 3 7 3 6 = 2 2 4 3

Then we package these vectors up as the change-of-basis matrix,

CB,C = 1 2 1 2 2332 1 3 1 4 1 0 2 3

Now consider a single (arbitrary) vector y = 2 6 3 4 . First,build the vector representation of y relative to B. This will require writing y as a linear combination of the vectors in B,

ρB y = ρB 2 6 3 4 = ρB (21) 1 2 1 2 + (6) 1 3 1 1 + (11) 2 3 3 4 + (7) 1 3 3 0 = 21 6 11 7

Now, applying Theorem CB we can convert the representation of y relative to B into a representation relative to C,

ρC y = CB,CρB y  Theorem CB = 1 2 1 2 2332 1 3 1 4 1 0 2 3 21 6 11 7 = 12 5 20 22  Definition MVP

We could continue further with this example, perhaps by computing the representation of y relative to the basis C directly as a check on our work (Exercise CB.C20). Or we could choose another vector to play the role of y and compute two different representations of this vector relative to the two bases B and C.

We saw in Theorem UMCOB that if each vector of an orthonormal basis is multiplied by a unitary matrix, the result is an orthonormal basis. Since a unitary matrix is invertible (Theorem UMI) we are not too surprised that this conversion preserves the linear independence and spanning properties — any nonsingular matrix would behave the same way. However, the preservation of the orthogonality of the basis is peculiar to the properties of a unitary matrix. We can now prove the converse of Theorem UMCOB.

Theorem CBOB
Change of Basis for Orthonormal Bases
Suppose that B and C are orthonormal bases for n. Then the change-of-basis matrix, CB,C, is a unitary matrix.

Proof   Let B = x1,x2,x3,,xn and C = y1,y2,y3,,yn. Then

yi = ρC ei  Definition VR = CB,CρB ei  Theorem CB = CB,Cxi  Definition VR

For convenience, now write U = CB,C. Choose any x n. then,

UUx = i=1n UUx,x i xi  Theorem COB = i=1n Ux, Ux i xi  Theorem AIP = i=1n Ux,Ux i xi  Theorem AA = i=1n Ux,y i xi  Theorem CB = i=1n x,Uy i xi  Theorem AIP = i=1n x,U1y i xi  Theorem UMI = i=1n x,x i xi  Theorem ICBM = x  Theorem COB = Inx  Theorem MMIM

Since x was arbitrary, Theorem EMMVP tells us that UU = I n, which is Definition UM.

Subsection MRS: Matrix Representations and Similarity

Here is the main theorem of this section. It looks a bit involved at first glance, but the proof should make you realize it is not all that complicated. In any event, we are more interested in a special case.

Theorem MRCB
Matrix Representation and Change of Basis
Suppose that T : UV is a linear transformation, B and C are bases for U, and D and E are bases for V . Then

MB,DT = C E,DMC,ET C B,C

Proof  

CE,DMC,ET C B,C = ME,DIV MC,ET M B,CIU  Definition CBM = ME,DIV MB,ETIU  Theorem MRCLT = ME,DIV MB,ET  Definition IDLT = MB,DIV T  Theorem MRCLT = MB,DT  Definition IDLT

We will be most interested in a special case of this theorem (Theorem SCB), but here’s an example that illustrates the full generality of Theorem MRCB.

Example MRCM
Matrix representations and change-of-basis matrices
Begin with two vector spaces, S2, the subspace of M22 containing all 2 × 2 symmetric matrices, and P3 (Example VSP), the vector space of all polynomials of degree 3 or less. Then define the linear transformation Q: S2P3 by

Q ab bc = (5a2b+6c)+(3ab+2c)x+(a+3bc)x2+(4a+2b+c)x3

Here are two bases for each vector space, one nice, one nasty. First for S2,

B = 5 3 32 , 2 3 3 0 , 12 24 C = 10 00 , 01 10 , 00 01

and then for P3,

D = 2 + x 2x2 + 3x3, 1 2x2 + 3x3, 3 x + x3, x2 + x3E = 1,x,x2,x3

We’ll begin with a matrix representation of Q relative to C and E. We first find vector representations of the elements of C relative to E,

ρE Q 10 00 = ρE 5 + 3x + x2 4x3 = 5 3 1 4 ρE Q 01 10 = ρE 2 x + 3x2 + 2x3 = 2 1 3 2 ρE Q 00 01 = ρE 6 + 2x x2 + x3 = 6 2 1 1

So

MC,EQ = 5 2 6 3 1 2 1 3 1 4 2 1

Now we construct two change-of-basis matrices. First, CB,C requires vector representations of the elements of B, relative to C. Since C is a nice basis, this is straightforward,

ρC 5 3 32 = ρC (5) 10 00 + (3) 01 10 + (2) 00 01 = 5 3 2 ρC 2 3 3 0 = ρC (2) 10 00 + (3) 01 10 + (0) 00 01 = 2 3 0 ρC 12 24 = ρC (1) 10 00 + (2) 01 10 + (4) 00 01 = 1 2 4

So

CB,C = 5 2 1 332 2 0 4

The other change-of-basis matrix we’ll compute is CE,D. However, since E is a nice basis (and D is not) we’ll turn it around and instead compute CD,E and apply Theorem ICBM to use an inverse to compute CE,D.

ρE 2 + x 2x2 + 3x3 = ρ E (2)1 + (1)x + (2)x2 + (3)x3 = 2 1 2 3 ρE 1 2x2 + 3x3 = ρ E (1)1 + (0)x + (2)x2 + (3)x3 = 1 0 2 3 ρE 3 x + x3 = ρ E (3)1 + (1)x + (0)x2 + (1)x3 = 3 1 0 1 ρE x2 + x3 = ρ E (0)1 + (0)x + (1)x2 + (1)x3 = 0 0 1 1

So, we can package these column vectors up as a matrix to obtain CD,E and then,

CE,D = CD,E 1  Theorem ICBM = 2 13 0 1 0 1 0 22 0 1 3 3 1 1 1 = 1 2 1 1 2 5 11 1 3 1 1 2 61 0

We are now in a position to apply Theorem MRCB. The matrix representation of Q relative to B and D can be obtained as follows,

MB,DQ = C E,DMC,EQC B,C  Theorem MRCB = 1 2 1 1 2 5 11 1 3 1 1 2 61 0 5 2 6 3 1 2 1 3 1 4 2 1 5 2 1 332 2 0 4 = 1 2 1 1 2 5 11 1 3 1 1 2 61 0 19 16 25 14 9 9 2 7 3 2814 4 = 3923 14 62 34 12 5332 5 4415 7

Now check our work by computing MB,DQ directly (Exercise CB.C21).

Here is a special case of the previous theorem, where we choose U and V to be the same vector space, so the matrix representations and the change-of-basis matrices are all square of the same size.

Theorem SCB
Similarity and Change of Basis
Suppose that T : V V is a linear transformation and B and C are bases of V . Then

MB,BT = C B,C1M C,CT C B,C

Proof   In the conclusion of Theorem MRCB, replace D by B, and replace E by C,

MB,BT = C C,BMC,CT C B,C  Theorem MRCB = CB,C1M C,CT C B,C  Theorem ICBM

This is the third surprise of this chapter. Theorem SCB considers the special case where a linear transformation has the same vector space for the domain and codomain (V ). We build a matrix representation of T using the basis B simultaneously for both the domain and codomain (MB,BT ), and then we build a second matrix representation of T, now using the basis C for both the domain and codomain (MC,CT ). Then these two representations are related via a similarity transformation (Definition SIM) using a change-of-basis matrix (CB,C)!

Example MRBE
Matrix representation with basis of eigenvectors
We return to the linear transfomation T : M22M22 of Example ELTBM defined by

T ab cd = 17a + 11b + 8c 11d57a + 35b + 24c 33d 14a + 10b + 6c 10d41a + 25b + 16c 23d

In Example ELTBM we showcased four eigenvectors of T. We will now put these four vectors in a set,

B = x1,x2,x3,x4 = 01 01 , 11 10 , 13 23 , 26 14

Check that B is a basis of M22 by first establishing the linear independence of B and then employing Theorem G to get the spanning property easily. Here is a second set of 2 × 2 matrices, which also forms a basis of M22 (Example BM),

C = y1,y2,y3,y4 = 10 00 , 01 00 , 00 10 , 00 01

We can build two matrix representations of T, one relative to B and one relative to C. Each is easy, but for wildly different reasons. In our computation of the matrix representation relative to B we borrow some of our work in Example ELTBM. Here are the representations, then the explanation.

ρB T x1 = ρB 2x1 = ρB 2x1 + 0x2 + 0x3 + 0x4 = 2 0 0 0 ρB T x2 = ρB 2x2 = ρB 0x1 + 2x2 + 0x3 + 0x4 = 0 2 0 0 ρB T x3 = ρB (1)x3 = ρB 0x1 + 0x2 + (1)x3 + 0x4 = 0 0 1 0 ρB T x4 = ρB (2)x4 = ρB 0x1 + 0x2 + 0x3 + (2)x4 = 0 0 0 2

So the resulting representation is

MB,BT = 20 0 0 02 0 0 001 0 00 0 2

Very pretty. Now for the matrix representation relative to C first compute,

ρC T y1 = ρC 1757 1441 = ρC (17) 10 00 + (57) 01 00 + (14) 00 10 + (41) 00 01 = 17 57 14 41 ρC T y2 = ρC 1135 1025 = ρC 11 10 00 + 35 01 00 + 10 00 10 + 25 00 01 = 11 35 10 25 ρC T y3 = ρC 824 616 = ρC 8 10 00 + 24 01 00 + 6 00 10 + 16 00 01 = 8 24 6 16 ρC T y4 = ρC 1133 1023 = ρC (11) 10 00 + (33) 01 00 + (10) 00 10 + (23) 00 01 = 11 33 10 23

So the resulting representation is

MC,CT = 1711 8 11 57352433 1410 6 10 41251623

Not quite as pretty. The purpose of this example is to illustrate Theorem SCB. This theorem says that the two matrix representations, MB,BT and MC,CT , of the one linear transformation, T, are related by a similarity transformation using the change-of-basis matrix CB,C. Lets compute this change-of-basis matrix. Notice that since C is such a nice basis, this is fairly straightforward,

ρC x1 = ρC 01 01 = ρC 0 10 00 + 1 01 00 + 0 00 10 + 1 00 01 = 0 1 0 1 ρC x2 = ρC 11 10 = ρC 1 10 00 + 1 01 00 + 1 00 10 + 0 00 01 = 1 1 1 0 ρC x3 = ρC 13 23 = ρC 1 10 00 + 3 01 00 + 2 00 10 + 3 00 01 = 1 3 2 3 ρC x4 = ρC 26 14 = ρC 2 10 00 + 6 01 00 + 1 00 10 + 4 00 01 = 2 6 1 4

So we have,

CB,C = 0112 1136 0121 1034

Now, according to Theorem SCB we can write,

MB,BT = C B,C1M C,CT C B,C 20 0 0 02 0 0 001 0 00 0 2 = 0112 1136 0121 1034 1 1711 8 11 57352433 1410 6 10 41251623 0112 1136 0121 1034

This should look and feel exactly like the process for diagonalizing a matrix, as was described in Section SD. And it is.

We can now return to the question of computing an eigenvalue or eigenvector of a linear transformation. For a linear transformation of the form T : V V , we know that representations relative to different bases are similar matrices. We also know that similar matrices have equal characteristic polynomials by Theorem SMEE. We will now show that eigenvalues of a linear transformation T are precisely the eigenvalues of any matrix representation of T. Since the choice of a different matrix representation leads to a similar matrix, there will be no “new” eigenvalues obtained from this second representation. Similarly, the change-of-basis matrix can be used to show that eigenvectors obtained from one matrix representation will be precisely those obtained from any other representation. So we can determine the eigenvalues and eigenvectors of a linear transformation by forming one matrix representation, using any basis we please, and analyzing the matrix in the manner of Chapter E.

Theorem EER
Eigenvalues, Eigenvectors, Representations
Suppose that T : V V is a linear transformation and B is a basis of V . Then v V is an eigenvector of T for the eigenvalue λ if and only if ρB v is an eigenvector of MB,BT for the eigenvalue λ.

Proof   ( ) Assume that v V is an eigenvector of T for the eigenvalue λ. Then

MB,BT ρ B v = ρB T v  Theorem FTMR = ρB λv  Definition EELT = λρB v  Theorem VRLT

which by Definition EEM says that ρB v is an eigenvector of the matrix MB,BT for the eigenvalue λ.

( ) Assume that ρB v is an eigenvector of MB,BT for the eigenvalue λ. Then

T v = ρB1 ρ B T v  Definition IVLT = ρB1 M B,BT ρ B v  Theorem FTMR = ρB1 λρ B v  Definition EEM = λρB1 ρ B v  Theorem ILTLT = λv  Definition IVLT

which by Definition EELT says v is an eigenvector of T for the eigenvalue λ.

Subsection CELT: Computing Eigenvectors of Linear Transformations

Knowing that the eigenvalues of a linear transformation are the eigenvalues of any representation, no matter what the choice of the basis B might be, we could now unambigously define items such as the characteristic polynomial of a linear transformation, rather than a matrix. We’ll say that again — eigenvalues, eigenvectors, and characteristic polynomials are intrinsic properties of a linear transformation, independent of the choice of a basis used to construct a matrix representation.

As a practical matter, how does one compute the eigenvalues and eigenvectors of a linear transformation of the form T : V V ? Choose a nice basis B for V , one where the vector representations of the values of the linear transformations necessary for the matrix representation are easy to compute. Construct the matrix representation relative to this basis, and find the eigenvalues and eigenvectors of this matrix using the techniques of Chapter E. The resulting eigenvalues of the matrix are precisely the eigenvalues of the linear transformation. The eigenvectors of the matrix are column vectors that need to be converted to vectors in V through application of ρB1.

Now consider the case where the matrix representation of a linear transformation is diagonalizable. The n linearly independent eigenvectors that must exist for the matrix (Theorem DC) can be converted (via ρB1) into eigenvectors of the linear transformation. A matrix representation of the linear transformation relative to a basis of eigenvectors will be a diagonal matrix — an especially nice representation! Though we did not know it at the time, the diagonalizations of Section SD were really finding especially pleasing matrix representations of linear transformations.

Here are some examples.

Example ELTT
Eigenvectors of a linear transformation, twice
Consider the linear transformation S : M22M22 defined by

S ab cd = b c 3d 14a 15b 13c + d 18a + 21b + 19c + 3d 6a 7b 7c 3d

To find the eigenvalues and eigenvectors of S we will build a matrix representation and analyze the matrix. Since Theorem EER places no restriction on the choice of the basis B, we may as well use a basis that is easy to work with. So set

B = x1,x2,x3,x4 = 10 00 , 01 00 , 00 10 , 00 01

Then to build the matrix representation of S relative to B compute,

ρB S x1 = ρB 0 14 18 6 = ρB 0x1 + (14)x2 + 18x3 + (6)x4 = 0 14 18 6 ρB S x2 = ρB 115 21 7 = ρB (1)x1 + (15)x2 + 21x3 + (7)x4 = 1 15 21 7 ρB S x3 = ρB 113 19 7 = ρB (1)x1 + (13)x2 + 19x3 + (7)x4 = 1 13 19 7 ρB S x4 = ρB 3 1 3 3 = ρB (3)x1 + 1x2 + 3x3 + (3)x4 = 3 1 3 3

So by Definition MR we have

M = MB,BS = 0 1 1 3 141513 1 18 21 19 3 6 7 7 3

Now compute eigenvalues and eigenvectors of the matrix representation of M with the techniques of Section EE. First the characteristic polynomial,

pM x = det M xI4 = x4 x3 10x2 + 4x + 24 = (x 3)(x 2)(x + 2)2

We could now make statements about the eigenvalues of M, but in light of Theorem EER we can refer to the eigenvalues of S and mildly abuse (or extend) our notation for multiplicities to write

αS 3 = 1 αS 2 = 1 αS 2 = 2

Now compute the eigenvectors of M,

λ = 3 M 3I4 = 3 1 1 3 141813 1 18 21 16 3 6 7 7 6  RREF 100 1 0103 001 3 000 0 M 3 = NM 3I4 = 1 3 3 1

λ = 2 M 2I4 = 2 1 1 3 141713 1 18 21 17 3 6 7 7 5  RREF 100 2 0104 001 3 000 0 M 2 = NM 2I4 = 2 4 3 1 λ = 2M (2)I4 = 2 1 1 3 141313 1 18 21 21 3 6 7 7 1  RREF 1001 011 1 000 0 000 0 M 2 = NM (2)I4 = 0 1 1 0 , 1 1 0 1

According to Theorem EER the eigenvectors just listed as basis vectors for the eigenspaces of M are vector representations (relative to B) of eigenvectors for S. So the application if the inverse function ρB1 will convert these column vectors into elements of the vector space M22 (2 × 2 matrices) that are eigenvectors of S. Since ρB is an isomorphism (Theorem VRILT), so is ρB1. Applying the inverse function will then preserve linear independence and spanning properties, so with a sweeping application of the Coordinatization Principle and some extensions of our previous notation for eigenspaces and geometric multiplicities, we can write,

ρB1 1 3 3 1 = (1)x1 + 3x2 + (3)x3 + 1x4 = 13 31 ρB1 2 4 3 1 = (2)x1 + 4x2 + (3)x3 + 1x4 = 24 31 ρB1 0 1 1 0 = 0x1 + (1)x2 + 1x3 + 0x4 = 01 1 0 ρB1 1 1 0 1 = 1x1 + (1)x2 + 0x3 + 1x4 = 11 0 1

So

S 3 = 13 31 S 2 = 24 31 S 2 = 01 1 0 , 11 0 1

with geometric multiplicities given by

γS 3 = 1 γS 2 = 1 γS 2 = 2

Suppose we now decided to build another matrix representation of S, only now relative to a linearly independent set of eigenvectors of S, such as

C = 13 31 , 24 31 , 01 1 0 , 11 0 1

At this point you should have computed enough matrix representations to predict that the result of representing S relative to C will be a diagonal matrix. Computing this representation is an example of how Theorem SCB generalizes the diagonalizations from Section SD. For the record, here is the diagonal representation,

MC,CS = 30 0 0 02 0 0 002 0 00 0 2

Our interest in this example is not necessarily building nice representations, but instead we want to demonstrate how eigenvalues and eigenvectors are an intrinsic property of a linear transformation, independent of any particular representation. To this end, we will repeat the foregoing, but replace B by another basis. We will make this basis different, but not extremely so,

D = y1,y2,y3,y4 = 10 00 , 11 00 , 11 10 , 11 11

Then to build the matrix representation of S relative to D compute,

ρD S y1 = ρD 0 14 18 6 = ρD 14y1 + (32)y2 + 24y3 + (6)y4 = 14 32 24 6 ρD S y2 = ρD 129 3913 = ρD 28y1 + (68)y2 + 52y3 + (13)y4 = 28 68 52 13 ρD S y3 = ρD 242 5820 = ρD 40y1 + (100)y2 + 78y3 + (20)y4 = 40 100 78 20 ρD S y4 = ρD 541 6123 = ρD 36y1 + (102)y2 + 84y3 + (23)y4 = 36 102 84 23

So by Definition MR we have

N = MD,DS = 14 28 40 36 3268100102 24 52 78 84 6 13 20 23

Now compute eigenvalues and eigenvectors of the matrix representation of N with the techniques of Section EE. First the characteristic polynomial,

pN x = det N xI4 = x4 x3 10x2 + 4x + 24 = (x 3)(x 2)(x + 2)2

Of course this is not news. We now know that M = MB,BS and N = MD,DS are similar matrices (Theorem SCB). But Theorem SMEE told us long ago that similar matrices have identical characteristic polynomials. Now compute eigenvectors for the matrix representation, which will be different than what we found for M,

λ = 3 N 3I4 = 11 28 40 36 3271100102 24 52 75 84 6 13 20 26  RREF 100 4 0106 001 4 000 0 N 3 = NN 3I4 = 4 6 4 1

λ = 2 N 2I4 = 12 28 40 36 3270100102 24 52 76 84 6 13 20 25  RREF 100 6 0107 001 4 000 0 N 2 = NN 2I4 = 6 7 4 1 λ = 2N (2)I4 = 16 28 40 36 3266100102 24 52 80 84 6 13 20 21  RREF 1013 01 2 3 00 0 0 00 0 0 N 2 = NN (2)I4 = 1 2 1 0 , 3 3 0 1

Employing Theorem EER we can apply ρD1 to each of the basis vectors of the eigenspaces of N to obtain eigenvectors for S that also form bases for eigenspaces of S,

ρD1 4 6 4 1 = (4)y1 + 6y2 + (4)y3 + 1y4 = 13 31 ρD1 6 7 4 1 = (6)y1 + 7y2 + (4)y3 + 1y4 = 24 31 ρD1 1 2 1 0 = 1y1 + (2)y2 + 1y3 + 0y4 = 01 1 0 ρD1 3 3 0 1 = 3y1 + (3)y2 + 0y3 + 1y4 = 12 1 1

The eigenspaces for the eigenvalues of algebraic multiplicity 1 are exactly as before,

S 3 = 13 31 S 2 = 24 31

However, the eigenspace for λ = 2 would at first glance appear to be different. Here are the two eigenspaces for λ = 2, first the eigenspace obtained from M = MB,BS, then followed by the eigenspace obtained from M = MD,DS.

S 2 = 01 1 0 , 11 0 1 S 2 = 01 1 0 , 12 1 1

Subspaces generally have many bases, and that is the situation here. With a careful proof of set equality, you can show that these two eigenspaces are equal sets. The key observation to make such a proof go is that

12 1 1 = 01 1 0 + 11 0 1

which will establish that the second set is a subset of the first. With equal dimensions, Theorem EDYES will finish the task. So the eigenvalues of a linear transformation are independent of the matrix representation employed to compute them!

Another example, this time a bit larger and with complex eigenvalues.

Example CELT
Complex eigenvectors of a linear transformation
Consider the linear transformation Q: P4P4 defined by

Q a + bx + cx2 + dx3 + ex4 = (46a 22b + 13c + 5d + e) + (117a + 57b 32c 15d 4e)x+ (69a 29b + 21c 7e)x2 + (159a + 73b 44c 13d + 2e)x3+ (195a 87b + 55c + 10d 13e)x4

Choose a simple basis to compute with, say

B = 1,x,x2,x3,x4

Then it should be apparent that the matrix representation of Q relative to B is

M = MB,BQ = 46 22 13 5 1 117 57 3215 4 69 29 21 0 7 159 73 4413 2 19587 55 10 13

Compute the characteristic polynomial, eigenvalues and eigenvectors according to the techniques of Section EE,

pQ x = x5 + 6x4 x3 88x2 + 252x 208 = (x 2)2(x + 4) x2 6x + 13 = (x 2)2(x + 4) x (3 + 2i) x (3 2i)

αQ 2 = 2 αQ 4 = 1 αQ 3 + 2i = 1 αQ 3 2i = 1 λ = 2 M (2)I5 = 48 22 13 5 1 117 55 3215 4 69 29 19 0 7 159 73 4415 2 19587 55 10 15  RREF 100 1 2 1 2 0105 25 2 00126 000 0 0 000 0 0 M 2 = NM (2)I5 = 1 2 5 2 2 1 0 , 1 2 5 2 6 0 1 = 1 5 4 2 0 , 1 5 12 0 2

λ = 4 M (4)I5 = 42 22 13 5 1 117 61 32154 69 29 25 0 7 159 73 44 9 2 19587 55 10 9  RREF 1000 1 01003 00101 00012 0000 0 M 4 = NM (4)I5 = 1 3 1 2 1 λ = 3 + 2i M (3 + 2i)I5 = 49 2i 22 13 5 1 117 54 2i 32 15 4 69 29 18 2i 0 7 159 73 44 16 2i 2 195 87 55 10 16 2i  RREF 10003 4 + i 4 0100 7 4 i 4 00101 2 + i 2 0001 7 4 i 4 0000 0 M 3 + 2i = NM (3 + 2i)I5 = 3 4 i 4 7 4 + i 4 1 2 i 2 7 4 + i 4 1 = 3 i 7 + i 2 2i 7 + i 4

λ = 3 2i M (3 2i)I5 = 49 + 2i 22 13 5 1 117 54 + 2i 32 15 4 69 29 18 + 2i 0 7 159 73 44 16 + 2i 2 195 87 55 10 16 + 2i  RREF 10003 4 i 4 0100 7 4 + i 4 00101 2 i 2 0001 7 4 + i 4 0000 0 M 3 2i = NM (3 2i)I5 = 3 4 + i 4 7 4 i 4 1 2 + i 2 7 4 i 4 1 = 3 + i 7 i 2 + 2i 7 i 4 It is straightforward to convert each of these basis vectors for eigenspaces of M back to elements of P4 by applying the isomorphism ρB1, ρB1 1 5 4 2 0 = 1 + 5x + 4x2 + 2x3 ρB1 1 5 12 0 2 = 1 + 5x + 12x2 + 2x4 ρB1 1 3 1 2 1 = 1 + 3x + x2 + 2x3 + x4 ρB1 3 i 7 + i 2 2i 7 + i 4 = (3 i) + (7 + i)x + (2 2i)x2 + (7 + i)x3 + 4x4 ρB1 3 + i 7 i 2 + 2i 7 i 4 = (3 + i) + (7 i)x + (2 + 2i)x2 + (7 i)x3 + 4x4

So we apply Theorem EER and the Coordinatization Principle to get the eigenspaces for Q,

Q 2 = 1 + 5x + 4x2 + 2x3,1 + 5x + 12x2 + 2x4 Q 4 = 1 + 3x + x2 + 2x3 + x4 Q 3 + 2i = (3 i) + (7 + i)x + (2 2i)x2 + (7 + i)x3 + 4x4 Q 3 2i = (3 + i) + (7 i)x + (2 + 2i)x2 + (7 i)x3 + 4x4

with geometric multiplicities

γQ 2 = 2 γQ 4 = 1 γQ 3 + 2i = 1 γQ 3 2i = 1

Subsection READ: Reading Questions

  1. The change-of-basis matrix is a matrix representation of which linear transformation?
  2. Find the change-of-basis matrix, CB,C, for the two bases of 2 B = 2 3 , 1 2 C = 1 0 , 1 1
  3. What is the third “surprise,” and why is it surprising?

Subsection EXC: Exercises

C20 In Example CBCV we computed the vector representation of y relative to C, ρC y, as an example of Theorem CB. Compute this same representation directly. In other words, apply Definition VR rather than Theorem CB.  
Contributed by Robert Beezer

C21 Perform a check on Example MRCM by computing MB,DQ directly. In other words, apply Definition MR rather than Theorem MRCB.  
Contributed by Robert Beezer Solution [1688]

C30 Find a basis for the vector space P3 composed of eigenvectors of the linear transformation T. Then find a matrix representation of T relative to this basis.

T : P3P3,T a + bx + cx2 + dx3 = (a+c+d)+(b+c+d)x+(a+b+c)x2+(a+b+d)x3

 
Contributed by Robert Beezer Solution [1689]

C40 Let S22 be the vector space of 2 × 2 symmetric matrices. Find a basis B for S22 that yields a diagonal matrix representation of the linear transformation R. (15 points)

R: S22S22,R ab bc = 5a + 2b 3c 12a + 5b 6c 12a + 5b 6c 6a 2b + 4c

 
Contributed by Robert Beezer Solution [1691]

C41 Let S22 be the vector space of 2 × 2 symmetric matrices. Find a basis for S22 composed of eigenvectors of the linear transformation Q: S22S22. (15 points)

Q ab bc = 25a + 18b + 30c 16a 11b 20c 16a 11b 20c 11a 9b 12c

 
Contributed by Robert Beezer Solution [1694]

T10 Suppose that T : V V is an invertible linear transformation with a nonzero eigenvalue λ. Prove that 1 λ is an eigenvalue of T1.  
Contributed by Robert Beezer Solution [1696]

T15 Suppose that V is a vector space and T : V V is a linear transformation. Prove that T is injective if and only if λ = 0 is not an eigenvalue of T.  
Contributed by Robert Beezer

Subsection SOL: Solutions

C21 Contributed by Robert Beezer Statement [1685]
Apply Definition MR,

ρD Q 5 3 32 = ρD 19 + 14x 2x2 28x3 = ρD (39)(2 + x 2x2 + 3x3) + 62(1 2x2 + 3x3) + (53)(3 x + x3) + (44)(x2 + x3) = 39 62 53 44 ρD Q 2 3 3 0 = ρD 16 + 9x 7x2 14x3 = ρD (23)(2 + x 2x2 + 3x3) + (34)(1 2x2 + 3x3) + (32)(3 x + x3) + (15)(x2 + x3) = 23 34 32 15 ρD Q 12 24 = ρD 25 + 9x + 3x2 + 4x3 = ρD (14)(2 + x 2x2 + 3x3) + (12)(1 2x2 + 3x3) + 5(3 x + x3) + (7)(x2 + x3) = 14 12 5 7

These three vectors are the columns of the matrix representation,

MB,DQ = 3923 14 62 34 12 5332 5 4415 7

which coincides with the result obtained in Example MRCM.

C30 Contributed by Robert Beezer Statement [1685]
With the domain and codomain being identical, we will build a matrix representation using the same basis for both the domain and codomain. The eigenvalues of the matrix representation will be the eigenvalues of the linear transformation, and we can obtain the eigenvectors of the linear transformation by un-coordinatizing (Theorem EER). Since the method does not depend on which basis we choose, we can choose a natural basis for ease of computation, say,

B = 1,x,x2,x3

The matrix representation is then,

MB,BT = 1011 0111 1110 1101

The eigenvalues and eigenvectors of this matrix were computed in Example ESMS4. A basis for 4, composed of eigenvectors of the matrix representation is,

C = 1 1 1 1 , 1 1 0 0 , 0 0 1 1 , 1 1 1 1

Applying ρB1 to each vector of this set, yields a basis of P3 composed of eigenvectors of T,

D = 1 + x + x2 + x3,1 + x, x2 + x3, 1 x + x2 + x3

The matrix representation of T relative to the basis D will be a diagonal matrix with the corresponding eigenvalues along the diagonal, so in this case we get

MD,DT = 300 0 010 0 001 0 0001

C40 Contributed by Robert Beezer Statement [1685]
Begin with a matrix representation of R, any matrix representation, but use the same basis for both instances of S22. We’ll choose a basis that makes it easy to compute vector representations in S22.

B = 10 00 , 01 10 , 00 01

Then the resulting matrix representation of R (Definition MR) is

MB,BR = 5 2 3 12 5 6 6 2 4

Now, compute the eigenvalues and eigenvectors of this matrix, with the goal of diagonalizing the matrix (Theorem DC),

λ = 2 MB,BR 2 = 1 2 1 λ = 1 MB,BR 1 = 1 0 2 , 1 3 0

The three vectors that occur as basis elements for these eigenspaces will together form a linearly independent set (check this!). So these column vectors may be employed in a matrix that will diagonalize the matrix representation. If we “un-coordinatize” these three column vectors relative to the basis B, we will find three linearly independent elements of S22 that are eigenvectors of the linear transformation R (Theorem EER). A matrix representation relative to this basis of eigenvectors will be diagonal, with the eigenvalues (λ = 2,1) as the diagonal elements. Here we go,

ρB1 1 2 1 = (1) 10 00 + (2) 01 10 + 1 00 01 = 12 2 1 ρB1 1 0 2 = (1) 10 00 + 0 01 10 + 2 00 01 = 10 0 2 ρB1 1 3 0 = 1 10 00 + 3 01 10 + 0 00 01 = 13 30

So the requested basis of S22, yielding a diagonal matrix representation of R, is

12 2 1 10 0 2 , 13 30

C41 Contributed by Robert Beezer Statement [1686]
Use a single basis for both the domain and codomain, since they are equal.

B = 10 00 , 01 10 , 00 01

The matrix representation of Q relative to B is

M = MB,BQ = 25 18 30 161120 11 9 12

We can analyze this matrix with the techniques of Section EE and then apply Theorem EER. The eigenvalues of this matrix are λ = 2,1,3 with eigenspaces

M 2 = 6 4 3 M 1 = 2 1 1 M 3 = 3 2 1

Because the three eigenvalues are distinct, the three basis vectors from the three eigenspaces for a linearly independent set (Theorem EDELI). Theorem EER says we can uncoordinatize these eigenvectors to obtain eigenvectors of Q. By Theorem ILTLI the resulting set will remain linearly independent. Set

C = ρB1 6 4 3 ,ρB1 2 1 1 ,ρB1 3 2 1 = 64 4 3 , 21 1 1 , 32 2 1

Then C is a linearly independent set of size 3 in the vector space M22, which has dimension 3 as well. By Theorem G, C is a basis of M22.

T10 Contributed by Robert Beezer Statement [1687]
Let v be an eigenvector of T for the eigenvalue λ. Then,

T1 v = 1 λλT1 v  λ0 = 1 λT1 λv  Theorem ILTLT = 1 λT1 T v  v eigenvector of T = 1 λIV v  Definition IVLT = 1 λv  Definition IDLT

which says that 1 λ is an eigenvalue of T1 with eigenvector v. Note that it is possible to prove that any eigenvalue of an invertible linear transformation is never zero. So the hypothesis that λ be nonzero is just a convenience for this problem.