Section MR  Matrix Representations

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 that linear transformations whose domain and codomain are vector spaces of columns vectors have a close relationship with matrices (Theorem MBLT, Theorem MLTCV). In this section, we will extend the relationship between matrices and linear transformations to the setting of linear transformations between abstract vector spaces.

Definition MR
Matrix Representation
Suppose that T : UV is a linear transformation, B = u1,u2,u3,,un is a basis for U of size n, and C is a basis for V of size m. Then the matrix representation of T relative to B and C is the m × n matrix,

MB,CT = ρ C T u1 ρC T u2 ρC T u3 ρC T un

Example OLTTR
One linear transformation, three representations
Consider the linear transformation

S : P3M22,S a + bx + cx2 + dx3 = 3a + 7b 2c 5d 8a + 14b 2c 11d 4a 8b + 2c + 6d12a + 22b 4c 17d

First, we build a representation relative to the bases,

B = 1 + 2x + x2 x3,1 + 3x + x2 + x3, 1 2x + 2x3,2 + 3x + 2x2 5x3 C = 11 12 , 23 25 , 11 0 2 , 14 24

We evaluate S with each element of the basis for the domain, B, and coordinatize the result relative to the vectors in the basis for the codomain, C.

ρC S 1 + 2x + x2 x3 = ρ C 20 45 2469 = ρC (90) 11 12 + 37 23 25 + (40) 11 0 2 + 4 14 24 = 90 37 40 4 ρC S 1 + 3x + x2 + x3 = ρ C 17 37 2057 = ρC (72) 11 12 + 29 23 25 + (34) 11 0 2 + 3 14 24 = 72 29 34 3 ρC S 1 2x + 2x3 = ρ C 2758 32 90 = ρC 114 11 12 + (46) 23 25 + 54 11 0 2 + (5) 14 24 = 114 46 54 5 ρC S 2 + 3x + 2x2 5x3 = ρ C 48 109 58167 = ρC (220) 11 12 + 91 23 25 + 96 11 0 2 + 10 14 24 = 220 91 96 10

Thus, employing Definition MR

MB,CS = 9072114220 37 29 46 91 4034 54 96 4 3 5 10

Often we use “nice” bases to build matrix representations and the work involved is much easier. Suppose we take bases

D = 1,x,x2,x3 E = 10 00 , 01 00 , 00 10 , 00 01

The evaluation of S at the elements of D is easy and coordinatization relative to E can be done on sight,

ρE S 1 = ρE 3 8 412 = ρE 3 10 00 + 8 01 00 + (4) 00 10 + 12 00 01 = 3 8 4 12 ρE S x = ρE 7 14 822 = ρE 7 10 00 + 14 01 00 + (8) 00 10 + 22 00 01 = 7 14 8 22 ρE S x2 = ρ E 22 2 4 = ρE (2) 10 00 + (2) 01 00 + 2 00 10 + (4) 00 01 = 2 2 2 4 ρE S x3 = ρ E 511 6 17 = ρE (5) 10 00 + (11) 01 00 + 6 00 10 + (17) 00 01 = 5 11 6 17

So the matrix representation of S relative to D and E is

MD,ES = 3 7 2 5 8 14211 48 2 6 1222417

One more time, but now let’s use bases

F = 1 + x x2 + 2x3, 1 + 2x + 2x3,2 + x 2x2 + 3x3,1 + x + 2x3 G = 1 1 12 , 12 0 2 , 2 1 23 , 11 02

and evaluate S with the elements of F, then coordinatize the results relative to G,

ρG S 1 + x x2 + 2x3 = ρ G 2 2 24 = ρG 2 1 1 12 = 2 0 0 0 ρG S 1 + 2x + 2x3 = ρ G 12 02 = ρG (1) 12 0 2 = 0 1 0 0 ρG S 2 + x 2x2 + 3x3 = ρ G 2 1 23 = ρG 2 1 23 = 0 0 1 0 ρG S 1 + x + 2x3 = ρ G 00 00 = ρG 0 11 02 = 0 0 0 0

So we arrive at an especially economical matrix representation,

MF,GS = 2 0 00 0100 0 0 10 0 0 00

We may choose to use whatever terms we want when we make a definition. Some are arbitrary, while others make sense, but only in light of subsequent theorems. Matrix representation is in the latter category. We begin with a linear transformation and produce a matrix. So what? Here’s the theorem that justifies the term “matrix representation.”

Theorem FTMR
Fundamental Theorem of Matrix Representation
Suppose that T : UV is a linear transformation, B is a basis for U, C is a basis for V and MB,CT is the matrix representation of T relative to B and C. Then, for any u U,

ρC T u = MB,CT ρ B u

or equivalently

T u = ρC1 M B,CT ρ B u

Proof   Let B = u1,u2,u3,,un be the basis of U. Since u U, there are scalars a1,a2,a3,,an such that

u = a1u1 + a2u2 + a3u3 + + anun

Then,

MB,CT ρ B u = ρC T u1 ρC T u2 ρC T u3 ρC T un ρB u  Definition MR = ρC T u1 ρC T u2 ρC T u3 ρC T un a1 a2 a3 an  Definition VR = a1ρC T u1 + a2ρC T u2 + + anρC T un  Definition MVP = ρC a1T u1 + a2T u2 + a3T u3 + + anT un  Theorem LTLC = ρC T a1u1 + a2u2 + a3u3 + + anun  Theorem LTLC = ρC T u

The alternative conclusion is obtained as

T u = IV T u  Definition IDLT = ρC1 ρ C T u  Definition IVLT = ρC1 ρ C T u  Definition LTC = ρC1 M B,CT ρ B u

This theorem says that we can apply T to u and coordinatize the result relative to C in V , or we can first coordinatize u relative to B in U, then multiply by the matrix representation. Either way, the result is the same. So the effect of a linear transformation can always be accomplished by a matrix-vector product (Definition MVP). That’s important enough to say again. The effect of a linear transformation is a matrix-vector product.

2006/11/15: Commutative diagrams not available in XML version. See PDF versions.

The alternative conclusion of this result might be even more striking. It says that to effect a linear transformation (T) of a vector (u), coordinatize the input (with ρB), do a matrix-vector product (with MB,CT ), and un-coordinatize the result (with ρC1). So, absent some bookkeeping about vector representations, a linear transformation is a matrix.

Here’s an example to illustrate how the “action” of a linear transformation can be effected by matrix multiplication.

Example ALTMM
A linear transformation as matrix multiplication
In Example OLTTR we found three representations of the linear transformation S. In this example, we will compute a single output of S in four different ways. First “normally,” then three times over using Theorem FTMR.

Choose p(x) = 3 x + 2x2 5x3, for no particular reason. Then the straightforward application of S to p(x) yields

S p(x) = S 3 x + 2x2 5x3 = 3(3) + 7(1) 2(2) 5(5) 8(3) + 14(1) 2(2) 11(5) 4(3) 8(1) + 2(2) + 6(5)12(3) + 22(1) 4(2) 17(5) = 23 61 3091

Now use the representation of S relative to the bases B and C and Theorem FTMR. Note that we will employ the following linear combination in moving from the second line to the third,

3 x + 2x2 5x3 = 48(1 + 2x + x2 x3) + (20)(1 + 3x + x2 + x3)+ (1)(1 2x + 2x3) + (13)(2 + 3x + 2x2 5x3)

S p(x) = ρC1 M B,CSρ B p(x) = ρC1 M B,CSρ B 3 x + 2x2 5x3 = ρC1 M B,CS 48 20 1 13 = ρC1 9072114220 37 29 46 91 4034 54 96 4 3 5 10 48 20 1 13 = ρC1 134 59 46 7 = (134) 11 12 + 59 23 25 + (46) 11 0 2 + 7 14 24 = 23 61 3091 Again, but now with “nice” bases like D and E, and the computations are more transparent. S p(x) = ρE1 M D,ESρ D p(x) = ρE1 M D,ESρ D 3 x + 2x2 5x3 = ρE1 M D,ESρ D 3(1) + (1)(x) + 2(x2) + (5)(x3) = ρE1 M D,ES 3 1 2 5 = ρE1 3 7 2 5 8 14211 48 2 6 1222417 3 1 2 5 = ρE1 23 61 30 91 = 23 10 00 + 61 01 00 + (30) 00 10 + 91 00 01 = 23 61 3091

OK, last time, now with the bases F and G. The coordinatizations will take some work this time, but the matrix-vector product (Definition MVP) (which is the actual action of the linear transformation) will be especially easy, given the diagonal nature of the matrix representation, MF,GS. Here we go,

S p(x) = ρG1 M F,GSρ F p(x) = ρG1 M F,GSρ F 3 x + 2x2 5x3 = ρG1 M F,GSρ F 32(1 + x x2 + 2x3) 7(1 + 2x + 2x3) 17(2 + x 2x2 + 3x3) 2(1 + x + 2x3) = ρG1 M F,GS 32 7 17 2 = ρG1 2 0 00 0100 0 0 10 0 0 00 32 7 17 2 = ρG1 64 7 17 0 = 64 1 1 12 + 7 12 0 2 + (17) 2 1 23 + 0 11 02 = 23 61 3091

This example is not meant to necessarily illustrate that any one of these four computations is simpler than the others. Instead, it is meant to illustrate the many different ways we can arrive at the same result, with the last three all employing a matrix representation to effect the linear transformation.

We will use Theorem FTMR frequently in the next few sections. A typical application will feel like the linear transformation T “commutes” with a vector representation, ρC, and as it does the transformation morphs into a matrix, MB,CT , while the vector representation changes to a new basis, ρB. Or vice-versa.

Subsection NRFO: New Representations from Old

In Subsection LT.NLTFO we built new linear transformations from other linear transformations. Sums, scalar multiples and compositions. These new linear transformations will have matrix represntations as well. How do the new matrix representations relate to the old matrix representations? Here are the three theorems.

Theorem MRSLT
Matrix Representation of a Sum of Linear Transformations
Suppose that T : UV and S : UV are linear transformations, B is a basis of U and C is a basis of V . Then

MB,CT+S = M B,CT + M B,CS

Proof   Let x be any vector in n. Define u U by u = ρB1 x, so x = ρB u. Then,

MB,CT+Sx = M B,CT+Sρ B u  Substitution = ρC T + S u  Theorem FTMR = ρC T u + S u  Definition LTA = ρC T u + ρC S u  Definition LT = MB,CT ρ B u + MB,CS ρ B u  Theorem FTMR = MB,CT + M B,CS ρ B u  Theorem MMDAA = MB,CT + M B,CS x  Substitution

Since the matrices MB,CT+S and MB,CT + M B,CS have equal matrix-vector products for every vector in n, by Theorem EMMVP they are equal matrices. (Now would be a good time to double-back and study the proof of Theorem EMMVP. You did promise, didn’t you?)

Theorem MRMLT
Matrix Representation of a Multiple of a Linear Transformation
Suppose that T : UV is a linear transformation, α , B is a basis of U and C is a basis of V . Then

MB,CαT = αM B,CT

Proof   Let x be any vector in n. Define u U by u = ρB1 x, so x = ρB u. Then,

MB,CαT x = M B,CαT ρ B u  Substitution = ρC αT u  Theorem FTMR = ρC αT u  Definition LTSM = αρC T u  Definition LT = α MB,CT ρ B u  Theorem FTMR = αMB,CT ρ B u  Theorem MMSMM = αMB,CT x  Substitution

Since the matrices MB,CαT and αMB,CT have equal matrix-vector products for every vector in n, by Theorem EMMVP they are equal matrices.

The vector space of all linear transformations from U to V is now isomorphic to the vector space of all m × n matrices.

Theorem MRCLT
Matrix Representation of a Composition of Linear Transformations
Suppose that T : UV and S : V W are linear transformations, B is a basis of U, C is a basis of V , and D is a basis of W. Then

MB,DST = M C,DSM B,CT

Proof   Let x be any vector in n. Define u U by u = ρB1 x, so x = ρB u. Then,

MB,DST x = M B,DST ρ B u  Substitution = ρD S T u  Theorem FTMR = ρD S T u  Definition LTC = MC,DSρ C T u  Theorem FTMR = MC,DS M B,CT ρ B u  Theorem FTMR = MC,DSM B,CT ρ B u  Theorem MMA = MC,DSM B,CT x  Substitution

Since the matrices MB,DST and MC,DSM B,CT have equal matrix-vector products for every vector in n, by Theorem EMMVP they are equal matrices.

This is the second great surprise of introductory linear algebra. Matrices are linear transformations (functions, really), and matrix multiplication is function composition! We can form the composition of two linear transformations, then form the matrix representation of the result. Or we can form the matrix representation of each linear transformation separately, then multiply the two representations together via Definition MM. In either case, we arrive at the same result.

Example MPMR
Matrix product of matrix representations
Consider the two linear transformations,

T : 2P 2 T a b = (a + 3b) + (2a + 4b)x + (a 2b)x2 S : P2M22 S a + bx + cx2 = 2a + b + 2c a + 4b c a + 3c 3a + b + 2c

and bases for 2, P2 and M22 (respectively),

B = 3 1 , 2 1 C = 1 2x + x2, 1 + 3x,2x + 3x2 D = 12 11 , 11 1 2 , 12 0 0 , 23 2 2

Begin by computing the new linear transformation that is the composition of T and S (Definition LTC, Theorem CLTLT), S T : 2M 22,

S T a b = S T a b = S (a + 3b) + (2a + 4b)x + (a 2b)x2 = 2(a + 3b) + (2a + 4b) + 2(a 2b) (a + 3b) + 4(2a + 4b) (a 2b) (a + 3b) + 3(a 2b) 3(a + 3b) + (2a + 4b) + 2(a 2b) = 2a + 6b6a + 21b 4a 9b a + 9b

Now compute the matrix representations (Definition MR) for each of these three linear transformations (T, S, S T), relative to the appropriate bases. First for T,

ρC T 3 1 = ρC 10x + x2 = ρC 28(1 2x + x2) + 28(1 + 3x) + (9)(2x + 3x2) = 28 28 9 ρC T 2 1 = ρC 1 + 8x = ρC 33(1 2x + x2) + 32(1 + 3x) + (11)(2x + 3x2) = 33 32 11

So we have the matrix representation of T,

MB,CT = 28 33 28 32 911

Now, a representation of S,

ρD S 1 2x + x2 = ρ D 28 2 3 = ρD (11) 12 11 + (21) 11 1 2 + 0 12 0 0 + (17) 23 2 2 = 11 21 0 17 ρD S 1 + 3x = ρD 111 1 0 = ρD 26 12 11 + 51 11 1 2 + 0 12 0 0 + (38) 23 2 2 = 26 51 0 38 ρD S 2x + 3x2 = ρ D 85 98 = ρD 34 12 11 + 67 11 1 2 + 1 12 0 0 + (46) 23 2 2 = 34 67 1 46

So we have the matrix representation of S,

MC,DS = 11 26 34 21 51 67 0 0 1 17 3846

Finally, a representation of S T,

ρD S T 3 1 = ρD 1239 3 12 = ρD 114 12 11 + 237 11 1 2 + (9) 12 0 0 + (174) 23 2 2 = 114 237 9 174 ρD S T 2 1 = ρD 1033 111 = ρD 95 12 11 + 202 11 1 2 + (11) 12 0 0 + (149) 23 2 2 = 95 202 11 149

So we have the matrix representation of S T,

MB,DST = 114 95 237 202 9 11 174149

Now, we are all set to verify the conclusion of Theorem MRCLT,

MC,DSM B,CT = 11 26 34 21 51 67 0 0 1 17 3846 28 33 28 32 911 = 114 95 237 202 9 11 174149 = MB,DST

We have intentionally used non-standard bases. If you were to choose “nice” bases for the three vector spaces, then the result of the theorem might be rather transparent. But this would still be a worthwhile exercise — give it a go.

A diagram, similar to ones we have seen earlier, might make the importance of this theorem clearer,

2006/11/15: Commutative diagrams not available in XML version. See PDF versions.

One of our goals in the first part of this book is to make the definition of matrix multiplication (Definition MVP, Definition MM) seem as natural as possible. However, many are brought up with an entry-by-entry description of matrix multiplication (Theorem ME) as the definition of matrix multiplication, and then theorems about columns of matrices and linear combinations follow from that definition. With this unmotivated definition, the realization that matrix multiplication is function composition is quite remarkable. It is an interesting exercise to begin with the question, “What is the matrix representation of the composition of two linear transformations?” and then, without using any theorems about matrix multiplication, finally arrive at the entry-by-entry description of matrix multiplication. Try it yourself (Exercise MR.T80).

Subsection PMR: Properties of Matrix Representations

It will not be a surprise to discover that the kernel and range of a linear transformation are closely related to the null space and column space of the transformation’s matrix representation. Perhaps this idea has been bouncing around in your head already, even before seeing the definition of a matrix representation. However, with a formal definition of a matrix representation (Definition MR), and a fundamental theorem to go with it (Theorem FTMR) we can be formal about the relationship, using the idea of isomorphic vector spaces (Definition IVS). Here are the twin theorems.

Theorem KNSI
Kernel and Null Space Isomorphism
Suppose that T : UV is a linear transformation, B is a basis for U of size n, and C is a basis for V . Then the kernel of T is isomorphic to the null space of MB,CT ,

KTNMB,CT

Proof   To establish that two vector spaces are isomorphic, we must find an isomorphism between them, an invertible linear transformation (Definition IVS). The kernel of the linear transformation T, KT, is a subspace of U, while the null space of the matrix representation, NMB,CT is a subspace of n. The function ρB is defined as a function from U to n, but we can just as well employ the definition of ρB as a function from KT to NMB,CT .

We must first insure that if we choose an input for ρB from KT that then the output will be an element of NMB,CT . So suppose that u KT. Then

MB,CT ρ B u = ρC T u  Theorem FTMR = ρC 0  Definition KLT = 0  Theorem LTTZZ

This says that ρB u NMB,CT , as desired.

The restriction in the size of the domain and codomain ρB will not affect the fact that ρB is a linear transformation (Theorem VRLT), nor will it affect the fact that ρB is injective (Theorem VRI). Something must be done though to verify that ρB is surjective. To this end, appeal to the definition of surjective (Definition SLT), and suppose that we have an element of the codomain, x NMB,CT n and we wish to find an element of the domain with x as its image. We now show that the desired element of the domain is u = ρB1 x. First, verify that u KT,

T u = T ρB1 x = ρC1 M B,CT ρ B ρB1 x  Theorem FTMR = ρC1 M B,CT I n x  Definition IVLT = ρC1 M B,CT x  Definition IDLT = ρC1 0 n  Definition KLT = 0V  Theorem LTTZZ  Second, verify that the proposed isomorphism, ρB, takes u to x, ρB u = ρB ρB1 x  Substitution = In x  Definition IVLT = x  Definition IDLT

With ρB demonstrated to be an injective and surjective linear transformation from KT to NMB,CT , Theorem ILTIS tells us ρB is invertible, and so by Definition IVS, we say KT and NMB,CT are isomorphic.

Example KVMR
Kernel via matrix representation
Consider the kernel of the linear transformation

T : M22P2,T ab cd = (2ab+c5d)+(a+4b+5b+2d)x+(3a2b+c8d)x2

We will begin with a matrix representation of T relative to the bases for M22 and P2 (respectively),

B = 1 2 11 , 1 3 14 , 1 2 02 , 2 5 24 C = 1 + x + x2,2 + 3x, 1 2x2

Then,

ρC T 1 2 11 = ρC 4 + 2x + 6x2 = ρC 2(1 + x + x2) + 0(2 + 3x) + (2)(1 2x2) = 2 0 2 ρC T 1 3 14 = ρC 18 + 28x2 = ρC (24)(1 + x + x2) + 8(2 + 3x) + (26)(1 2x2) = 24 8 26 ρC T 1 2 02 = ρC 10 + 5x + 15x2 = ρC 5(1 + x + x2) + 0(2 + 3x) + (5)(1 2x2) = 5 0 5 ρC T 2 5 24 = ρC 17 + 4x + 26x2 = ρC (8)(1 + x + x2) + (4)(2 + 3x) + (17)(1 2x2) = 8 4 17

So the matrix representation of T (relative to B and C) is

MB,CT = 2 24 5 8 0 8 0 4 226517

We know from Theorem KNSI that the kernel of the linear transformation T is isomorphic to the null space of the matrix representation MB,CT and by studying the proof of Theorem KNSI we learn that ρB is an isomorphism between these null spaces. Rather than trying to compute the kernel of T using definitions and techniques from Chapter LT we will instead analyze the null space of MB,CT using techniques from way back in Chapter V. First row-reduce MB,CT ,

2 24 5 8 0 8 0 4 226517  RREF 105 22 0101 2 0000

So, by Theorem BNS, a basis for NMB,CT is

5 2 0 1 0 , 2 1 2 0 1

We can now convert this basis of NMB,CT into a basis of KT by applying ρB1 to each element of the basis,

ρB1 5 2 0 1 0 = (5 2) 1 2 11 + 0 1 3 14 + 1 1 2 02 + 0 2 5 24 = 3 23 5 2 1 2 ρB1 2 1 2 0 1 = (2) 1 2 11 + (1 2) 1 3 14 + 0 1 2 02 + 1 2 5 24 = 1 21 2 1 2 0

So the set

3 23 5 2 1 2 , 1 21 2 1 2 0

is a basis for KT Just for fun, you might evaluate T with each of these two basis vectors and verify that the output is the zero polynomial (Exercise MR.C10).

An entirely similar result applies to the range of a linear transformation and the column space of a matrix representation of the linear transformation.

Theorem RCSI
Range and Column Space Isomorphism
Suppose that T : UV is a linear transformation, B is a basis for U of size n, and C is a basis for V of size m. Then the range of T is isomorphic to the column space of MB,CT ,

TCMB,CT

Proof   To establish that two vector spaces are isomorphic, we must find an isomorphism between them, an invertible linear transformation (Definition IVS). The range of the linear transformation T, T, is a subspace of V , while the column space of the matrix representation, CMB,CT is a subspace of m. The function ρC is defined as a function from V to m, but we can just as well employ the definition of ρC as a function from T to CMB,CT .

We must first insure that if we choose an input for ρC from T that then the output will be an element of CMB,CT . So suppose that v T. Then there is a vector u U, such that T u = v. Consider

MB,CT ρ B u = ρC T u  Theorem FTMR = ρC v  Definition RLT

This says that ρC v CMB,CT , as desired.

The restriction in the size of the domain and codomain will not affect the fact that ρC is a linear transformation (Theorem VRLT), nor will it affect the fact that ρC is injective (Theorem VRI). Something must be done though to verify that ρC is surjective. This all gets a bit confusing, since the domain of our isomorphism is the range of the linear transformation, so think about your objects as you go. To establish that ρC is surjective, appeal to the definition of a surjective linear transformation (Definition SLT), and suppose that we have an element of the codomain, y CMB,CT m and we wish to find an element of the domain with y as its image. Since y CMB,CT , there exists a vector, x n with MB,CT x = y. We now show that the desired element of the domain is v = ρC1 y. First, verify that v T by applying T to u = ρB1 x,

T u = T ρB1 x = ρC1 M B,CT ρ B ρB1 x  Theorem FTMR = ρC1 M B,CT I n x  Definition IVLT = ρC1 M B,CT x  Definition IDLT = ρC1 y  Definition CSM = v  Substitution  Second, verify that the proposed isomorphism, ρC, takes v to y, ρC v = ρC ρC1 y  Substitution = Im y  Definition IVLT = y  Definition IDLT

With ρC demonstrated to be an injective and surjective linear transformation from T to CMB,CT , Theorem ILTIS tells us ρC is invertible, and so by Definition IVS, we say T and CMB,CT are isomorphic.

Example RVMR
Range via matrix representation
In this example, we will recycle the linear transformation T and the bases B and C of Example KVMR but now we will compute the range of T,

T : M22P2,T ab cd = (2ab+c5d)+(a+4b+5b+2d)x+(3a2b+c8d)x2

With bases B and C,

B = 1 2 11 , 1 3 14 , 1 2 02 , 2 5 24 C = 1 + x + x2,2 + 3x, 1 2x2

we obtain the matrix representation

MB,CT = 2 24 5 8 0 8 0 4 226517

We know from Theorem RCSI that the range of the linear transformation T is isomorphic to the column space of the matrix representation MB,CT and by studying the proof of Theorem RCSI we learn that ρC is an isomorphism between these subspaces. Notice that since the range is a subspace of the codomain, we will employ ρC as the isomorphism, rather than ρB, which was the correct choice for an isomophism between the null spaces of Example KVMR.

Rather than trying to compute the range of T using definitions and techniques from Chapter LT we will instead analyze the column space of MB,CT using techniques from way back in Chapter M. First row-reduce MB,CT t,

2 0 2 24826 5 0 5 8 417  RREF 10 1 0125 4 00 0 00 0

Now employ Theorem CSRST and Theorem BRS (there are other methods we could choose here to compute the column space, such as Theorem BCS) to obtain the basis for CMB,CT ,

1 0 1 , 0 1 25 4

We can now convert this basis of CMB,CT into a basis of T by applying ρC1 to each element of the basis,

ρC1 1 0 1 = (1 + x + x2) (1 2x2) = 2 + x + 3x2 ρC1 0 1 25 4 = (2 + 3x) 25 4 (1 2x2) = 33 4 + 3x + 31 2 x2

So the set

2 + 3x + 3x2,33 4 + 3x + 31 2 x2

is a basis for T.

Theorem KNSI and Theorem RCSI can be viewed as further formal evidence for the Coordinatization Principle, though they are not direct consequences.

Subsection IVLT: Invertible Linear Transformations

We have seen, both in theorems and in examples, that questions about linear transformations are often equivalent to questions about matrices. It is the matrix representation of a linear transformation that makes this idea precise. Here’s our final theorem that solidifies this connection.

Theorem IMR
Invertible Matrix Representations
Suppose that T : UV is an invertible linear transformation, B is a basis for U and C is a basis for V . Then the matrix representation of T relative to B and C, MB,CT is an invertible matrix, and

MC,BT1 = MB,CT 1

Proof   This theorem states that the matrix representation of T1 can be found by finding the matrix inverse of the matrix representation of T (with suitable bases in the right places). It also says that the matrix representation of T is an invertible matrix. We can establish the invertibility, and precisely what the inverse is, by appealing to the definition of a matrix inverse, Definition MI. To this end, let B = u1,u2,u3,,un and C = v1,v2,v3,,vn. Then

MC,BT1 MB,CT = M B,BT1 T  Theorem MRCLT = MB,BIU  Definition IVLT = ρB IU u1 ρB IU u2 ρB IU un Definition MR = ρB u1 ρB u2 ρB u3 ρB un  Definition IDLT = e1|e2|e3||en  Definition VR = In  Definition IM  and MB,CT M C,BT1 = MC,CTT1  Theorem MRCLT = MC,CIV  Definition IVLT = ρC IV v1 ρC IV v2 ρC IV vn Definition MR = ρC v1 ρC v2 ρC v3 ρC vn  Definition IDLT = e1|e2|e3||en  Definition VR = In  Definition IM

So by Definition MI, the matrix MB,CT has an inverse, and that inverse is MC,BT1.

Example ILTVR
Inverse of a linear transformation via a representation
Consider the linear transformation

R: P3M22,R a + bx + cx2 + x3 = a + b c + 2d2a + 3b 2c + 3d a + b + 2d a + b + 2c 5d

If we wish to quickly find a formula for the inverse of R (presuming it exists), then choosing “nice” bases will work best. So build a matrix representation of R relative to the bases B and C,

B = 1,x,x2,x3 C = 10 00 , 01 00 , 00 10 , 00 01

Then,

ρC R 1 = ρC 1 2 11 = 1 2 1 1 ρC R x = ρC 13 11 = 1 3 1 1 ρC R x2 = ρ C 12 0 2 = 1 2 0 2 ρC R x3 = ρ C 2 3 25 = 2 3 2 5

So a representation of R is

MB,CR = 1 11 2 2 32 3 1 1 0 2 11 2 5

The matrix MB,CR is invertible (as you can check) so we know by Theorem IMR that R is invertible. Furthermore,

MC,BR1 = MB,CR 1 = 1 11 2 2 32 3 1 1 0 2 11 2 5 1 = 2072 3 8 3 1 1 1 0 1 0 6 2 1 1

We can use this representation of the inverse linear transformation, in concert with Theorem FTMR, to determine an explicit formula for the inverse itself,

R1 ab cd = ρB1 M C,BR1 ρC ab cd  Theorem FTMR = ρB1 M B,CR 1ρ C ab cd  Theorem IMR = ρB1 M B,CR 1 a b c d  Definition VR = ρB1 2072 3 8 3 1 1 1 0 1 0 6 2 1 1 a b c d  Definition MI = ρB1 20a 7b 2c + 3d 8a + 3b + c d a + c 6a + 2b + c d  Definition MVP = (20a 7b 2c + 3d) + (8a + 3b + c d)x + (a + c)x2 + (6a + 2b + c d)x3  Definition VR

You might look back at Example AIVLT, where we first witnessed the inverse of a linear transformation and recognize that the inverse (S) was built from using the method of this example on a matrix representation of T.

Theorem IMILT
Invertible Matrices, Invertible Linear Transformation
Suppose that A is a square matrix of size n and T : nn is the linear transformation defined by T x = Ax. Then A is invertible matrix if and only if T is an invertible linear transformation.

Proof   Choose bases B = C = e1,e2,e3,,en consisting of the standard unit vectors as a basis of n (Theorem SUVB) and build a matrix representation of T relative to B and C. Then

ρC T ei = ρC Aei = ρC Ai = Ai

So then the matrix representation of T, relative to B and C, is simply MB,CT = A. This is the basic observation that makes the rest of this proof go.

( ) Suppose T is invertible. Then T is injective by Theorem ILTIS and

n A = dim NA  Definition NOM = dim NMB,CT = dim ker T  Theorem KNSI = dim 0  Theorem KILT = 0

Then Theorem RNNM tells us that A is nonsingular, and therefore A is invertible (Theorem NI).

( ) Suppose A is a nonsingular matrix, then A is invertible (Theorem NI) and has zero nullity (Theorem RNNM). So

n T = dim KT  Definition NOLT = dim NMB,CT  Theorem KNSI = dim NA = dim 0  Theorem NMTNS = 0

So T has zero nullity, and therefore has a trivial kernel and by Theorem KILT T is injective. Furthermore, by Theorem RPNDD,

r T = dim n n T = n 0 = n

So T has full rank and therefore the range of T is all of n and by Theorem RSLT T is surjective. Finally, with T known to be injective and surjective, Theorem ILTIS says T is invertible.

This theorem looks like more work than you would imagine it to be. But by now, the connections between matrices and linear transformations should be starting to become more transparent, and you may have already recognized the invertiblity of a matrix as being tantamount to the invertiblity of the associated matrix representation. See Exercise MR.T60 as well.

We can update the NMEx series of theorems, yet again.

Theorem NME9
Nonsingular Matrix Equivalences, Round 9
Suppose that A is a square matrix of size n. The following are equivalent.

  1. A is nonsingular.
  2. A row-reduces to the identity matrix.
  3. The null space of A contains only the zero vector, NA = 0.
  4. The linear system SA,b has a unique solution for every possible choice of b.
  5. The columns of A are a linearly independent set.
  6. A is invertible.
  7. The column space of A is n, CA = n.
  8. The columns of A are a basis for n.
  9. The rank of A is n, r A = n.
  10. The nullity of A is zero, n A = 0.
  11. The determinant of A is nonzero, det A0.
  12. λ = 0 is not an eigenvalue of A.
  13. The linear transformation T : nn defined by T x = Ax is invertible.

Proof   By Theorem IMILT the new addition to this list is equivalent to the statement that A is invertible so we can expand Theorem NME8.

Subsection READ: Reading Questions

  1. Why does Theorem FTMR deserve the moniker “fundamental”?
  2. Find the matrix representation, MB,CT of the linear transformation
    T : 22,T x1 x2 = 2x1 x2 3x1 + 2x2

    relative to the bases

    B = 2 3 , 1 2 C = 1 0 , 1 1
  3. What is the second “surprise,” and why is it surprising?

Subsection EXC: Exercises

C10 Example KVMR concludes with a basis for the kernel of the linear transformation T. Compute the value of T for each of these two basis vectors. Did you get what you expected?  
Contributed by Robert Beezer

C20 Compute the matrix representation of T relative to the bases B and C.

T : P33,T a + bx + cx2 + dx3 = 2a 3b + 4c 2d a + b c + d 3a + 2c 3d B = 1,x,x2,x3 C = 1 0 0 , 1 1 0 , 1 1 1

 
Contributed by Robert Beezer Solution [1590]

C21 Find a matrix representation of the linear transformation T relative to the bases B and C.

T : P22,T p(x) = p(1) p(3) B = 2 5x + x2,1 + x x2,x2 C = 3 4 , 2 3

 
Contributed by Robert Beezer Solution [1591]

C22 Let S22 be the vector space of 2 × 2 symmetric matrices. Build the matrix representation of the linear transformation T : P2S22 relative to the bases B and C and then use this matrix representation to compute T 3 + 5x 2x2.

B = 1,1 + x,1 + x + x2 C = 10 00 , 01 10 , 00 01 T a + bx + cx2 = 2a b + ca + 3b c a + 3b c a c

 
Contributed by Robert Beezer Solution [1592]

C25 Use a matrix representation to determine if the linear transformation T : P3M22 surjective.

T a + bx + cx2 + dx3 = a + 4b + c + 2d4a b + 6c d a + 5b 2c + 2d a + 2c + 5d

 
Contributed by Robert Beezer Solution [1594]

C30 Find bases for the kernel and range of the linear transformation S below.

S : M22P2,S ab cd = (a+2b+5c4d)+(3ab+8c+2d)x+(a+b+4c2d)x2

 
Contributed by Robert Beezer Solution [1596]

C40 Let S22 be the set of 2 × 2 symmetric matrices. Verify that the linear transformation R is invertible and find R1.

R: S22P2,R ab bc = (ab)+(2a3b2c)x+(ab+c)x2

 
Contributed by Robert Beezer Solution [1598]

C41 Prove that the linear transformation S is invertible. Then find a formula for the inverse linear transformation, S1, by employing a matrix inverse. (15 points)

S : P1M1,2,S a + bx = 3a + b2a + b

 
Contributed by Robert Beezer Solution [1600]

C42 The linear transformation R: M12M21 is invertible. Use a matrix representation to determine a formula for the inverse linear transformation R1: M 21M12.

R ab = a + 3b 4a + 11b

 
Contributed by Robert Beezer Solution [1602]

C50 Use a matrix representation to find a basis for the range of the linear transformation L. (15 points)

L: M22P2,T ab cd = (a+2b+4c+d)+(3a+c2d)x+(a+b+3c+3d)x2

 
Contributed by Robert Beezer Solution [1604]

C51 Use a matrix representation to find a basis for the kernel of the linear transformation L. (15 points)

L: M22P2,T ab cd = (a+2b+4c+d)+(3a+c2d)x+(a+b+3c+3d)x2

 
Contributed by Robert Beezer

C52 Find a basis for the kernel of the linear transformation T : P2M22.

T a + bx + cx2 = a + 2b 2c 2a + 2b a + b 4c3a + 2b + 2c

 
Contributed by Robert Beezer Solution [1606]

M20 The linear transformation D performs differentiation on polynomials. Use a matrix representation of D to find the rank and nullity of D.

D: PnPn,D p(x) = p(x)

 
Contributed by Robert Beezer Solution [1609]

T60 Create an entirely different proof of Theorem IMILT that relies on Definition IVLT to establish the invertibility of T, and that relies on Definition MI to establish the invertibility of A.  
Contributed by Robert Beezer

T80 Suppose that T : UV and S : V W are linear transformations, and that B, C and D are bases for U, V , and W. Using only Definition MR define matrix representations for T and S. Using these two definitions, and Definition MR, derive a matrix representation for the composition S T in terms of the entries of the matrices MB,CT and MC,DS. Explain how you would use this result to motivate a definition for matrix multiplication that is strikingly similar to Theorem EMP.  
Contributed by Robert Beezer Solution [1611]

Subsection SOL: Solutions

C20 Contributed by Robert Beezer Statement [1582]
Apply Definition MR,

ρC T 1 = ρC 2 1 3 = ρC 1 1 0 0 + (2) 1 1 0 + 3 1 1 1 = 1 2 3 ρC T x = ρC 3 1 0 = ρC (4) 1 0 0 + 1 1 1 0 + 0 1 1 1 = 4 1 0 ρC T x2 = ρ C 4 1 2 = ρC 5 1 0 0 + (3) 1 1 0 + 2 1 1 1 = 5 3 2 ρC T x3 = ρ C 2 1 3 = ρC (3) 1 0 0 + 4 1 1 0 + (3) 1 1 1 = 3 4 3

These four vectors are the columns of the matrix representation,

MB,CT = 1 4 5 3 2 1 3 4 3 0 2 3

C21 Contributed by Robert Beezer Statement [1582]
Applying Definition MR,

ρC T 2 5x + x2 = ρ C 2 4 = ρC 2 3 4 + (4) 2 3 = 2 4 ρC T 1 + x x2 = ρ C 1 5 = ρC 13 3 4 + (19) 2 3 = 13 19 ρC T x2 = ρ C 1 9 = ρC (15) 3 4 + 23 2 3 = 15 23

So the resulting matrix representation is

MB,CT = 2 13 15 419 23

C22 Contributed by Robert Beezer Statement [1583]
Input to T the vectors of the basis B and coordinatize the outputs relative to C,

ρC T 1 = ρC 21 11 = ρC 2 10 00 + 1 01 10 + 1 00 01 = 2 1 1 ρC T 1 + x = ρC 14 41 = ρC 1 10 00 + 4 01 10 + 1 00 01 = 1 4 1 ρC T 1 + x + x2 = ρ C 23 30 = ρC 2 10 00 + 3 01 10 + 0 00 01 = 2 3 0

Applying Definition MR we have the matrix representation

MB,CT = 212 143 110

To compute T 3 + 5x 2x2 employ Theorem FTMR,

T 3 + 5x 2x2 = ρ C1 M B,CT ρ B 3 + 5x 2x2 = ρC1 M B,CT ρ B (2)(1) + 7(1 + x) + (2)(1 + x + x2) = ρC1 212 143 110 2 7 2 = ρC1 1 20 5 = (1) 10 00 + 20 01 10 + 5 00 01 = 120 20 5

You can, of course, check your answer by evaluating T 3 + 5x 2x2 directly.

C25 Contributed by Robert Beezer Statement [1584]
Choose bases B and C for the matrix representation,

B = 1,x,x2,x3 C = 10 00 , 01 00 , 00 10 , 00 01

Input to T the vectors of the basis B and coordinatize the outputs relative to C,

ρC T 1 = ρC 14 1 1 = ρC (1) 10 00 + 4 01 00 + 1 00 10 + 1 00 01 = 1 4 1 1 ρC T x = ρC 41 5 0 = ρC 4 10 00 + (1) 01 00 + 5 00 10 + 0 00 01 = 4 1 5 0 ρC T x2 = ρ C 1 6 22 = ρC 1 10 00 + 6 01 00 + (2) 00 10 + 2 00 01 = 1 6 2 2 ρC T x3 = ρ C 21 2 5 = ρC 2 10 00 + (1) 01 00 + 2 00 10 + 5 00 01 = 2 1 2 5

Applying Definition MR we have the matrix representation

MB,CT = 1 4 1 2 4 1 6 1 1 5 2 2 1 0 2 5

Properties of this matrix representation will translate to properties of the linear transformation The matrix representation is nonsingular since it row-reduces to the identity matrix (Theorem NMRRI) and therefore has a column space equal to 4 (Theorem CNMB). The column space of the matrix representation is isomorphic to the range of the linear transformation (Theorem RCSI). So the range of T has dimension 4, equal to the dimension of the codomain M22. By Theorem ROSLT, T is surjective.

C30 Contributed by Robert Beezer Statement [1584]
These subspaces will be easiest to construct by analyzing a matrix representation of S. Since we can use any matrix representation, we might as well use natural bases that allow us to construct the matrix representation quickly and easily,

B = 10 00 , 01 00 , 00 10 , 00 01 C = 1,x,x2

then we can practically build the matrix representation on sight,

MB,CS = 1 2 54 318 2 1 1 42

The first step is to find bases for the null space and column space of the matrix representation. Row-reducing the matrix representation we find,

103 0 0112 000 0

So by Theorem BNS and Theorem BCS, we have

NMB,CS = 3 1 1 0 , 0 2 0 1 CMB,CS = 1 3 1 , 2 1 1

Now, the proofs of Theorem KNSI and Theorem RCSI tell us that we can apply ρB1 and ρC1 (respectively) to “un-coordinatize” and get bases for the kernel and range of the linear transformation S itself,

KS = 31 1 0 , 02 01 S = 1 + 3x + x2,2 x + x2

C40 Contributed by Robert Beezer Statement [1585]
The analysis of R will be easiest if we analyze a matrix representation of R. Since we can use any matrix representation, we might as well use natural bases that allow us to construct the matrix representation quickly and easily,

B = 10 00 , 01 10 , 00 01 C = 1,x,x2

then we can practically build the matrix representation on sight,

MB,CR = 11 0 232 11 1

This matrix representation is invertible (it has a nonzero determinant of 1, Theorem SMZD, Theorem NI) so Theorem IMR tells us that the linear transformation S is also invertible. To find a formula for R1 we compute,

R1 a + bx + cx2 = ρ B1 M C,BR1 ρC a + bx + cx2  Theorem FTMR = ρB1 M B,CR 1ρ C a + bx + cx2  Theorem IMR = ρB1 M B,CR 1 a b c  Definition VR = ρB1 5 12 4 12 1 0 1 a b c  Definition MI = ρB1 5a b 2c 4a b 2c a + c  Definition MVP = 5a b 2c4a b 2c 4a b 2c a + c  Definition VR

C41 Contributed by Robert Beezer Statement [1585]
First, build a matrix representation of S (Definition MR). We are free to choose whatever bases we wish, so we should choose ones that are easy to work with, such as

B = 1,x C = 10 , 01

The resulting matrix representation is then

MB,CT = 31 21

this matrix is invertible, since it has a nonzero determinant, so by Theorem IMR the linear transformation S is invertible. We can use the matrix inverse and Theorem IMR to find a formula for the inverse linear transformation,

S1 ab = ρB1 M C,BS1 ρC ab  Theorem FTMR = ρB1 M B,CS 1ρ C ab  Theorem IMR = ρB1 M B,CS 1 a b  Definition VR = ρB1 31 21 1 a b = ρB1 1 1 2 3 a b  Definition MI = ρB1 a b 2a + 3b  Definition MVP = (a b) + (2a + 3b)x  Definition VR

C42 Contributed by Robert Beezer Statement [1586]
Choose bases B and C for M12 and M21 (respectively),

B = 10 , 01 C = 1 0 , 0 1

The resulting matrix representation is

MB,CR = 1 3 411

This matrix is invertible (its determinant is nonzero, Theorem SMZD), so by Theorem IMR, we can compute the matrix representation of R1 with a matrix inverse (Theorem TTMI),

MC,BR1 = 1 3 411 1 = 11 3 4 1

To obtain a general formula for R1, use Theorem FTMR,

R1 x y = ρB MC,BR1 ρC x y = ρB 11 3 4 1 x y = ρB 11x + 3y 4x y = 11x + 3y 4x y

C50 Contributed by Robert Beezer Statement [1586]
As usual, build any matrix representation of L, most likely using a “nice” bases, such as

B = 10 00 , 01 00 , 00 10 , 00 01 C = 1,x,x2

Then the matrix representation (Definition MR) is,

MB,CL = 1 24 1 3 012 113 3

Theorem RCSI tells us that we can compute the column space of the matrix representation, then use the isomorphism ρC1 to convert the column space of the matrix representation into the range of the linear transformation. So we first analyze the matrix representation,

1 24 1 3 012 113 3  RREF 1001 0101 001 1

With three nonzero rows in the reduced row-echelon form of the matrix, we know the column space has dimension 3. Since P2 has dimension 3 (Theorem DP), the range must be all of P2. So any basis of P2 would suffice as a basis for the range. For instance, C itself would be a correct answer.

A more laborious approach would be to use Theorem BCS and choose the first three columns of the matrix representation as a basis for the range of the matrix representation. These could then be “un-coordinatized” with ρC1 to yield a (“not nice”) basis for P2.

C52 Contributed by Robert Beezer Statement [1587]
Choose bases B and C for the matrix representation,

B = 1,x,x2 C = 10 00 , 01 00 , 00 10 , 00 01

Input to T the vectors of the basis B and coordinatize the outputs relative to C,

ρC T 1 = ρC 1 2 13 = ρC 1 10 00 + 2 01 00 + (1) 00 10 + 3 00 01 = 1 2 1 3 ρC T x = ρC 22 12 = ρC 2 10 00 + 2 01 00 + 1 00 10 + 2 00 01 = 2 2 1 2 ρC T x2 = ρ C 20 42 = ρC (2) 10 00 + 0 01 00 + (4) 00 10 + 2 00 01 = 2 0 4 2

Applying Definition MR we have the matrix representation

MB,CT = 1 22 2 2 0 114 3 2 2

The null space of the matrix representation is isomorphic (via ρB) to the kernel of the linear transformation (Theorem KNSI). So we compute the null space of the matrix representation by first row-reducing the matrix to,

10 2 012 00 0 00 0

Employing Theorem BNS we have

NMB,CT = 2 2 1

We only need to uncoordinatize this one basis vector to get a basis for KT,

KT = ρB1 2 2 1 = 2 + 2x + x2

M20 Contributed by Robert Beezer Statement [1588]
Build a matrix representation (Definition MR) with the set

B = 1,x,x2,,xn

employed as a basis of both the domain and codomain. Then

ρB D 1 = ρB 0 = 0 0 0 0 0 ρB D x = ρB 1 = 1 0 0 0 0 ρB D x2 = ρ B 2x = 0 2 0 0 0 ρB D x3 = ρ B 3x2 = 0 0 3 0 0 ρB D xn = ρ B nxn1 = 0 0 0 n 0

and the resulting matrix representation is

MB,BD = 010000 002000 000300 00000n 000000

This (n + 1) × (n + 1) matrix is very close to being in reduced row-echelon form. Multiply row i by 1 i, for 1 i n, to convert it to reduced row-echelon form. From this we can see that matrix representation MB,BD has rank n and nullity 1. Applying Theorem RCSI and Theorem KNSI tells us that the linear transformation D will have the same values for the rank and nullity, as well.

T80 Contributed by Robert Beezer Statement [1589]
Suppose that B = u1,u2,u3,,um, C = v1,v2,v3,,vn and D = w1,w2,w3,,wp. For convenience, set M = MB,CT , mij = Mij, 1 i n, 1 j m, and similarly, set N = MC,DS, nij = Nij, 1 i p, 1 j n. We want to learn about the matrix representation of S T : V W relative to B and D. We will examine a single (generic) entry of this representation.

MB,DST ij = ρD S T uj i  Definition MR = ρD S T uj i  Definition LTC = ρD S k=1nm kjvk i  Definition MR = ρD k=1nm kjS vk i  Theorem LTLC = ρD k=1nm kj =1pn kw i  Definition MR = ρD k=1n =1pm kjnkw i  Property DVA = ρD =1p k=1nm kjnkw i  Property C = ρD =1p k=1nm kjnk w i  Property DSA = k=1nm kjnik  Definition VR = k=1nn ikmkj  Property MCCN = k=1n M C,DS ik MB,CT kj  Property MCCN This formula for the entry of a matrix should remind you of Theorem EMP. However, while the theorem presumed we knew how to multiply matrices, the solution before us never uses any understanding of matrix products. It uses the definitions of vector and matrix representations, properties of linear transformations and vector spaces. So if we began a course by first discussing vector space, and then linear transformations between vector spaces, we could carry matrix representations into a motivation for a definition of matrix multiplication that is grounded in function composition. That is worth saying again — a definition of matrix representations of linear transformations results in a matrix product being the representation of a composition of linear transformations.

This exercise is meant to explain why many authors take the formula in Theorem EMP as their definition of matrix multiplication, and why it is a natural choice when the proper motivation is in place. If we first defined matrix multiplication in the style of Theorem EMP, then the above argument, followed by a simple application of the definition of matrix equality (Definition ME), would yield Theorem MRCLT.