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Section MR  Matrix Representations

From A First Course in Linear Algebra
Version 2.01
© 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 : U↦V 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

(This definition contains Notation MR.) △

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

S : P3↦M22,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 1 2 , 23 2 5 , −1−1 0 −2 , −1−4 −2 −4

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 1 2 + 37 23 2 5 + (−40) −1−1 0 −2 + 4 −1−4 −2 −4 = −90 37 −40 4 ρC S 1 + 3x + x2 + x3 = ρ C 17 37 −2057 = ρC (−72) 11 1 2 + 29 23 2 5 + (−34) −1−1 0 −2 + 3 −1−4 −2 −4 = −72 29 −34 3 ρC S −1 − 2x + 2x3 = ρ C −27−58 32 −90 = ρC 114 11 1 2 + (−46) 23 2 5 + 54 −1−1 0 −2 + (−5) −1−4 −2 −4 = 114 −46 54 −5 ρC S 2 + 3x + 2x2 − 5x3 = ρ C 48 109 −58167 = ρC (−220) 11 1 2 + 91 23 2 5 + −96 −1−1 0 −2 + 10 −1−4 −2 −4 = −220 91 −96 10

Thus, employing Definition MR

MB,CS = −90−72114−220 37 29 −46 91 −40 −34 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 0 0 , 01 0 0 , 00 1 0 , 00 0 1

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 0 0 + 8 01 0 0 + (−4) 00 1 0 + 12 00 0 1 = 3 8 −4 12 ρE S x = ρE 7 14 −822 = ρE 7 10 0 0 + 14 01 0 0 + (−8) 00 1 0 + 22 00 0 1 = 7 14 −8 22 ρE S x2 = ρ E −2−2 2 −4 = ρE (−2) 10 0 0 + (−2) 01 0 0 + 2 00 1 0 + (−4) 00 0 1 = −2 −2 2 −4 ρE S x3 = ρ E −5−11 6 −17 = ρE (−5) 10 0 0 + (−11) 01 0 0 + 6 00 1 0 + (−17) 00 0 1 = −5 −11 6 −17

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

MD,ES = 3 7 −2 −5 8 14−2−11 −4−8 2 6 1222−4−17

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 0 2

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 1−2 0 −2 = 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 0 0 = ρG 0 11 0 2 = 0 0 0 0

So we arrive at an especially economical matrix representation,

MF,GS = 2 0 00 0 −100 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 : U↦V 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 = ρC−1 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 = ρC−1 ∘ ρ C T u  Definition IVLT = ρC−1 ρ C T u  Definition LTC = ρC−1 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.

PIC
Diagram FTMR. Fundamental Theorem of Matrix Representations

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 ρC−1). So, absent some bookkeeping about vector representations, a linear transformation is a matrix. To adjust the diagram, we “reverse” the arrow on the right, which means inverting the vector representation ρC on V . Now we can go directly across the top of the diagram, computing the linear transformation between the abstract vector spaces. Or, we can around the other three sides, using vector representation, a matrix-vector product, followed by un-coordinatization.

PIC
Diagram FTMRA. Fundamental Theorem of Matrix Representations (Alternate)

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) = ρC−1 M B,CSρ B p(x) = ρC−1 M B,CSρ B 3 − x + 2x2 − 5x3 = ρC−1 M B,CS 48 −20 −1 −13 = ρC−1 −90−72114−220 37 29 −46 91 −40 −34 54 −96 4 3 −5 10 48 −20 −1 −13 = ρC−1 −134 59 −46 7 = (−134) 11 1 2 + 59 23 2 5 + (−46) −1−1 0 −2 + 7 −1−4 −2 −4 = 23 61 −3091 Again, but now with “nice” bases like D and E, and the computations are more transparent. S p(x) = ρE−1 M D,ESρ D p(x) = ρE−1 M D,ESρ D 3 − x + 2x2 − 5x3 = ρE−1 M D,ESρ D 3(1) + (−1)(x) + 2(x2) + (−5)(x3) = ρE−1 M D,ES 3 −1 2 −5 = ρE−1 3 7 −2 −5 8 14−2−11 −4−8 2 6 1222−4−17 3 −1 2 −5 = ρE−1 23 61 −30 91 = 23 10 0 0 + 61 01 0 0 + (−30) 00 1 0 + 91 00 0 1 = 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) = ρG−1 M F,GSρ F p(x) = ρG−1 M F,GSρ F 3 − x + 2x2 − 5x3 = ρG−1 M F,GSρ F 32(1 + x − x2 + 2x3) − 7(−1 + 2x + 2x3) − 17(2 + x − 2x2 + 3x3) − 2(1 + x + 2x3) = ρG−1 M F,GS 32 −7 −17 −2 = ρG−1 2 0 00 0 −100 0 0 10 0 0 00 32 −7 −17 −2 = ρG−1 64 7 −17 0 = 64 1 1 −12 + 7 −12 0 2 + (−17) 2 1 −23 + 0 11 0 2 = 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 : U↦V and S : U↦V 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 ℂn. Define u ∈ U by u = ρB−1 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 ℂn, 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 to come back to this theorem sometime, didn’t you?) ■

Theorem MRMLT
Matrix Representation of a Multiple of a Linear Transformation
Suppose that T : U↦V is a linear transformation, α ∈ ℂ, 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 ℂn. Define u ∈ U by u = ρB−1 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 ℂn, 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 : U↦V 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,DS∘T = M C,DSM B,CT
□

Proof   Let x be any vector in ℂn. Define u ∈ U by u = ρB−1 x, so x = ρB u. Then,

MB,DS∘T x = M B,DS∘T 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,DS∘T and MC,DSM B,CT have equal matrix-vector products for every vector in ℂn, 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 : ℂ2↦P 2 T a b = (−a + 3b) + (2a + 4b)x + (a − 2b)x2 S : P2↦M22 S a + bx + cx2 = 2a + b + 2c a + 4b − c −a + 3c 3a + b + 2c

and bases for ℂ2, P2 and M22 (respectively),

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

Begin by computing the new linear transformation that is the composition of T and S (Definition LTC, Theorem CLTLT), S ∘ T : ℂ2↦M 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 −9−11

Now, a representation of S,

ρD S 1 − 2x + x2 = ρ D 2−8 2 3 = ρD (−11) 1−2 1 −1 + (−21) 1−1 1 2 + 0 −12 0 0 + (17) 2−3 2 2 = −11 −21 0 17 ρD S −1 + 3x = ρD 111 1 0 = ρD 26 1−2 1 −1 + 51 1−1 1 2 + 0 −12 0 0 + (−38) 2−3 2 2 = 26 51 0 −38 ρD S 2x + 3x2 = ρ D 85 9 8 = ρD 34 1−2 1 −1 + 67 1−1 1 2 + 1 −12 0 0 + (−46) 2−3 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 −38−46

Finally, a representation of S ∘ T,

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

So we have the matrix representation of S ∘ T,

MB,DS∘T = 114 95 237 202 −9 −11 −174−149

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 −38−46 28 33 28 32 −9−11 = 114 95 237 202 −9 −11 −174−149 = MB,DS∘T

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,

PIC
Diagram MRCLT. Matrix Representation and Composition of Linear Transformations

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 : U↦V 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 ,

KT≅NMB,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 ℂn. The function ρB is defined as a function from U to ℂn, 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 ⊆ ℂn 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 = ρB−1 x. First, verify that u ∈KT,

T u = T ρB−1 x = ρC−1 M B,CT ρ B ρB−1 x  Theorem FTMR = ρC−1 M B,CT I ℂn x  Definition IVLT = ρC−1 M B,CT x  Definition IDLT = ρC−1 0 ℂn  Definition KLT = 0V  Theorem LTTZZ  Second, verify that the proposed isomorphism, ρB, takes u to x, ρB u = ρB ρB−1 x  Substitution = Iℂn 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 : M22↦P2,T ab c d = (2a−b+c−5d)+(a+4b+5b+2d)x+(3a−2b+c−8d)x2

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

B = 1 2 −1−1 , 1 3 −1−4 , 1 2 0 −2 , 2 5 −2−4 C = 1 + x + x2,2 + 3x, − 1 − 2x2

Then,

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

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 −2−26−5−17 → RREF 105 22 0 101 2 0 000

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 ρB−1 to each element of the basis,

ρB−1 −5 2 0 1 0 = (−5 2) 1 2 −1−1 + 0 1 3 −1−4 + 1 1 2 0 −2 + 0 2 5 −2−4 = −3 2−3 5 2 1 2 ρB−1 −2 −1 2 0 1 = (−2) 1 2 −1−1 + (−1 2) 1 3 −1−4 + 0 1 2 0 −2 + 1 2 5 −2−4 = −1 2−1 2 1 2 0

So the set

−3 2−3 5 2 1 2 , −1 2−1 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 : U↦V 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 ,

ℛT≅CMB,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, ℛT, is a subspace of V , while the column space of the matrix representation, CMB,CT is a subspace of ℂm. The function ρC is defined as a function from V to ℂm, but we can just as well employ the definition of ρC as a function from ℛT to CMB,CT .

We must first insure that if we choose an input for ρC from ℛT that then the output will be an element of CMB,CT . So suppose that v ∈ℛT. 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 ⊆ ℂm and we wish to find an element of the domain with y as its image. Since y ∈CMB,CT , there exists a vector, x ∈ ℂn with MB,CT x = y. We now show that the desired element of the domain is v = ρC−1 y. First, verify that v ∈ℛT by applying T to u = ρB−1 x,

T u = T ρB−1 x = ρC−1 M B,CT ρ B ρB−1 x  Theorem FTMR = ρC−1 M B,CT I ℂn x  Definition IVLT = ρC−1 M B,CT x  Definition IDLT = ρC−1 y  Definition CSM = v  Substitution  Second, verify that the proposed isomorphism, ρC, takes v to y, ρC v = ρC ρC−1 y  Substitution = Iℂm y  Definition IVLT = y  Definition IDLT

With ρC demonstrated to be an injective and surjective linear transformation from ℛT to CMB,CT , Theorem ILTIS tells us ρC is invertible, and so by Definition IVS, we say ℛT 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 : M22↦P2,T ab c d = (2a−b+c−5d)+(a+4b+5b+2d)x+(3a−2b+c−8d)x2

With bases B and C,

B = 1 2 −1−1 , 1 3 −1−4 , 1 2 0 −2 , 2 5 −2−4 C = 1 + x + x2,2 + 3x, − 1 − 2x2

we obtain the matrix representation

MB,CT = 2 −24 5 −8 0 8 0 4 −2−26−5−17

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 −248−26 5 0 −5 −8 4−17 → RREF 10 −1 0 1−25 4 0 0 0 0 0 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 ℛT by applying ρC−1 to each element of the basis,

ρC−1 1 0 −1 = (1 + x + x2) − (−1 − 2x2) = 2 + x + 3x2 ρC−1 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 ℛT. ⊠

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 : U↦V is a linear transformation, B is a basis for U and C is a basis for V . Then T is an invertible linear transfromation if and only if the matrix representation of T relative to B and C, MB,CT is an invertible matrix. When T is invertible,

MC,BT−1 = MB,CT −1
□

Proof   ( ⇐) Suppose T is invertible, so the inverse linear transformation T−1: V ↦U exists (Definition IVLT). Both linear transformations have matrix representations relative to the bases of U and V , namely MB,CT and MC,BT−1 (Definition MR). Then

MC,BT−1 MB,CT = M B,BT−1 ∘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,BT−1 = MC,CT∘T−1  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

These two equations show that MB,CT and MC,BT−1 are inverse matrices (Definition MI) and establish that when T is invertible, then MC,BT−1 = M B,CT −1.

( ⇐) Suppose now that MB,CT is an invertible matrix and hence nonsingular (Theorem NI). We compute the nullity of T,

n T = dim KT  Definition KLT = dim NMB,CT  Theorem KNSI = n MB,CT  Definition NOM = 0  Theorem RNNM

So the kernel of T is trivial, and by Theorem KILT, T is injective.

We now compute the rank of T,

r T = dim ℛT  Definition RLT = dim CMB,CT  Theorem RCSI = r MB,CT  Definition ROM = dim V  Theorem RNNM

Since the dimension of the range of T equals the dimension of the codomain V , by Theorem EDYES, ℛT = V . Which says that T is surjective by Theorem RSLT.

Because T is both injective and surjective, by Theorem ILTIS, T is invertible. ■

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. The next example shows how to apply this theorem to the problem of actually building a formula for the inverse of an invertible linear transformation.

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

R: P3↦M22,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 0 0 , 01 0 0 , 00 1 0 , 00 0 1

Then,

ρC R 1 = ρC 1 2 1 −1 = 1 2 1 −1 ρC R x = ρC 13 1 1 = 1 3 1 1 ρC R x2 = ρ C −1−2 0 2 = −1 −2 0 2 ρC R x3 = ρ C 2 3 2 −5 = 2 3 2 −5

So a representation of R is

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

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

MC,BR−1 = MB,CR −1 = 1 1−1 2 2 3−2 3 1 1 0 2 −11 2 −5 −1 = 20−7−2 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,

R−1 ab c d = ρB−1 M C,BR−1 ρC ab c d  Theorem FTMR = ρB−1 M B,CR −1ρ C ab c d  Theorem IMR = ρB−1 M B,CR −1 a b c d  Definition VR = ρB−1 20−7−2 3 −8 3 1 −1 −1 0 1 0 −6 2 1 −1 a b c d  Definition MI = ρB−1 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 Example ILTVR with a matrix representation of T.

Theorem IMILT
Invertible Matrices, Invertible Linear Transformation
Suppose that A is a square matrix of size n and T : ℂn↦ℂn 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 ℂn (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. with this observation, the proof becomes a specialization of Theorem IMR,

T  is invertible⇔MB,CT   is invertible⇔A  is invertible ■

This theorem may seem gratuitous. Why state such a special case of Theorem IMR? Because it adds another condition to our NMEx series of theorems, and in some ways it is the most fundamental expression of what it means for a matrix to be nonsingular — the associated linear transformation is invertible. This is our final update.

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 ℒSA,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 ℂn, CA = ℂn.
  8. The columns of A are a basis for ℂn.
  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 A≠0.
  12. λ = 0 is not an eigenvalue of A.
  13. The linear transformation T : ℂn↦ℂn 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 : ℂ2↦ℂ2,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 : P3↦ℂ3,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 [1653]

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

T : P2↦ℂ2,T p(x) = p(1) p(3) <