The matrix of a linear operator

Section 5.2 The matrix of a linear operator

Recall that a linear transformation

T : V \to V

is referred to as a linear operator. Recall also that two matrices

A

and

B

are similar if there exists an invertible matrix

P

such that

B = P A P^{- 1},

and that similar matrices have a lot of properties in common. In particular, if

A

is similar to

B,

then

A

and

B

have the same trace, determinant, and eigenvalues. One way to understand this is the realization that two matrices are similar if they are representations of the same operator, with respect to different bases.

🔗

Since the domain and codomain of a linear operator are the same, we can consider the matrix

M_{D B} (T)

where

B

and

D

are the same ordered basis. This leads to the next definition.

🔗

Definition 5.2.1.

Let

T : V \to V

be a linear operator, and let

B = {b_{1}, b_{2}, \dots, b_{n}}

be an ordered basis of

V .

The matrix

M_{B} (T) = M_{B B} (T)

is called the

B

-matrix of

T .

🔗

The following result collects several useful properties of the

B

-matrix of an operator. Most of these were already encountered for the matrix

M_{D B} (T)

of a transformation, although not all were stated formally.

🔗

Theorem 5.2.2.

Let

T : V \to V

be a linear operator, and let

B = {b_{1}, b_{2}, \dots, b_{n}}

be a basis for

V .

Then

$C_{B} (T (v)) = M_{B} (T) C_{B} (v)$ for all $v \in V .$

🔗
If $S : V \to V$ is another operator, then $M_{B} (S T) = M_{B} (S) M_{B} (T) .$

🔗
$T$ is an isomorphism if and only if $M_{B} (T)$ is invertible.

🔗
If $T$ is an isomorphism, then $M_{B} (T^{- 1}) = [M_{B} (T)]^{- 1} .$

🔗
$M_{B} (T) = [\begin{matrix} C_{B} (T (b_{1})) & \dots & C_{B} (T (b_{n})) \end{matrix}] .$

🔗

🔗

Example 5.2.3.

Find the

B

-matrix of the operator

T : P_{2} (R) \to P_{2} (R)

given by

T (p (x)) = p (0) (1 + x^{2}) + p (1) x,

with respect to the ordered basis

B = {1 - x, x + 3 x^{2}, 2 - x^{2}} .

🔗

Solution.

We compute

\begin{aligned} T (1 - x) & = 1 (1 + x^{2}) + 0 (x) = 1 + x^{2} \\ T (x + 3 x^{2}) & = 0 (1 + x^{2}) + 4 x = 4 x \\ T (2 - x^{2}) & = 2 (1 + x^{2}) + 1 (x) = 2 + x + 2 x^{2} . \end{aligned}

We now need to write each of these in terms of the basis

B .

We can do this by working out how to write each polynomial above in terms of

B .

Or we can be systematic.

🔗

Let

P = [\begin{matrix} 1 & 0 & 2 \\ - 1 & 1 & 0 \\ 0 & 3 & - 1 \end{matrix}]

be the matrix whose columns are given by the coefficient representations of the polynomials in

B

with respect to the standard basis

B_{0} = {1, x, x^{2}} .

For

T (1 - x) = 1 + x^{2}

we need to solve the equation

a (1 - x) + b (x + 3 x^{2}) + c (2 - x^{2}) = 1 + x^{2}

for scalars

a, b, c .

But this is equivalent to the system

\begin{aligned} a + 2 c & = 1 \\ - a + b & = 0 \\ 3 b - c & = 1, \end{aligned}

which, in turn, is equivalent to the matrix equation

[\begin{matrix} 1 & 0 & 2 \\ - 1 & 1 & 0 \\ 0 & 3 & - 1 \end{matrix}] [\begin{matrix} a \\ b \\ c \end{matrix}] = [\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}];

that is,

P C_{B} (1 + x^{2}) = C_{B_{0}} (1 + x^{2}) .

Thus,

C_{B} (1 + x^{2}) = [\begin{matrix} a \\ b \\ c \end{matrix}] = P^{- 1} C_{B_{0}} (1 + x^{2}) = P^{- 1} [\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}] .

Similarly,

C_{B} (4 x) = P^{- 1} [\begin{matrix} 0 \\ 4 \\ 0 \end{matrix}] = P^{- 1} C_{B_{0}} (4 x),

and

C_{B} (2 + x + 2 x^{2}) = P^{- 1} [\begin{matrix} 2 \\ 1 \\ 2 \end{matrix}] = P^{- 1} C_{B_{0}} (2 + x + 2 x^{2}) .

Using the computer, we find:

🔗

xxxxxxxxxx
 
import sympy as sy
sy.init_printing()
P = sy.Matrix(3,3,[1,0,2,-1,1,0,0,3,-1])
M = sy.Matrix(3,3,[1,0,2,0,4,1,1,0,2])
P**-1, P**-1*M

That is,

\begin{aligned} M_{B} (T) & = [\begin{array}{c} C_{B} (T (1 - x)) & C_{B} (T (x + 3 x^{2})) & C_{B} (T (2 - x^{2})) \end{array}] \\ = [\begin{array}{c} P^{- 1} C_{B_{0}} T (1 - x) & P^{- 1} C_{B_{0}} T (x + 3 x^{2}) & P^{- 1} C_{B_{0}} (T (2 - x^{2})) \end{array}] \\ = P^{- 1} [\begin{array}{c} C_{B_{0}} (1 + x^{2}) & C_{B_{0}} (4 x) & C_{B_{0}} (2 + x + 2 x^{2}) \end{array}] \\ = [\begin{array}{c} 1 / 7 & - 6 / 7 & 3 / 7 \\ 1 / 7 & 1 / 7 & 2 / 7 \\ 3 / 7 & 3 / 7 & - 1 / 7 \end{array}] [\begin{array}{c} 1 & 0 & 2 \\ 0 & 4 & 1 \\ 1 & 0 & 2 \end{array}] \\ = [\begin{array}{c} 3 / 7 & - 24 / 7 & 0 \\ 3 / 7 & 4 / 7 & 1 \\ 2 / 7 & 12 / 7 & 1 \end{array}] . \end{aligned}

🔗

Let’s confirm that this works. Suppose we have

\begin{aligned} p (x) & = C_{B}^{- 1} [\begin{array}{c} a \\ b \\ c \end{array}] = a (1 - x) + b (x + 3 x^{2}) + c (2 - x^{2}) \\ = (a + 2 c) + (- a + b) x + (3 b - c) x^{2} . \end{aligned}

Then

T (p (x)) = (a + 2 c) (1 + x^{2}) + (4 b + c) x,

and we find

C_{B} (T (p (x))) = P^{- 1} [\begin{matrix} a + 2 c \\ 4 b + c \\ a + 2 c \end{matrix}] = [\begin{matrix} \frac{3}{7} a - \frac{24}{7} b \\ \frac{3}{7} a + \frac{4}{7} b + c \\ \frac{2}{7} a + \frac{12}{7} b + c \end{matrix}] .

On the other hand,

M_{B} (T) = [\begin{matrix} 3 / 7 & - 24 / 7 & 0 \\ 3 / 7 & 4 / 7 & 1 \\ 2 / 7 & 12 / 7 & 1 \end{matrix}] [\begin{matrix} a \\ b \\ c \end{matrix}] = [\begin{matrix} \frac{3}{7} a - \frac{24}{7} b \\ \frac{3}{7} a + \frac{4}{7} b + c \\ \frac{2}{7} a + \frac{12}{7} b + c \end{matrix}] .

The results agree, but possibly leave us a little confused.

🔗

In general, given an ordered basis

B = {b_{1}, b_{2}, \dots, b_{n}}

for a vector space

V

with standard basis

B_{0} = {e_{1}, e_{2}, \dots, e_{n}},

if we let

P = [\begin{matrix} C_{B_{0}} (b_{1}) & \dots & C_{B_{0}} (b_{n}) \end{matrix}],

then

\begin{aligned} M_{B} (T) & = [\begin{array}{c} C_{B} (T (b_{1})) & \dots & C_{B} (T (b_{n}) \end{array}] \\ = P^{- 1} [\begin{array}{c} C_{B_{0}} (T (b_{1})) & \dots & C_{B_{0}} (T (b_{n}) \end{array}], \end{aligned}

since multiplying by

P^{- 1}

converts vectors written in terms of

B_{0}

to vectors written in terms of

B .

🔗

As we saw above, this gives us the result, but doesn’t shed much light on the problem, unless we have an easy way to write vectors in terms of the basis

B .

Let’s revisit the problem. Instead of using the given basis

B,

let’s use the standard basis

B_{0} = {1, x, x^{2}} .

We quickly find

T (1) = 1 + x + x^{2}, T (x) = x, and T (x^{2}) = x,

so with respect to the standard basis,

M_{B_{0}} (T) = [\begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 0 & 0 \end{matrix}] .

Now, recall that

M_{B} (T) = P^{- 1} [\begin{matrix} C_{B_{0}} T (1 - x) & C_{B_{0}} (T (x + 3 x^{2}) & C_{B_{0}} (T (2 - x^{2})) \end{matrix}]

and note that for any polynomial

p (x),

C_{B_{0}} (T (p (x))) = M_{B_{0}} (T) C_{B_{0}} (p (x)) .

But

[\begin{matrix} C_{B_{0}} (1 - x) & C_{B_{0}} (x + 3 x^{2}) & C_{B_{0}} (2 - x^{2}) \end{matrix}] = P,

so we get

\begin{aligned} M_{B} (T) & = P^{- 1} [\begin{array}{c} C_{B_{0}} T (1 - x) & C_{B_{0}} (T (x + 3 x^{2}) & C_{B_{0}} (T (2 - x^{2})) \end{array}] \\ = P^{- 1} [\begin{array}{c} M_{B_{0}} (T) C_{B_{0}} (1 - x) & M_{B_{0}} (T) C_{B_{0}} (x + 3 x^{2}) & M_{B_{0}} (T) C_{B_{0}} (2 - x^{2}) \end{array}] \\ = P^{- 1} M_{B_{0}} (T) [\begin{array}{c} C_{B_{0}} (1 - x) & C_{B_{0}} (x + 3 x^{2}) & C_{B_{0}} (2 - x^{2}) \end{array}] \\ = P^{- 1} M_{B_{0}} (T) P . \end{aligned}

🔗

Now we have a much more efficient method for arriving at the matrix

M_{B} (T) .

The matrix

M_{B_{0}} (T)

is easy to determine, the matrix

P

is easy to determine, and with the help of the computer, it’s easy to compute

P^{- 1} M_{B_{0}} P = M_{B} (T) .

🔗

xxxxxxxxxx
 
M0 = sy.Matrix(3,3,[1,0,0,1,1,1,1,0,0])
P**-1*M0*P

Exercise 5.2.4.

Determine the matrix of the operator

T : R^{3} \to R^{3}

given by

T (x, y, z) = (3 x - 2 y + 4 z, x - 5 y, 2 y - 7 z)

with respect to the ordered basis

B = {(1, 2, 0), (0, - 1, 2), (1, 2, 1)} .

(You may want to use the Sage cell below for computational assistance.)

🔗

xxxxxxxxxx
 
​

The matrix

P

used in the above examples is known as a change matrix. If the columns of

P

are the coefficient vectors of

B = {b_{1}, b_{2}, \dots, b_{n}}

with respect to another basis

D,

then we have

\begin{aligned} P & = [\begin{array}{c} C_{D} (b_{1}) & \dots & C_{D} (b_{n}) \end{array}] \\ = [\begin{array}{c} C_{D} (1_{V} (b_{1})) & \dots & C_{D} (1_{V} (b_{n})) \end{array}] \\ = M_{D B} (1_{V}) . \end{aligned}

In other words,

P

is the matrix of the identity transformation

1_{V} : V \to V,

where we use the basis

B

for the domain, and the basis

D

for the codomain.

🔗

Definition 5.2.5.

The change matrix with respect to ordered bases

B, D

V

is denoted

P_{D \leftarrow B},

and defined by

P_{D \leftarrow B} = M_{D B} (1_{V}) .

🔗

Theorem 5.2.6.

Let

B = {b_{1}, b_{2}, \dots, b_{n}}

and

D

be two ordered bases of

V .

Then

P_{D \leftarrow B} = [\begin{matrix} C_{D} (b_{1}) & \dots & C_{D} (b_{n}) \end{matrix}],

and satisfies

C_{D} (v) = P_{D \leftarrow B} C_{B} (v)

for all

v \in V .

🔗

The matrix

P_{D \leftarrow B}

is invertible, and

(P_{D \leftarrow B})^{- 1} = P_{B \leftarrow D} .

Moreover, if

E

is a third ordered basis, then

P_{E \leftarrow D} P_{D \leftarrow B} = P_{E \leftarrow B} .

🔗

Exercise 5.2.7.

Prove Theorem 5.2.6.

🔗

Hint.

The identity operator does nothing. Convince yourself

M_{D B} (1_{V})

amounts to taking the vectors in

B

and writing them in terms of the vectors in

D .

🔗

Example 5.2.8.

Let

B = {1, x, x^{2}}

and let

D = {1 + x, x + x^{2}, 2 - 3 x + x^{2}}

be ordered bases of

P_{2} (R) .

Find the change matrix

P_{D \leftarrow B} .

🔗

Solution.

Finding this matrix requires us to first write the vectors in

B

in terms of the vectors in

D .

However, it’s much easier to do this the other way around. We easily find

P_{B \leftarrow D} = [\begin{matrix} 1 & 0 & 2 \\ 1 & 1 & - 3 \\ 0 & 1 & 1 \end{matrix}],

and by Theorem 5.2.6, we have

P_{D \leftarrow B} = (P_{B \leftarrow D})^{- 1} = \frac{1}{6} [\begin{matrix} 4 & 2 & - 2 \\ - 1 & 1 & 5 \\ 1 & - 1 & 1 \end{matrix}] .

🔗

Note that the change matrix notation is useful for linear transformations between different vector spaces as well. Recall Theorem 5.1.6, which gave the result

M_{D_{0} B_{0}} (T) = Q M_{D_{1} B_{1}} P^{- 1},

where (using our new notation)

P = P_{B_{0} \leftarrow B_{1}}

and

Q = P_{D_{0} \leftarrow D_{1}} .

In this notation, we have

M_{D_{0} B_{0}} (T) = P_{D_{0} \leftarrow D_{1}} M_{D_{1} B_{1}} (T) P_{B_{1} \leftarrow B_{0}},

which seems more intiutive.

🔗

The above results give a straightforward procedure for determining the matrix of any operator, with respect to any basis, if we let

D

be the standard basis. The importance of these results is not just their computational simplicity, however. The most important outcome of the above is that if

M_{B} (T)

and

M_{D} (T)

give the matrix of

T

with respect to two different bases, then

M_{B} (T) = (P_{D \leftarrow B})^{- 1} M_{D} (T) P_{D \leftarrow B},

so that the two matrices are similar.

🔗

Recall from Theorem 4.1.10 that similar matrices have the same determinant, trace, and eigenvalues. This means that we can unambiguously define the determinant and trace of an operator, and that we can compute eigenvalues of an operator using any matrix representation of that operator.

🔗

Exercises Exercises

1.

Let

{\vec{b}}_{1} = [\begin{matrix} 1 \\ 1 \end{matrix}]

and

{\vec{b}}_{2} = [\begin{matrix} - 1 \\ - 2 \end{matrix}] .

The set

B = {{\vec{b}}_{1}, {\vec{b}}_{2}}

is a basis for

R^{2} .

🔗

Let

T : R^{2} \to R^{2}

be a linear operator such that

T ({\vec{b}}_{1}) = 8 {\vec{b}}_{1} + 3 {\vec{b}}_{2}

and

T ({\vec{b}}_{2}) = 7 {\vec{b}}_{1} + 3 {\vec{b}}_{2} .

🔗

Find the matrix $M_{B} (T)$ of $T$ relative to the basis $B .$
🔗

🔗
Find the matrix $M_{E} (T)$ of $T$ relative to the standard basis $E$ for $R^{2} .$
🔗

🔗

🔗

2.

Let

f : R^{2} \to R^{2}

be the linear operator defined by

f (\vec{x}) = [\begin{array}{cc} - 5 & 0 \\ 2 & 5 \end{array}] \vec{x} .

Let

\begin{array}{lcl} B & = & {⟨ 1, - 2 ⟩, ⟨ - 3, 5 ⟩}, \\ C & = & {⟨ - 1, - 2 ⟩, ⟨ - 1, - 1 ⟩}, \end{array}

be two different bases for

R^{2} .

🔗

(a)

Find the matrix

M_{B} (f)

for

f

relative to the basis

B .

🔗

(b)

Find the matrix

M_{C} (f)

for

f

relative to the basis

C .

🔗

(c)

Find the transition matrix

P_{B \leftarrow C}

such that

C_{B} (v) = P_{B \leftarrow C} C_{C} (v) .

🔗

(d)

Find the transition matrix

P_{C \leftarrow B}

such that

C_{C} (v) = P_{C \leftarrow B} C_{B} (v) .

🔗

Reminder:

P_{C \leftarrow B} = P_{B \leftarrow C}^{- 1}

🔗

On paper, confirm that

P_{C \leftarrow B} M_{B} (f) P_{B \leftarrow C} = M_{C} (f) .

🔗

You have attempted 1 of 5 activities on this page.

🔗

Prev Top Next