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2.2 The Matrix Representation of a Linear Transformation

Until now, we have studied linear transformations by examining their ranges and null spaces. In this section, we embark on one of the most useful approaches to the analysis of a linear transformation on a finite-dimensional vector space: the representation of a linear transformation by a matrix. In fact, we develop a one-to-one correspondence between matrices and linear transformations that allows us to utilize properties of one to study properties of the other.

We first need the concept of an ordered basis for a vector space.

Definitions.

Let V be a finite-dimensional vector space. An ordered basis for V is a basis for V endowed with a specific order; that is, an ordered basis for V is a finite sequence of linearly independent vectors in V that generates V.

Example 1

In $F^{3}, β = {e_{1}, e_{2}, e_{3}}$ $F^{3}, β = {e_{1}, e_{2}, e_{3}}$ can be considered an ordered basis. Also $γ = {e_{2}, e_{1}, e_{3}}$ $γ = {e_{2}, e_{1}, e_{3}}$ is an ordered basis, but $β \neq γ$ $β \neq γ$ as ordered bases.

For the vector space $F^{n}$ $F^{n}$ , we call ${e_{1}, e_{2}, \dots, e_{n}}$ ${e_{1}, e_{2}, \dots, e_{n}}$ the standard ordered basis for $F^{n}$ $F^{n}$ . Similarly, for the vector space $P_{n} (F)$ $P_{n} (F)$ , we call ${1, x, \dots, x^{n}}$ ${1, x, \dots, x^{n}}$ the standard ordered basis for $P_{n} (F)$ $P_{n} (F)$ .

Now that we have the concept of ordered basis, we can identify abstract vectors in an n-dimensional vector space with n-tuples. This identification is provided through the use of coordinate vectors, as introduced next.

Definitions.

Let $β = {u_{1}, u_{2}, \dots, u_{n}}$ $β = {u_{1}, u_{2}, \dots, u_{n}}$ be an ordered basis for a finite-dimensional vector space V. For $x \in V$ $x \in V$ , let $a_{1}, a_{2}, \dots, a_{n}$ $a_{1}, a_{2}, \dots, a_{n}$ be the unique scalars such that

x = \sum_{i = 1}^{n} a_{i} u_{i} .

$x = \sum_{i = 1}^{n} a_{i} u_{i} .$

We define the coordinate vector of x relative to $β$ $β$ , denoted ${[x]}_{β}$ ${[x]}_{β}$ , by

{[x]}_{β} = (\begin{matrix} a_{1} \\ a_{2} \\ ⋮ \\ a_{n} \end{matrix}) .

${[x]}_{β} = (\begin{matrix} a_{1} \\ a_{2} \\ ⋮ \\ a_{n} \end{matrix}) .$

Notice that ${[u_{i}]}_{β} = e_{i}$ ${[u_{i}]}_{β} = e_{i}$ in the preceding definition. It is left as an exercise to show that the correspondence $x \to {[x]}_{β}$ $x \to {[x]}_{β}$ provides us with a linear transformation from V to $F^{n}$ $F^{n}$ . We study this transformation in Section 2.4 in more detail.

Example 2

Let $V = P_{2} (R)$ $V = P_{2} (R)$ , and let $β = {1, x, x^{2}}$ $β = {1, x, x^{2}}$ be the standard ordered basis for V. If $f (x) = 4 + 6 x - 7 x^{2}$ $f (x) = 4 + 6 x - 7 x^{2}$ , then

{[f]}_{β} = (\begin{matrix} 4 \\ 6 \\ - 7 \end{matrix}) .

${[f]}_{β} = (\begin{matrix} 4 \\ 6 \\ - 7 \end{matrix}) .$

Let us now proceed with the promised matrix representation of a linear transformation. Suppose that V and W are finite-dimensional vector spaces with ordered bases $β = {v_{1}, v_{2}, \dots, v_{n}}$ $β = {v_{1}, v_{2}, \dots, v_{n}}$ and $γ = {w_{1}, w_{2}, \dots, w_{m}}$ $γ = {w_{1}, w_{2}, \dots, w_{m}}$ , respectively. Let $T : V \to W$ $T : V \to W$ be linear. Then for each $j = 1, 2, \dots, n$ $j = 1, 2, \dots, n$ there exist unique scalars $a_{i j} \in F, i = 1, 2, \dots, m$ $a_{i j} \in F, i = 1, 2, \dots, m$ such that

T (v_{j}) = \sum_{i = 1}^{m} a_{i j} w_{i} for j = 1, 2, \dots, n .

$T (v_{j}) = \sum_{i = 1}^{m} a_{i j} w_{i} for j = 1, 2, \dots, n .$

Definitions.

Using the notation above, we call the $m \times n$ $m \times n$ matrix A defined by $A_{i j} = a_{i j}$ $A_{i j} = a_{i j}$ the matrix representation of T in the ordered bases $β$ $β$ and $γ$ $γ$ and write $A = [T]_{β}^{γ}$ $A = [T]_{β}^{γ}$ . If $V = W$ $V = W$ and $β = γ$ $β = γ$ , then we write $A = {[T]}_{β}$ $A = {[T]}_{β}$ .

Notice that the jth column of A is simply ${[T (v_{j})]}_{γ}$ ${[T (v_{j})]}_{γ}$ . Also observe that if $U : V \to W$ $U : V \to W$ is a linear transformation such that $[U]_{β}^{γ} = [T]_{β}^{γ}$ $[U]_{β}^{γ} = [T]_{β}^{γ}$ , then $U = T$ $U = T$ by the corollary to Theorem 2.6 (p. 73).

We illustrate the computation of $[T]_{β}^{γ}$ $[T]_{β}^{γ}$ in the next several examples.

Example 3

Let $T : R^{2} \to R^{3}$ $T : R^{2} \to R^{3}$ be the linear transformation defined by

T (a_{1}, a_{2}) = (a_{1} + 3 a_{2}, 0, 2 a_{1} - 4 a_{2}) .

$T (a_{1}, a_{2}) = (a_{1} + 3 a_{2}, 0, 2 a_{1} - 4 a_{2}) .$

Let $β$ $β$ and $γ$ $γ$ be the standard ordered bases for $R^{2}$ $R^{2}$ and $R^{3}$ $R^{3}$ , respectively. Now

T (1, 0) = (1, 0, 2) = 1 e_{1} + 0 e_{2} + 2 e_{3}

$T (1, 0) = (1, 0, 2) = 1 e_{1} + 0 e_{2} + 2 e_{3}$

and

T (0, 1) = (3, 0, - 4) = 3 e_{1} + 0 e_{2} - 4 e_{3} .

$T (0, 1) = (3, 0, - 4) = 3 e_{1} + 0 e_{2} - 4 e_{3} .$

Hence

[T]_{β}^{γ} = (\begin{matrix} 1 & 3 \\ 0 & 0 \\ 2 & - 4 \end{matrix}) .

$[T]_{β}^{γ} = (\begin{matrix} 1 & 3 \\ 0 & 0 \\ 2 & - 4 \end{matrix}) .$

If we let $γ^{'} = {e_{3}, e_{2}, e_{1}}$ $γ^{'} = {e_{3}, e_{2}, e_{1}}$ , then

[T]_{β}^{γ'} = (\begin{matrix} 2 & - 4 \\ 0 & 0 \\ 1 & 3 \end{matrix}) .

$[T]_{β}^{γ'} = (\begin{matrix} 2 & - 4 \\ 0 & 0 \\ 1 & 3 \end{matrix}) .$

Example 4

Let $T : P_{3} (R) \to P_{2} (R)$ $T : P_{3} (R) \to P_{2} (R)$ be the linear transformation defined by $T(f(x)) = f^{'} (x)$ $T(f(x)) = f^{'} (x)$ . Let $β$ $β$ and $γ$ $γ$ be the standard ordered bases for $P_{3} (R)$ $P_{3} (R)$ and $P_{2} (R)$ $P_{2} (R)$ , respectively. Then

\begin{matrix} T (1) & = & 0 \cdot 1 & + & 0 \cdot x & + & 0 \cdot x^{2} \\ T (x) & = & 1 \cdot 1 & + & 0 \cdot x & + & 0 \cdot x^{2} \\ T (x^{2}) & = & 0 \cdot 1 & + & 2 \cdot x & + & 0 \cdot x^{2} \\ T (x^{3}) & = & 0 \cdot 1 & + & 0 \cdot x & + & 3 \cdot x^{2} . \end{matrix}

$\begin{matrix} T (1) & = & 0 \cdot 1 & + & 0 \cdot x & + & 0 \cdot x^{2} \\ T (x) & = & 1 \cdot 1 & + & 0 \cdot x & + & 0 \cdot x^{2} \\ T (x^{2}) & = & 0 \cdot 1 & + & 2 \cdot x & + & 0 \cdot x^{2} \\ T (x^{3}) & = & 0 \cdot 1 & + & 0 \cdot x & + & 3 \cdot x^{2} . \end{matrix}$

[T]_{β}^{γ} = (\begin{matrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{matrix}) .

$[T]_{β}^{γ} = (\begin{matrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{matrix}) .$

Note that when $T (x^{j})$ $T (x^{j})$ is written as a linear combination of the vectors of $γ$ $γ$ , its coefficients give the entries of column $j + 1$ $j + 1$ of $[T]_{β}^{γ}$ $[T]_{β}^{γ}$ .

Let V and W be finite-dimensional vector spaces with ordered bases $β = {v_{1}, v_{2}, \dots, v_{n}}$ $β = {v_{1}, v_{2}, \dots, v_{n}}$ and $γ = {w_{1}, w_{2}, \dots, w_{m}}$ $γ = {w_{1}, w_{2}, \dots, w_{m}}$ , respectively. Then

T_{0} (v_{j}) = 0 = 0 w_{1} + 0 w_{2} + \dots + 0 w_{m} .

$T_{0} (v_{j}) = 0 = 0 w_{1} + 0 w_{2} + \dots + 0 w_{m} .$

Hence ${[T_{0}]}_{β}^{γ} = O$ ${[T_{0}]}_{β}^{γ} = O$ , the $m \times n$ $m \times n$ zero matrix. Also,

I_{V} (v_{j}) = v_{j} = 0 v_{1} + 0 v_{2} + \dots + 0 v_{j - 1} + 1 v_{j} + 0 v_{j + 1} + \dots + 0 v_{n} .

$I_{V} (v_{j}) = v_{j} = 0 v_{1} + 0 v_{2} + \dots + 0 v_{j - 1} + 1 v_{j} + 0 v_{j + 1} + \dots + 0 v_{n} .$

Hence the jth column of $I_{V}$ $I_{V}$ is $e_{j}$ $e_{j}$ , that is,

{[I_{V}]}_{β} = (\begin{matrix} 1 & 0 & \dots & 0 \\ 0 & 1 & \dots & 0 \\ ⋮ & ⋮ & ⋮ \\ 0 & 0 & \dots & 1 \end{matrix}) .

${[I_{V}]}_{β} = (\begin{matrix} 1 & 0 & \dots & 0 \\ 0 & 1 & \dots & 0 \\ ⋮ & ⋮ & ⋮ \\ 0 & 0 & \dots & 1 \end{matrix}) .$

The preceding matrix is called the $n \times n$ $n \times n$ identity matrix and is defined next, along with a very useful notation, the Kronecker delta.

Definitions.

We define the Kronecker delta $δ_{i j}$ $δ_{i j}$ by $δ_{i j} = 1$ $δ_{i j} = 1$ if $i = j$ $i = j$ and $δ_{i j} = 0$ $δ_{i j} = 0$ if $i \neq j$ $i \neq j$ . The $n \times n$ $n \times n$ identity matrix $I_{n}$ $I_{n}$ is defined by ${(I_{n})}_{i j} = δ_{i j}$ ${(I_{n})}_{i j} = δ_{i j}$ . When the context is clear, we sometimes omit the subscript n from $I_{n}$ $I_{n}$ .

For example,

I_{1} = (1), I_{2} = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}), and I_{3} = (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}) .

$I_{1} = (1), I_{2} = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}), and I_{3} = (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}) .$

Thus, the matrix representation of a zero transformation is a zero matrix, and the matrix representation of an identity transformation is an identity matrix.

Now that we have defined a procedure for associating matrices with linear transformations, we show in Theorem 2.8 that this association “preserves” addition and scalar multiplication. To make this more explicit, we need some preliminary discussion about the addition and scalar multiplication of linear transformations.

Definitions.

Let $T, U : V \to W$ $T, U : V \to W$ be arbitrary functions, where V and W are vector spaces over F, and let $a \in F$ $a \in F$ . We define $T + U : V \to W$ $T + U : V \to W$ by $(T + U) (x) = T (x) + U (x)$ $(T + U) (x) = T (x) + U (x)$ for all $x \in V$ $x \in V$ , and $a T : V \to W$ $a T : V \to W$ by $(a T) (x) = a T (x)$ $(a T) (x) = a T (x)$ for all $x \in V$ $x \in V$ .

Of course, these are just the usual definitions of addition and scalar multiplication of functions. We are fortunate, however, to have the result that both sums and scalar multiples of linear transformations are also linear.

Theorem 2.7

Let V and W be vector spaces over a field F, and let $T, U : V \to W$ $T, U : V \to W$ be linear.

(a) For all $a \in F, a T + U$ $a \in F, a T + U$ is linear.
(b) Using the operations of addition and scalar multiplication in the preceding definition, the collection of all linear transformations from V to W is a vector space over F.

Proof.

(a) Let $x, y \in V$ $x, y \in V$ and $c \in F$ $c \in F$ . Then

\begin{matrix} (a T + U) (c x + y) & = & a T (c x + y) + U (c x + y) \\ = & a [T (c x + y)] + c U (x) + U (y) \\ = & a [c T (x) + T (y)] + c U (x) + U (y) \\ = & a c T (x) + c U (x) + a T (y) + U (y) \\ = & c (a T + U) (x) + (a T + U) (y) . \end{matrix}

$\begin{matrix} (a T + U) (c x + y) & = & a T (c x + y) + U (c x + y) \\ = & a [T (c x + y)] + c U (x) + U (y) \\ = & a [c T (x) + T (y)] + c U (x) + U (y) \\ = & a c T (x) + c U (x) + a T (y) + U (y) \\ = & c (a T + U) (x) + (a T + U) (y) . \end{matrix}$

So $a T + U$ $a T + U$ is linear.

(b) Noting that $T_{0}$ $T_{0}$ , the zero transformation, plays the role of the zero vector, it is easy to verify that the axioms of a vector space are satisfied, and hence that the collection of all linear transformations from V into W is a vector space over F.

Definitions.

Let V and W be vector spaces over F. We denote the vector space of all linear transformations from V into W by L(V, M). In the case that $V = W$ $V = W$ , we write L(V) instead of L(V, V).

In Section 2.4, we see a complete identification of L(V, W) with the vector space $M_{m \times n} (F)$ $M_{m \times n} (F)$ , where n and m are the dimensions of V and W, respectively. This identification is easily established by the use of the next theorem.

Theorem 2.8.

Let V and W be finite-dimensional vector spaces with ordered bases $β$ $β$ and $γ$ $γ$ , respectively, and let $T, U : V \to W$ $T, U : V \to W$ be linear transformations. Then

(a) ${[T + U]}_{β}^{γ} = {[T]}_{β}^{γ} + {[U]}_{β}^{γ}$ ${[T + U]}_{β}^{γ} = {[T]}_{β}^{γ} + {[U]}_{β}^{γ}$ and
(b) ${[a T]}_{β}^{γ} = a {[T]}_{β}^{γ}$ ${[a T]}_{β}^{γ} = a {[T]}_{β}^{γ}$ for all scalars a.

Proof.

Let $β = {v_{1}, v_{2}, \dots, v_{n}}$ $β = {v_{1}, v_{2}, \dots, v_{n}}$ and $γ = {w_{1}, w_{2}, \dots, w_{m}}$ $γ = {w_{1}, w_{2}, \dots, w_{m}}$ . There exist unique scalars $a_{i j}$ $a_{i j}$ and $b_{i j} (1 \leq i \leq m, 1 \leq j \leq n)$ $b_{i j} (1 \leq i \leq m, 1 \leq j \leq n)$ such that

T (v_{j}) = \sum_{i = 1}^{m} a_{i j} w_{i} and     U (v_{j}) = \sum_{i = 1}^{m} b_{i j} w_{i} for 1 \leq j \leq n .

$T (v_{j}) = \sum_{i = 1}^{m} a_{i j} w_{i} and U (v_{j}) = \sum_{i = 1}^{m} b_{i j} w_{i} for 1 \leq j \leq n .$

Hence

(T + U) (v_{j}) = \sum_{i = 1}^{m} (a_{i j} + b_{i j}) w_{i} .

$(T + U) (v_{j}) = \sum_{i = 1}^{m} (a_{i j} + b_{i j}) w_{i} .$

Thus

{({[T + U]}_{β}^{γ})}_{i j} = a_{i j} + b_{i j} = {({[T]}_{β}^{γ} + {[U]}_{β}^{γ})}_{i j} .

${({[T + U]}_{β}^{γ})}_{i j} = a_{i j} + b_{i j} = {({[T]}_{β}^{γ} + {[U]}_{β}^{γ})}_{i j} .$

So (a) is proved, and the proof of (b) is similar.

Example 5

Let $T : R^{2} \to R^{3}$ $T : R^{2} \to R^{3}$ and $U : R^{2} \to R^{3}$ $U : R^{2} \to R^{3}$ be the linear transformations respectively defined by

T (a_{1}, a_{2}) = (a_{1} + 3 a_{2}, 0, 2 a_{1} - 4 a_{2}) and U (a_{1}, a_{2}) = (a_{1} - a_{2}, 2 a_{1}, 3 a_{1} + 2 a_{2}) .

$T (a_{1}, a_{2}) = (a_{1} + 3 a_{2}, 0, 2 a_{1} - 4 a_{2}) and U (a_{1}, a_{2}) = (a_{1} - a_{2}, 2 a_{1}, 3 a_{1} + 2 a_{2}) .$

Let $β$ $β$ and $γ$ $γ$ be the standard ordered bases of $R^{2}$ $R^{2}$ and $R^{3}$ $R^{3}$ , respectively. Then

{[T]}_{β}^{γ} = (\begin{matrix} 1 & 3 \\ 0 & 0 \\ 2 & - 4 \end{matrix}),

${[T]}_{β}^{γ} = (\begin{matrix} 1 & 3 \\ 0 & 0 \\ 2 & - 4 \end{matrix}),$

(as computed in Example 3), and

{[U]}_{β}^{γ} = (\begin{matrix} 1 & - 1 \\ 2 & 0 \\ 3 & 2 \end{matrix}) .

${[U]}_{β}^{γ} = (\begin{matrix} 1 & - 1 \\ 2 & 0 \\ 3 & 2 \end{matrix}) .$

If we compute $T + U$ $T + U$ using the preceding definitions, we obtain

(T + U) (a_{1}, a_{2}) = (2 a_{1} + 2 a_{2}, 2 a_{1}, 5 a_{1} - 2 a_{2}) .

$(T + U) (a_{1}, a_{2}) = (2 a_{1} + 2 a_{2}, 2 a_{1}, 5 a_{1} - 2 a_{2}) .$

{[T + U]}_{β}^{γ} = (\begin{matrix} 2 & 2 \\ 2 & 0 \\ 5 & - 2 \end{matrix}),

${[T + U]}_{β}^{γ} = (\begin{matrix} 2 & 2 \\ 2 & 0 \\ 5 & - 2 \end{matrix}),$

which is simply ${[T]}_{β}^{γ} + {[U]}_{β}^{γ}$ ${[T]}_{β}^{γ} + {[U]}_{β}^{γ}$ , illustrating Theorem 2.8.

Exercises

Label the following statements as true or false. Assume that V and W are finite-dimensional vector spaces with ordered bases $β$ $β$ and $γ$ $γ$ , respectively, and $T, U : V \to W$ $T, U : V \to W$ are linear transformations.
1. (a) For any scalar $a, a T + U$ $a, a T + U$ is a linear transformation from V to W.
2. (b) ${[T]}_{β}^{γ} = {[U]}_{β}^{γ}$ ${[T]}_{β}^{γ} = {[U]}_{β}^{γ}$ implies that $T = U$ $T = U$ .
3. (c) If $m = dim (V)$ $m = dim (V)$ and $n = \dim (W)$ $n = \dim (W)$ , then ${[T]}_{β}^{γ}$ ${[T]}_{β}^{γ}$ is an $m \times n$ $m \times n$ matrix.
4. (d) ${[T + U]}_{β}^{γ} = {[T]}_{β}^{γ} + {[U]}_{β}^{γ}$ ${[T + U]}_{β}^{γ} = {[T]}_{β}^{γ} + {[U]}_{β}^{γ}$ .
5. (e) $L (V, W)$ $L (V, W)$ is a vector space.
6. (f) $L (V, W) = L (W, V)$ $L (V, W) = L (W, V)$ .
Let $β$ $β$ and $γ$ $γ$ be the standard ordered bases for $R^{n}$ $R^{n}$ and $R^{m}$ $R^{m}$ , respectively. For each linear transformation $T : R^{n} \to R^{m}$ $T : R^{n} \to R^{m}$ , compute ${[T]}_{β}^{γ}$ ${[T]}_{β}^{γ}$ .
1. (a) $T : R^{2} \to R^{3}$ $T : R^{2} \to R^{3}$ defined by $T (a_{1}, a_{2}) = (2 a_{1} - a_{2}, 3 a_{1} + 4 a_{2}, a_{1})$ $T (a_{1}, a_{2}) = (2 a_{1} - a_{2}, 3 a_{1} + 4 a_{2}, a_{1})$ .
2. (b) $T : R^{3} \to R^{2}$ $T : R^{3} \to R^{2}$ defined by $T (a_{1}, a_{2}, a_{3}) = (2 a_{1} + 3 a_{2} - a_{3}, a_{1} + a_{3})$ $T (a_{1}, a_{2}, a_{3}) = (2 a_{1} + 3 a_{2} - a_{3}, a_{1} + a_{3})$ .
3. (c) $T : R^{3} \to R$ $T : R^{3} \to R$ defined by $T (a_{1}, a_{2}, a_{3}) = 2 a_{1} + a_{2} - 3 a_{3}$ $T (a_{1}, a_{2}, a_{3}) = 2 a_{1} + a_{2} - 3 a_{3}$ .
4. (d) $T : R^{3} \to R^{3}$ $T : R^{3} \to R^{3}$ defined by
  
  $T (a_{1}, a_{2}, a_{3}) = (2 a_{2} + a_{3}, - a_{1} + 4 a_{2} + 5 a_{3}, a_{1} + a_{3}) .$ $T (a_{1}, a_{2}, a_{3}) = (2 a_{2} + a_{3}, - a_{1} + 4 a_{2} + 5 a_{3}, a_{1} + a_{3}) .$
5. (e) $T : R^{n} \to R^{n}$ $T : R^{n} \to R^{n}$ defined by $T (a_{1}, a_{2}, \dots, a_{n}) = (a_{1}, a_{1}, \dots, a_{1})$ $T (a_{1}, a_{2}, \dots, a_{n}) = (a_{1}, a_{1}, \dots, a_{1})$ .
6. (f) $T : R^{n} \to R^{n}$ $T : R^{n} \to R^{n}$ defined by $T (a_{1}, a_{2}, \dots, a_{n}) = (a_{n}, a_{n - 1}, \dots, a_{1})$ $T (a_{1}, a_{2}, \dots, a_{n}) = (a_{n}, a_{n - 1}, \dots, a_{1})$ .
7. (g) $T : R^{n} \to R$ $T : R^{n} \to R$ defined by $T (a_{1}, a_{2}, \dots, a_{n}) = a_{1} + a_{n}$ $T (a_{1}, a_{2}, \dots, a_{n}) = a_{1} + a_{n}$ .
Let $T : R^{2} \to R^{3}$ $T : R^{2} \to R^{3}$ be defined by $T (a_{1}, a_{2}) = (a_{1} - a_{2}, a_{1}, 2 a_{1} + a_{2})$ $T (a_{1}, a_{2}) = (a_{1} - a_{2}, a_{1}, 2 a_{1} + a_{2})$ . Let $β$ $β$ be the standard ordered basis for $R^{2}$ $R^{2}$ and $γ = {(1, 1, 0), (0, 1, 1), (2, 2, 3)}$ $γ = {(1, 1, 0), (0, 1, 1), (2, 2, 3)}$ . Compute ${[T]}_{β}^{γ}$ ${[T]}_{β}^{γ}$ . If $α = {(1, 2), (2, 3)}$ $α = {(1, 2), (2, 3)}$ , compute ${[T]}_{α}^{γ}$ ${[T]}_{α}^{γ}$ .
Define

$T : M_{2 \times 2} (R) \to P_{2} (R) by T (\begin{matrix} a & b \\ c & d \end{matrix}) = (a + b) + (2 d) x + b x^{2} .$ $T : M_{2 \times 2} (R) \to P_{2} (R) by T (\begin{matrix} a & b \\ c & d \end{matrix}) = (a + b) + (2 d) x + b x^{2} .$

Let

$β = {(\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}), (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix})} and γ = {1, x, x^{2}} .$ $β = {(\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}), (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix})} and γ = {1, x, x^{2}} .$

Compute ${[T]}_{β}^{γ}$ ${[T]}_{β}^{γ}$ .
Let

$\begin{array}{l} α = {(\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}), (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix})}, \\ β = {1, x, x^{2}}, \end{array}$ $\begin{array}{l} α = {(\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}), (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix})}, \\ β = {1, x, x^{2}}, \end{array}$

and

$γ = {1} .$ $γ = {1} .$
1. (a) Define $T : M_{2 \times 2} (F) \to M_{2 \times 2} (F)$ $T : M_{2 \times 2} (F) \to M_{2 \times 2} (F)$ by $T (A) = A^{t}$ $T (A) = A^{t}$ . Compute ${[T]}_{^{α}}$ ${[T]}_{^{α}}$ .
2. (b) Define
  
  $T : P_{2} (R) \to M_{2 \times 2} (R) by T (f (x)) = (\begin{matrix} f^{'} (0) & 2 f (1) \\ 0 & {f^{'}}^{'} (3) \end{matrix}),$ $T : P_{2} (R) \to M_{2 \times 2} (R) by T (f (x)) = (\begin{matrix} f^{'} (0) & 2 f (1) \\ 0 & {f^{'}}^{'} (3) \end{matrix}),$
  
  where ′ denotes differentiation. Compute ${[T]}_{β}^{α}$ ${[T]}_{β}^{α}$ .
3. (c) Define $T : M_{2 \times 2} (F) \to F$ $T : M_{2 \times 2} (F) \to F$ by $T (A) = tr (A)$ $T (A) = tr (A)$ . Compute ${[T]}_{α}^{γ}$ ${[T]}_{α}^{γ}$ .
4. (d) Define $T : P_{2} (R) \to R$ $T : P_{2} (R) \to R$ by $T (f (x)) = f (2)$ $T (f (x)) = f (2)$ . Compute ${[T]}_{β}^{γ}$ ${[T]}_{β}^{γ}$ .
5. (e) If
  
  $A = (\begin{matrix} 1 & - 2 \\ 0 & 4 \end{matrix}),$ $A = (\begin{matrix} 1 & - 2 \\ 0 & 4 \end{matrix}),$
  
  compute ${[A]}_{α}$ ${[A]}_{α}$ .
6. (f) If $f (x) = 3 - 6 x + x^{2}$ $f (x) = 3 - 6 x + x^{2}$ , compute ${[f (x)]}_{β}$ ${[f (x)]}_{β}$ .
7. (g) For $a \in F$ $a \in F$ , compute ${[a]}_{γ}$ ${[a]}_{γ}$ .
Complete the proof of part (b) of Theorem 2.7.
Prove part (b) of Theorem 2.8.
^† Let V be an n-dimensional vector space with an ordered basis $β$ $β$ . Define $T : V \to F^{n}$ $T : V \to F^{n}$ by $T (x) = {[x]}_{β}$ $T (x) = {[x]}_{β}$ . Prove that T is linear.
Let V be the vector space of complex numbers over the field R. Define $T : V \to V$ $T : V \to V$ by $T (z) = \bar{z}$ $T (z) = \bar{z}$ , where $\bar{z}$ $\bar{z}$ is the complex conjugate of z. Prove that T is linear, and compute ${[T]}_{β}$ ${[T]}_{β}$ , where $β = {1, i}$ $β = {1, i}$ . (Recall by Exercise 39 of Section 2.1 that T is not linear if V is regarded as a vector space over the field C.)
Let V be a vector space with the ordered basis $β = {v_{1}, v_{2}, \dots v_{n}}$ $β = {v_{1}, v_{2}, \dots v_{n}}$ . Define $v_{0} = 0$ $v_{0} = 0$ . By Theorem 2.6 (p. 73), there exists a linear transformation $T : V \to V$ $T : V \to V$ such that $T (v_{j}) = v_{j} + v_{j - 1}$ $T (v_{j}) = v_{j} + v_{j - 1}$ for $j = 1, 2, \dots, n$ $j = 1, 2, \dots, n$ Compute ${[T]}_{β}$ ${[T]}_{β}$ .
Let V be an n-dimensional vector space, and let $T : V \to V$ $T : V \to V$ be a linear transformation. Suppose that W is a T-invariant subspace of V (see the exercises of Section 2.1) having dimension k. Show that there is a basis $β$ $β$ for V such that ${[T]}_{β}$ ${[T]}_{β}$ has the form

$(\begin{matrix} A & B \\ O & C \end{matrix}),$ $(\begin{matrix} A & B \\ O & C \end{matrix}),$

where A is a $k \times k$ $k \times k$ matrix and O is the $(n - k) \times k$ $(n - k) \times k$ zero matrix.
^† Let $β = {v_{1}, v_{2}, \dots, v_{n}}$ $β = {v_{1}, v_{2}, \dots, v_{n}}$ be a basis for a vector space V and $T : V \to V$ $T : V \to V$ be a linear transformation. Prove that ${[T]}_{β}$ ${[T]}_{β}$ is upper triangular if and only if $T (v_{j}) \in span ({v_{1}, v_{2}, \dots, v_{j}})$ $T (v_{j}) \in span ({v_{1}, v_{2}, \dots, v_{j}})$ for $j = 1, 2, \dots, n$ $j = 1, 2, \dots, n$ . Visit goo.gl/k9ZrQb for a solution.
Let V be a finite-dimensional vector space and T be the projection on W along $W^{'}$ $W^{'}$ , where W and $W^{'}$ $W^{'}$ are subspaces of V. (See the definition in the exercises of Section 2.1 on page 76.) Find an ordered basis ft for V such that ${[T]}_{β}$ ${[T]}_{β}$ is a diagonal matrix.
Let V and W be vector spaces, and let T and U be nonzero linear transformations from V into W. If $R (T) \cap R (U) = {0}$ $R (T) \cap R (U) = {0}$ , prove that {T, U} is a linearly independent subset of L(V, W).
Let $V = P (R)$ $V = P (R)$ and $for j \geq 1$ $for j \geq 1$ define $T_{j} (f (x)) = f^{(j)} (x)$ $T_{j} (f (x)) = f^{(j)} (x)$ , where $f^{(j)} (x)$ $f^{(j)} (x)$ is the jth derivative of f(x). Prove that the set ${T_{1}, T_{2}, \dots, T_{n}}$ ${T_{1}, T_{2}, \dots, T_{n}}$ is a linearly independent subset of L(V) for any positive integer n.
Let V and W be vector spaces, and let S be a subset of V. Define $S^{0} = {T \in L (V, W) : T (x) = 0 for all x \in S}$ $S^{0} = {T \in L (V, W) : T (x) = 0 for all x \in S}$ . Prove the following statements.
1. (a) $S^{0}$ $S^{0}$ is a subspace of L(V, W).
2. (b) If $S_{1}$ $S_{1}$ and $S_{2}$ $S_{2}$ are subsets of V and $S_{1} \subseteq S_{2}$ $S_{1} \subseteq S_{2}$ , then $S_{2}^{0} \subseteq S_{1}^{0}$ $S_{2}^{0} \subseteq S_{1}^{0}$ .
3. (c) If $V_{1}$ $V_{1}$ and $V_{2}$ $V_{2}$ are subspaces of V, then ${(V_{1} \cup V_{2})}^{0} = {(V_{1} + V_{2})}^{0} = V_{1}^{0} \cap V_{2}^{0}$ ${(V_{1} \cup V_{2})}^{0} = {(V_{1} + V_{2})}^{0} = V_{1}^{0} \cap V_{2}^{0}$ .
Let V and W be vector spaces such that $dim (V) = dim (W)$ $dim (V) = dim (W)$ , and let $T : V \to W$ $T : V \to W$ be linear. Show that there exist ordered bases $β$ $β$ and $γ$ $γ$ for V and W, respectively, such that ${[T]}_{β}^{γ}$ ${[T]}_{β}^{γ}$ is a diagonal matrix.

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Table of Contents for 2.2 The Matrix Representation of a Linear Transformation

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Table of Contents for
2.2 The Matrix Representation of a Linear Transformation