Matrix of a linear transformation

In this section, we relate linear transformation over finite dimensional vector spaces with matrices. For this, we ask the reader to recall the results on ordered basis, studied in Section 3.4.

Let $V$ and $W$ be finite dimensional vector spaces over the set ${\mathbb{F}}$ with respective dimensions $m$ and $n.$ Also, let $T: V {\longrightarrow}W$ be a linear transformation. Suppose ${\cal B}_1=( {\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_n )$ is an ORDERED BASIS of $V.$ In the last section, we saw that a linear transformation is determined by its image on a basis of the domain space. We therefore look at the images of the vectors ${\mathbf v}_j \in {\cal B}_1$ for $1 \leq j \leq n.$

Now for each $j, \; 1 \leq j \leq n,$ the vectors $T({\mathbf v}_j) \in W.$ We now express these vectors in terms of an ordered basis ${\cal B}_2= ( {\mathbf w}_1, {\mathbf w}_2, \ldots, {\mathbf w}_m)$ of $W.$ So, for each $j, \; 1 \leq j \leq n,$ there exist unique scalars $a_{1j}, a_{2j}, \ldots, a_{mj} \in {\mathbb{F}}$ such that

$$T({\mathbf v}_1) = a_{11} {\mathbf w}_1 + a_{21} {\mathbf w}_2 + \cdots + a_{m1} {\mathbf w}_m$$

$$T({\mathbf v}_2) = a_{12} {\mathbf w}_1 + a_{22} {\mathbf w}_2 + \cdots + a_{m2} {\mathbf w}_m$$

$$\vdots$$

$$T({\mathbf v}_n) = a_{1n} {\mathbf w}_1 + a_{2n} {\mathbf w}_2 + \cdots + a_{mn} {\mathbf w}_m.$$

Or in short, $T({\mathbf v}_j) = \sum\limits_{i=1}^m a_{ij} {\mathbf w}_i$ for $1 \leq j \leq n.$ In other words, for each $j, \; 1 \leq j \leq n,$ the coordinates of $T({\mathbf v}_j)$ with respect to the ordered basis ${\cal B}_2$ is the column vector $[a_{1j}, a_{2j}, \ldots, a_{mj}]^t.$ Equivalently,

$$[T({\mathbf v}_j)]_{{\cal B}_2} = \begin{bmatrix} a_{1j} \\ a_{2j} \\ \vdots \\ a_{mj} \end{bmatrix}.$$

Let $[{\mathbf x}]_{{\cal B}_1} = [x_1, x_2, \ldots, x_n]^t$ be the coordinates of a vector ${\mathbf x}\in V.$ Then

$$T({\mathbf x}) = T ( \sum_{j=1}^n x_j {\mathbf v}_j) = \sum_{j=1}^n x_j T({\mathbf v}_j)$$

$$= \sum_{j=1}^n x_j ( \sum_{i=1}^m a_{ij} {\mathbf w}_i)$$

$$= \sum_{i=1}^m ( \sum\limits_{j=1}^n a_{ij} x_j) {\mathbf w}_i.$$

Define a matrix $A$ by $A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & ... ... \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix}.$ Then the coordinates of the vector $T({\mathbf x})$ with respect to the ordered basis ${\cal B}_2$ is

$$[T({\mathbf x})]_{{\cal B}_2} = \begin{bmatrix}\sum_{j=1}^n a_{1j} x_j \\ \sum_{j=1}^n a_{2j} x_... ... & a_{mn} \end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}$$

$$= A \; [{\mathbf x}]_{{\cal B}_1}.$$

The matrix $A$ is called the matrix of the linear transformation $T$ with respect to the ordered bases ${\cal B}_1$ and ${\cal B}_2,$ and is denoted by $T[{\cal B}_1, {\cal B}_2].$

We thus have the following theorem.

THEOREM 4.2.1   Let $V$ and $W$ be finite dimensional vector spaces with dimensions $n$ and $m,$ respectively. Let $T: V {\longrightarrow}W$ be a linear transformation. If ${\cal B}_1$ is an ordered basis of $V$ and ${\cal B}_2$ is an ordered basis of $W,$ then there exists an $m \times n$ matrix $A = T[{\cal B}_1, {\cal B}_2]$ such that

$$[T({\mathbf x})]_{{\cal B}_2} = A \; \; [x]_{{\cal B}_1}.$$

Remark 4.2.2   Let ${\cal B}_1=( {\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_n )$ be an ordered basis of $V$ and ${\cal B}_2= ( {\mathbf w}_1, {\mathbf w}_2, \ldots, {\mathbf w}_m)$ be an ordered basis of $W.$ Let $T : V \longrightarrow W$ be a linear transformation with $A =T[{\cal B}_1, {\cal B}_2].$ Then the first column of $A$ is the coordinate of the vector $T({\mathbf v}_1)$ in the basis ${\cal B}_2.$ In general, the $i^{\mbox{th}}$ column of $A$ is the coordinate of the vector $T({\mathbf v}_i)$ in the basis ${\cal B}_2.$

We now give a few examples to understand the above discussion and the theorem.

EXAMPLE 4.2.3  

1. Let $T: {\mathbb{R}}^2 {\longrightarrow}{\mathbb{R}}^2$ be a linear transformation, given by
   $$T(\;(x,y) \;) = (x+y, x-y).$$
   We obtain $T[{\cal B}_1, {\cal B}_2],$ the matrix of the linear transformation $T$ with respect to the ordered bases
   $${\cal B}_1 = \bigl((1,0), (0,1)\bigr) \;\; {\mbox{ and }} \;\; {\cal B}_2 = \bigl( (1,1), (1,-1)\bigr) \;\; {\mbox{ of }} \;\; {\mathbb{R}}^2.$$
   For any vector
   $$(x,y) \in {\mathbb{R}}^2, \; \; [(x,y)]_{{\cal B}_1} = \begin{bmatrix}x \\ y \end{bmatrix}$$
   as $(x,y) = x (1,0) + y (0,1).$ Also, by definition of the linear transformation $T,$ we have
   $$T(\;(1,0)\;) = (1,1) = 1 \cdot (1,1) + 0 \cdot (1,-1). \; {\mbox{ So, }} [T(\;(1,0)\;)]_{{\cal B}_2} = (1,0)^t$$
   and
   $$T(\;(0, 1)\;) = (1,-1) = 0 \cdot (1,1) + 1 \cdot (1,-1).$$
   That is, $\; [T(\;(0,1)\;)]_{{\cal B}_2} = (0,1)^t.$ So the $T[{\cal B}_1, {\cal B}_2] = \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}.$ Observe that in this case,
   $$[T(\;(x,y) \;)]_{{\cal B}_2} = [(x+y, x-y)]_{{\cal B}_2} = x (1,1) + y (1,-1) = \begin{bmatrix}x \\ y \end{bmatrix}, \; {\mbox{ and }}$$
   $$T[{\cal B}_1, {\cal B}_2] \; [(x,y)]_{{\cal B}_1} = \begin{bmatri... ...d{bmatrix} = \begin{bmatrix}x \\ y \end{bmatrix}=[T(\;(x,y) \;)]_{{\cal B}_2}.$$

2. Let ${\cal B}_1=\bigl( (1,0,0), (0,1,0), (0,0,1) \bigr),$ $\;{\cal B}_2 = \bigl((1,0,0), (1,1,0), (1,1,1)\bigr)$ be two ordered bases of ${\mathbb{R}}^3.$ Define
   $$T: {\mathbb{R}}^3 {\longrightarrow}{\mathbb{R}}^3 \;\;{\mbox{ by }} \;\; T({\mathbf x}) = {\mathbf x}.$$
   Then
   

   $$T((1,0,0)) = 1 \cdot (1,0,0) + 0 \cdot (1,1,0) + 0 \cdot (1,1,1),$$

   $$T((0,1,0)) = -1 \cdot (1,0,0) + 1 \cdot (1,1,0) + 0 \cdot (1,1,1), {\mbox{ and }}$$

   $$T((0,0,1)) = 0 \cdot (1,0,0) + (-1) \cdot (1,1,0) + 1 \cdot (1,1,1).$$

   
   Thus, we have
   

   $$T[{\cal B}_1, {\cal B}_2] = [ [T((1,0,0))]_{{\cal B}_2}, \; [T((0,1,0))]_{{\cal B}_2}, \; [T((0,0,1))]_{{\cal B}_2} ]$$

   $$= [ (1,0,0)^t, \; (-1,1,0)^t, \; (0,-1,1)^t ]$$

   $$= \begin{bmatrix}1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix}.$$

   
   Similarly check that $T[{\cal B}_1, {\cal B}_1] = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}.$
3. Let $T: {\mathbb{R}}^3 {\longrightarrow}{\mathbb{R}}^2$ be define by $T((x,y,z)) = (x+y-z, x+z).$ Let ${\cal B}_1= \bigl( (1,0,0), (0,1,0), (0,0,1)\bigr)$ and ${\cal B}_2 = \bigl( (1,0), (0,1)\bigr)$ be the ordered bases of the domain and range space, respectively. Then
   $$T[{\cal B}_1, {\cal B}_2] = \begin{bmatrix}1 & 1 & -1 \\ 1 & 0 & 1 \end{bmatrix}.$$
   Check that that $[T(x,y,z)]_{{\cal B}_2} = T[{\cal B}_1, {\cal B}_2] \; \; [(x,y,z)]_{{\cal B}_1}.$

EXERCISE 4.2.4   Recall the space ${\cal P}_n({\mathbb{R}})$ ( the vector space of all polynomials of degree less than or equal to $n$ ). We define a linear transformation $D: {\cal P}_n({\mathbb{R}}) {\longrightarrow}{\cal P}_n({\mathbb{R}})$ by

$$D(a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n) = a_1 + 2 a_2 x + \cdots + n a_n x^{n-1}.$$

Find the matrix of the linear transformation $D.$

However, note that the image of the linear transformation is contained in ${\cal P}_{n-1}({\mathbb{R}}).$

Remark 4.2.5  

1. Observe that
   $$T[{\cal B}_1, {\cal B}_2] = [ [T({\mathbf v}_1)]_{{\cal B}_2}, [T({\mathbf v}_2)]_{{\cal B}_2}, \ldots, [T({\mathbf v}_n)]_{{\cal B}_2} ].$$

2. It is important to note that
   $$[T({\mathbf x})]_{{\cal B}_2} = T[{\cal B}_1, {\cal B}_2] \; [{\mathbf x}]_{{\cal B}_1}.$$
   That is, we multiply the matrix of the linear transformation with the coordinates $[{\mathbf x}]_{{\cal B}_1},$ of the vector ${\mathbf x}\in V$ to obtain the coordinates of the vector $T({\mathbf x}) \in W.$
3. If $A$ is an $m \times n$ matrix, then $A$ induces a linear transformation $T_A : {\mathbb{R}}^n {\longrightarrow}{\mathbb{R}}^m,$ defined by
   $$T_A({\mathbf x}) = A {\mathbf x}.$$
   We sometimes write $A$ for $T_A.$ Suppose that the standard bases for ${\mathbb{R}}^n$ and ${\mathbb{R}}^m$ are the ordered bases ${\cal B}_1$ and ${\cal B}_2,$ respectively. Then observe that
   $$T[{\cal B}_1, {\cal B}_2] = A.$$