31 - Linear Maps - Matrix Representation

Matrix for $T$ with respect to bases $\mathcal{V}$ and $\mathcal{W}$

Let $V$ be a $n$-dimensional vector space over $\mathbf{F}$ with an ordered basis $\mathcal{V}$ of vectors $v_1,\cdots ,v_n$
Let $W$ be a $m$-dimensional vector space over $\mathbf{F}$ with an ordered basis $\mathcal{W}$ of vectors $w_1,\cdots ,w_m$
Hence every vector in $V$ and $W$ is represented by a coordinate vector in ${\mathbf{F}}^n$ and ${\mathbf{F}}^m$

Let $T:V\to W$ be a linear transformation
The matrix for $T$ with respect to the bases $\mathcal{V}$ and $\mathcal{W}$ is the matrix that takes the coordinate vector of $v$ to the coordinate vector of $Tv$

In other words this is the $m\times n$ matrix $A=(a_{ij})$ where

$$T(v_i)=\sum _{k=1}^m{a}_{ki}{w}_k$$

This is written as ${}_{\mathcal{W}}{T}_{\mathcal{V}}$ for this matrix $A$

Refer to the example below to get a better understanding!

Note about the property of the matrix $1\le i\le n$ then $T(v_i)$ can be uniquely expressed as a linear combination of $\mathcal{W}$

It is well defined and for each

Specifically, $a_{1i},\cdots ,a_{mi}$ are the coordinates of $T(v_i)$
These are the entries in the $i$th column of $A$
The coordinate column vector of $v_i$ is ${\mathbf{e}}_i^T$ and the $i$th column of $A$ is $A{\mathbf{e}}_i^T$
So the entries of $A$ are the coordinates of the image of the basis $\mathcal{V}$

Order of the vectors in the basis is very crucial, if the order changes so does the matrix

Matrix for $T$ with respect to this basis $V$

Same as above but with $V=W$
So we use the same ordered basis for both the domain and codomain of $T:V\to V$

Finding Matrix with respect to a basis (1)

Let $T:{\mathbb{R}}^3\to {\mathbb{R}}^3$ be defined by $T(x,y,z)=(0,x,y)$
Define $\mathcal{E}$ as the Canonical Basis in ${\mathbb{R}}^3$
Define $\mathcal{F}$ as the ordered basis

$${\mathbf{f}}_1=(1,1,1),{\mathbf{f}}_2=(1,1,0),{\mathbf{f}}_3=(0,1,1)$$

Then

$$\begin{array}{rl}T(1,0,0)& =(0,1,0)=-{\mathbf{f}}_1+{\mathbf{f}}_2+{\mathbf{f}}_3\\ T(0,1,0)& =(0,0,1)={\mathbf{f}}_1-{\mathbf{f}}_2\\ T(0,0,1)& =(0,0,0)=0\end{array}$$

Hence

$${}_{\mathcal{F}}{T}_{\mathcal{E}}=\left(\begin{array}{ccc}-1& 1& 0\\ 1& -1& 0\\ 1& 0& 0\end{array}\right)$$