How to find the matrix representation of a linear tranformation

Question

How does one find the matrix representation of a linear transformation $T:V\to W$ with respect to the basis $B$ for $V$ and $D$ for $W$?

Answer 1

Let $\mathcal B = \{b_1, \dots, b_m\}$ and $\mathcal D = \{d_1, \dots, d_n\}$. Then the matrix $[T]_{\mathcal D\leftarrow \mathcal B}$ is given by $$[T]_{\mathcal D\leftarrow \mathcal B} = \begin{bmatrix}[T(b_1)]_{\mathcal D} & \cdots & [T(b_m)]_{\mathcal D}\end{bmatrix}$$ where $[T(b_i)]_{\mathcal D}$ is the column whose coordinates are those of the image of $b_i$ under $T$ represented in the basis $\mathcal D$. Why does this work?

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Example. Let's say I have $\mathcal B = \{(1,1),(1,2)\}\subset \Bbb R^2$ and $\mathcal D = \{(1,0),(1,1)\}\subset \Bbb R^2$. Consider the linear operator, $P$, which projects vectors in $\Bbb R^2$ onto the line spanned by $(0,1)$. Then first we find $P(b_1)$ and $P(b_2)$.

$$P(b_1) = \frac{(1,1)\cdot(0,1)}{1^2}(0,1) = (0,1) \\ P(b_2) = \frac{(1,2)\cdot (0,1)}{1^2}(0,1) = (0,2)$$

Then we express these vectors in the basis $\mathcal D$:

$$P(b_1) = (0,1) = -(1,0) + (1,1) \\ P(b_2) = (0,2) = -2(1,0) + 2(1,1)$$

Now we can immediately construct the matrix $[P]_{\mathcal D \leftarrow \mathcal B}$:

$$[P]_{\mathcal D \leftarrow \mathcal B} = \begin{bmatrix} -1 & -2 \\ 1 & 2\end{bmatrix}$$

Answer 2

Find the decomposition as a linear combination of the image of each of the vector of the base.

Answer 3

The $i$th column of the matrix gives the coefficients of $T(v_i)$ (where $v_i$ is the $i$th basis vector in $B$) when it is written in terms of the basis $D$. That is, the $ji$ entry is the coefficient of $w_j$ (the $j$th basis vector in $D$) when $T(v_i)$ is written in terms of the basis $D$.

Answer 4

if $\dim V=n$ and $\dim W = m$

consider the $n$ vectors $b_j$ constituting the basis $B$, and consider the $n$ vectors obtained as linear transformation of them $T(b_j)$;

then the columns of the matrix associated to the linear transformation through the bases $B$ and $D$ are the vectors of $\mathbb{R}^m$ made up of the components of these vectors $T(b_j)$ with respect to the basis $D$ of $W$ made up of the $m$ vectors $d_i$.

So $M_T^{B,D}$ has $j$-th column equal to $(T(b_j)\cdot d_1,...,T(b_j)\cdot d_m)$ or considering the standard basis $E$ of $\mathbb{R}^m$,it can be defined as $\sum\limits_{i=1}^m=(T(b_j)\cdot d_i)e_i $.

All in all $\{m_{T_{ij}}^{B,D}\}_{1\le i \le m, \;1 \le j \le n}=(T(b_j)\cdot d_i)$