# How to find matrix of a linear transformation.

Let $T : \Bbb R^2 \to \Bbb R^3$ given by $$T((x_1, x_2)^T) = (x + 2y, 2x - 5y , 7y )^T$$.

We have $T((0, 1)^T) = (2, -5, 7)^T$ and $T((1, 0)^T) = (1, 2, 0)^T$

So the martix form of this linear transformation should be $$T (\mathbf x) = \begin{bmatrix} 1 & 2 \\ 2 & -5 \\ 0 & 7\end{bmatrix}\mathbf x$$,

where $\mathbf x \in \Bbb R^2$.

This was easy.

I understand this type of examples where the basis vectors are vectors in traditional sense.

Now consider $T : \Bbb P_n \to \Bbb P_n$ given by $T(f(t)) = \dfrac{df(t)}{dt}$, where $\Bbb P_n$ is space of polynomials with degree $n$.

The basis of this space is $B = \{1, t, t^2 , ... , t^n \}$

We have $T(t^k) = kt^{k - 1}$ for all $t^k \in B - \{1\}$ and $T(1) = 0$.

In the first example we had transformed bases $\mathbf a_2 = (2, -5, 7)^T$ and $\mathbf a_1 = (1, 2, 0)^T$ and we made matrix by just arranging those bases in a matrix like $[\mathbf a_1 \ \ \mathbf a_2]$.

In the polynomial example what should I do ? $\begin{bmatrix} 0 & 1 & 2t & ... & nt^{n-1} \end{bmatrix}$, huh ?

• You express the image of the basis elements under the transformation using linear combinations of the basis elements (and write down the scalar in the combination inside the matrix). Commented Oct 1, 2017 at 16:10

Let $\Bbb R^2$ have standard basis $\{e_1,e_2\}$ and $\Bbb R^3$ have standard basis $\{f_1,f_2,f_3\}$.

Then, $T(e_1) = 1f_1+2f_2+0f_3$ and $T(e_2) = 2f_1+(-5)f_2+7(f_3)$.

Therefore, the matrix representation for $T$ is $\begin{bmatrix}1&2\\2&-5\\0&7\end{bmatrix}$.

Let $\Bbb P^n$ have standard basis $\{1,t,t^2,\cdots,t^n\}$.

Then, $T(1) = 0 \cdot 1 + 0 \cdot t + 0 \cdot t^2 + \cdots + 0 \cdot t^{n-1} + 0 \cdot t^n$, and so on.

Also, $T(t^n) = 0 \cdot 1 + 0 \cdot t + 0 \cdot t^2 + \cdots + n \cdot t^{n-1} + 0 \cdot t^n$.

Therefore, the matrix representation is:

$$\begin{bmatrix} 0&1&0&\cdots&0\\ 0&0&2&\cdots&0\\ 0&0&0&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots&n\\ 0&0&0&\cdots&0 \end{bmatrix}$$

• Oh ok I understand it now. Commented Oct 1, 2017 at 16:27