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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 ?

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  • $\begingroup$ 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). $\endgroup$
    – Kenny Lau
    Commented Oct 1, 2017 at 16:10

1 Answer 1

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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:

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

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

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