0
$\begingroup$

Consider the linear trasnformation $L:\mathbb{P^3}\rightarrow \mathbb{P^2}$ given by the formula $L(p(t))=p'(t)-tp''(t)$ Find the matrix of the linear transformation $L$ with respect to the bases $S=\{1+t^3,t+t^2,t^2−t^3,t^3\}$ and $T=\{1,t,t^2\}$. Here's what I did to attempt the problem. I found the transformation of the standard bases to be :

$L(e_1)=\begin{bmatrix}0\\0\\0\\0\end{bmatrix}$

$L(e_2)=\begin{bmatrix}0\\0\\0\\0\end{bmatrix}$

$L(e_3)=\begin{bmatrix}0\\0\\0\\0\end{bmatrix}$

$L(e_4)=\begin{bmatrix}0\\-3\\0\\0\end{bmatrix}$

So I know that the matrix $A$ representing the linear transformation is :

$A=\begin{bmatrix}0&0&0&0\\0&0&0&-3\\0&0&0&0\\0&0&0&0\end{bmatrix}$

I don't know what to do after this step. How do I find the matrix with respect to both bases $S$ and $T$? I apologize if I did something horrible wrong or if the answer is obvious!

$\endgroup$
11
  • 1
    $\begingroup$ Did you mean $p''(t)$ when you wrote $t''(p)$? $\endgroup$ Oct 15 '19 at 17:12
  • 1
    $\begingroup$ and $L(e_1)$ when you wrote $T(e_1)$? $\endgroup$ Oct 15 '19 at 17:13
  • 1
    $\begingroup$ That's the standard basis for $\Bbb R^4$; what's $\Bbb P^3$? $\endgroup$ Oct 15 '19 at 17:25
  • 2
    $\begingroup$ Oh, I get it now, $T$ is the basis for the codomain, so find what $L$ does to each element of $S$ and express the results as linear combinations of elements of $T$ $\endgroup$ Oct 15 '19 at 17:28
  • 1
    $\begingroup$ Do you know what $L(s)$ is for each $s\in S$? $\endgroup$ Oct 15 '19 at 17:36
1
$\begingroup$

$S=\{1+t^3,t+t^2,t^2−t^3,t^3\}$ is a basis of $\Bbb P^3$, the domain of $L$,

and $T=\{1,t,t^2\}$ is a basis of $\Bbb P^2$, the codomain of $L$.

$L$ maps the elements of $S$ to $-3t^2, 1, 3t^2, -3t^2$, respectively,

so $A=\begin{bmatrix}0&1&0&0\\0&0&0&0\\-3&0&3&-3\end{bmatrix}.$

Note that $A$ has the proper size for mapping a $4$-dimensional space to a $3$-dimensional space.

$\endgroup$
1
  • $\begingroup$ I was going about solving the question in a completely wrong way! Thank you! $\endgroup$ Oct 16 '19 at 10:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.