0
$\begingroup$

I have been attempting quite a few of these "find the matrix from the linear transformation" problems and I have been stuck on this problem for some time. For the standard bases what I have is

\begin{bmatrix} p(1) & 0 & 0 \\ 0 & p\prime(1) & 0 \\ \\ 0 & 0 & p\prime\prime(-1) \end{bmatrix}

and here's what I have for the other basis

\begin{bmatrix} p(1) & p(1) & p(1) \\ 0 & p\prime(1) & p\prime(1) \\ \\ 0 & 0 & p\prime\prime(-1) \end{bmatrix}

Thanks in advance!

Let $T:P_3(\mathbb{R})\rightarrow\mathbb{R^3}$ be the linear transformation given by $$T(p)=(p(1), p\prime(1), p\prime\prime(-1))$$.

Find the matrix of $T$ relative to the standard basis of $P_3(\mathbb{R})$ and the basis $$v_1=(1,0,0), v_2=(1,1,0), v_3=(1,1,1)$$ of $\mathbb{R^3}$

$\endgroup$
8
  • $\begingroup$ Why do you tell us what the standard basis of $P_3(R)$ actually is, and for each vector $p$ in that basis, tell us what you computed as $T(p)$?That wy we'll know a bit more of what you actually know how to do. (You can edit your question by clicking on "edit" just below the question, to show your partial work.) $\endgroup$ Oct 27, 2017 at 23:42
  • $\begingroup$ What are you getting stuck on? $\endgroup$
    – copper.hat
    Oct 27, 2017 at 23:50
  • $\begingroup$ What is $P_3(\mathbb{R})$? Is it the vector space (algebra actually) of the polynomials of degree less than $3$, at most $3$, or something else? Telling us the "standard basis" would have clarified the question $\endgroup$
    – Tancredi
    Oct 27, 2017 at 23:54
  • $\begingroup$ Agree with john hughes on this that it would be more helpful if you show what you've done. Best $\endgroup$
    – vkan
    Oct 28, 2017 at 0:16
  • $\begingroup$ I have edited the post. $\endgroup$ Oct 28, 2017 at 0:42

1 Answer 1

0
$\begingroup$

Hint: would suggest by writing $p=ax^2+bx+c$ and seeing what happens to it by applying the transformation T.

(If the non-standard basis of $\mathbb{R}^3$ is confusing then perhaps do the problem with respect to the standard basis first).

$\endgroup$

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.