Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In my assignment I have a vector space of polynomials with the degree at most 3 described by this linear transformation: $$ \phi:\mathbb{R}_3[x]\to \mathbb{R}^3,\quad \phi(p) = [p(-1),p(0),p(1)]^T $$

I have to write the transformation matrix in the basis $\{1,x,x^2,x^3\}$ for $\mathbb{R}_3[x]$ and standard basis $\mathbb{R}^3$.

My solution

If we write the polynomial of degree $3$ in vector form we get $[c_1,c_2,c_3,c_4]^T$. The linear transformation is then of the form:

$$\phi([c_1,c_2,c_3,c_4]^T) = [c_1 - c_2 + c_3 + c_4, c_1, c_1 + c_2 + c_3 + c_4]$$

If I understand correctly then to get the transformation matrix for the basis $\{1,x,x^2,x^3\}$ I apply the transformation to the standard basis of $\mathbb{R}^4$: $[1,0,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,1]$. So I get the vectors, which, joined together, will form the new transformation matrix = A. I get
$$\begin{bmatrix} 1&-1&1&-1\\1&0&0&0\\1&1&1&1\end{bmatrix}$$

To get the transformation matrix of $\mathbb{R}^3$ I use the formula $D = PAP^{-1}$, where $P$ will be the identity matrix of $\mathbb{R}^3$ and $A$ is the matrix which I got before.

Is my approach correct?

share|cite|improve this question
up vote 1 down vote accepted

Your $A$ is correct, but the "$D=PAP^{-1}$" part is problematic. The matrix $A$ is 3-by-4. It is not a square matrix and the product $PAP^{-1}$ does not make any sense. Also, if I understand correctly, the assignment problem only asks you to find the transformation matrix with respect to the cannonical bases of $\mathbb{R}_3[x]$ and $\mathbb{R}^3$. So you have already solved the problem after you determined $A$. There is no need to perform any change of bases.

By the way, I cannot find any question mark in what you have written. So, what is your question?

share|cite|improve this answer
Thank you for your answer. :) I only wanted to make sure my solution was correct, that is why there wasn't a question anywhere. As for the other part of the question (transformation to R^3) I only read the instructions of the exercise wrong. – Trom Nov 23 '12 at 15:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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