# Matrix of a transformation from polynomials to Euclidean space, defined by evaluation

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?

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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.