About eigenvalues of a linear operator Let $T$ be a linear operator in the space of polynomials of degree less or equal to n, $T(p(x))=xp´(x)$. I need to calculate the eigenvalues of this operator. 
So far, I calculated the matrix of the transformation in the canonical basis:
$$
        \begin{matrix}
        0 & 0 & 0 & ... & 0 \\
        0 & 1 & 0 & ... & 0 \\
        0 & 0 & 2 &...  &0  \\
         &   &   & ... &  \\
        0 & 0 & 0 &...  &n  \\
        \end{matrix}
$$
And if I calculate the characteristic polynomial of this matrix, then $0,1,2,3,...,n$ are all eigenvalues. I don´t know how to calculate the associated eigenspace for EACH of these values. Am I missing something? Thak you!

For that matrix, an eigenvector for the $i$'th diagonal eigenvalue is $E_{i}$, where this is the vector with the $i$'th entry $1$ and all other entries $0$.
It looks like in your case all eigenvalues are distinct, and hence each eigenspace is $1$ dimensional and generated by each eigenvector we found.
