# 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!

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So, just to recap: you found a basis composed of eigenvectors of $T$, correct? –  Jonathan Y. Dec 9 '13 at 15:50
Not quite. I don´t know the English word of it. In Spanish we call it canonical basis, is the basis of the space of polinomials ${1,x,x^2,...,x^n}$ It think that those are in fact eigenvectors, but I was not thinking of that... –  Lessa121 Dec 9 '13 at 15:54
Standard basis, I think. Regardless, you've got it! (and the way you could've seen it is the representing matrix was diagonal.) –  Jonathan Y. Dec 9 '13 at 16:51
That's right. Thanks! –  Lessa121 Dec 9 '13 at 17:16

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.