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Let $V = R[X]_{≤3}$ and $α ∈ R$. Define the linear image $ L : V → V$ given by $L(P(X)) = αP(X) + (X + 1)P'(X)$.

Proof that $L$ diagonalizable and determine the matrix $L$ with respect to a basis of eigenvectors.

I have found this matrix:

$$L=\begin{bmatrix}\alpha&1&0&0\\\ 0&1+\alpha&2&0\\0&0&2+\alpha&3\\0&0&0&3+\alpha\end{bmatrix}.$$ I used the standard basis {${1,x,x^2,x^3}$}

Then you know the eigenvalues are $\alpha, 1+\alpha, 2+\alpha$ and $3+\alpha$ with respectively the eigenspaces $(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)$ and because every $d(\lambda)=m(\lambda)$ we know that $L$ is diagonalizable.

First is this correct? If so, How do I construct the matrix $L$ with respect to a basis of eigenvectors. Is it possible that this is just the matrix with on the diagonal the eigenvalues?

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    $\begingroup$ The matrix in the standard basis is correct. You deduce indeed, because it is triangular, that the eigenvalues are $\alpha$, $1+\alpha$, $2+\alpha$ and $3+\alpha$. They are all disctincts, so the matrix is diagonalizable. But you are wrong about the eigenspaces : the vectors of the standard basis are not eigenvectors, otherwise the matrix would be diagonal. $\endgroup$ – TheSilverDoe Aug 26 '20 at 15:49
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Let's try and find the eigenvectors of the representing matrix with respect to $\alpha$. We need to find the null space of $$ L-\alpha I= \begin{bmatrix} 0&1&0&0\\ 0&1&2&0\\ 0&0&2&3\\ 0&0&0&3 \end{bmatrix} $$ and we find $[1\ 0\ 0\ 0]^T$. This yields the polynomial $1$ as an eigenvector for $L$.

With respect to $1+\alpha$, we need the null space of $$ L-(1+\alpha) I= \begin{bmatrix} -1&1&0&0\\ 0&0&2&0\\ 0&0&1&3\\ 0&0&0&2 \end{bmatrix} $$ and we find $[1\ 1\ 0\ 0]^T$. This yields $1+x$ as an eigenvector for $L$.

You may find also the other eigenvectors. However, you don't need it. The matrix with respect to a basis of eigenvectors is $$ \begin{bmatrix} \alpha & 0 & 0 & 0 \\ 0 & 1+\alpha & 0 & 0 \\ 0 & 0 & 2+\alpha & 0 \\ 0 & 0 & 0 & 3+\alpha \end{bmatrix} $$ (or any permutation of the eigenvalues along the diagonal).

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