# Determine the matrix $L$ with respect to a basis of eigenvectors.

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?

• 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. – TheSilverDoe Aug 26 '20 at 15:49

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).