Proving that a linear transformation is diagonalisable

Given that $$V = \mathbb{R}[X]_{\leq 2}$$ and $$\alpha \in \mathbb{R}$$ prove that the linear transformation $$L: V \to V$$ given by $$L(P(X)) = \alpha P(X) + (X+1)P'(X)$$ is diagonalisable and determine the matrix representation with respect to a basis of eigenvectors.

I know that $$L$$ is diagonalisable if $$\#Spec(L) = \dim(V) = 3$$, meaning that $$L$$ must have 3 distinct eigenvalues. I imagine I have to find the eigenvalues of $$L$$, but, I'm not really sure where to go from here. Any help is appreciated.

Note that $$\operatorname{Id}_n\colon\mathbb R^n\longrightarrow\mathbb R^n$$ has a single eigenvalue, but it is diagonalizable nevertheless. So, no, you don't need to have $$3$$ distinct eigenvalues.

On the other hand:

• $$L(1)=\alpha\times1$$;
• $$L(X)=(\alpha+1)X+1$$;
• $$L(X^2)=2X+(\alpha+2)X^2$$.

Therefore, the matrix of $$L$$ with respect to the basis $$\{1,X,X^2\}$$ is$$\begin{bmatrix}\alpha&1&0\\0&\alpha+1&2\\0&0&\alpha+2\end{bmatrix}.$$Can you take it from here?

• I would then use the matrix representation to find the eigenvalues and then find the matrix with respect to the values of the eigenvectors correct? – SES Aug 23 at 13:55
• Yes, that would be correct. – José Carlos Santos Aug 23 at 14:14
• I suppose that it is correct now. – José Carlos Santos Aug 23 at 14:28
• I have checked my solution : I maintain it is the good one – Jean Marie Aug 23 at 14:29
• I will no disagree. After all, it is equal to mine. – José Carlos Santos Aug 23 at 14:30

The image of basis $$(1,X,X^2)$$ is $$(\alpha, (\alpha+1)X+1,(\alpha+2) X^2+2X)$$ therefore with matrix representation :

$$M=\begin{pmatrix}\alpha&1&0\\0&\alpha+1&2\\0&0&\alpha+2\end{pmatrix}$$

As $$M$$ is triangular, its diagonal entries are its eigenvalues :

$$\alpha<\alpha+1<\alpha+2$$

Being all distinct, whatever $$\alpha$$, this matrix is always diagonalizable.

• Your answer is different than Jose Carlos Santos'. One of them must be wrong. – Leo Aug 23 at 14:10
• Sorry, the $m_{11}$ entry is $\alpha$... – Jean Marie Aug 23 at 14:12

We may as well replace $$1$$ with an arbitrary constant $$-r$$. Suppose $$Q$$ is an eigenvector of $$L$$, say of eigenvalue $$\lambda$$. Rearranging gives $$(X - r) Q'(X) = (\lambda - \alpha) Q(X) .$$ But this is a separable ordinary differential equation. Since polynomials are analytic, it suffices to work on some nonempty open interval not containing $$r$$ (or more to the point, where $$X - r$$ is nonvanishing). Then, dividing gives $$\frac{\lambda - \alpha}{X - r} = \frac{Q'(X)}{Q(X)} = (\log Q)'(X),$$ and integrating gives that up to a constant multiple we have $$P(X) = (X - r)^{\lambda - \alpha}.$$ This function is an element of $$R[X]_{\leq 2}$$ precisely when $$\lambda - \alpha \in \{0, 1, 2\}$$. (Note that this solves efficiently the corresponding problem for the operator $$L$$ defined on the polynomial space $$R[X]_{\leq d}$$ for large $$d$$.)

• [+1] Good idea to connect this issue with a differential equation. – Jean Marie Aug 23 at 16:27

One can also approach this directly using the definition of eigenvector and basic facts about polynomials. If $$Q$$ is an eigenvector of $$L$$, by definition we have $$L(Q(X)) = \alpha Q(X) + (X + 1) Q'(X) = \lambda Q(X)$$ for some constant $$\lambda$$, and rearranging gives $$(X + 1) Q'(X) = (\lambda - \alpha) Q(X) .$$ If $$Q'(X) = 0$$, then $$Q$$ is constant and $$\lambda = \alpha$$, giving one eigenvalue. If $$Q'(X) \neq 0$$, then $$Q(X)$$ is divisible by $$(X + 1)$$, so $$Q(X) = R(X) (X + 1)$$ for some $$R$$ with $$\deg R \leq 1$$. Substituting into the previous display equation and cancelling gives $$R(X) + R'(X) (X + 1) = (\lambda - \alpha) R(X) ,$$ and rearranging gives $$(X + 1) R'(X) = (\lambda - (\alpha + 1)) R(X) .$$

Now, either $$R'(X) = 0$$, in which case $$R$$ is constant and $$\lambda = \alpha + 1$$, or $$R$$ a (now constant) multiple of $$X + 1$$, in which case $$Q(X) = R(X) (X + 1)$$ is, up to a nonzero multiple, $$(X + 1)^2$$, and substituting again gives that $$\lambda = \alpha + 2$$. We've now found that $$L$$ has distinct eigenvalues, namely, $$\alpha, \alpha + 1, \alpha + 2$$, and that with respect to the basis $$(1, X + 1, (X + 1)^2)$$, $$L$$ has the matrix representation $$[L] = \pmatrix{\alpha\\&\alpha + 1\\&&\alpha + 2} .$$