Polynomial endomorphism diagonalization

Question

I have this linear map : $$ \phi : E_3 \rightarrow E_3 $$ with $E_3 = \{P \in \mathbb{R}[X] | deg(P) \leq 3\}$

$$ \phi[(P)](X) = (X^2 - 1)P''(X) + 2XP'(X) $$

And i am asked to come up with a basis of $E_3$ that diagonalizes this map with the condition : this base must contain polynomials with dominant terms equal to $1$.

Here's how i proceeded :

First, i've written the matrix form of this map with respect to the canonical basis : $\{1,X,X^2,X^3\}$.

$$ M_3 = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ -2 & 0 & 6 & 0 \\ 0 & -6 & 0 & 12\\ \end{pmatrix} $$

Then i've determined the eigenvalues : $\{0,2,6,12\}$.

Now i'm looking for the eigenvectors :

I started with the vector associated to $\lambda = 0$ : $V_0 = (\frac{1}{3}\alpha,0,\alpha,0)$

And by setting $\alpha = 1$ i have a polynomial : $P_{0}(x) = \frac{1}{3} + X^3$ with a dominant term equal to $1$.

Am I on the right path ? Should i do the same for the other eigenvalues ?

Thank you :)

Answer 1

You have:

• $\phi[(1)](X)=0$;
• $\phi[(X)](X)=2X$;
• $\phi[(X^2)](X)=6X^2-2$;
• $\phi[(X^3)](X)=12X^3-6X$.

Therefore, the matrix of $\phi$ with respect to the given basis is$$M=\begin{bmatrix}0 & 0 & -2 & 0 \\ 0 & 2 & 0 & -6 \\ 0 & 0 & 6 & 0 \\ 0 & 0 & 0 & 12\end{bmatrix}.$$So (as you wrote), the eigenvalues of $M$ are $0$, $2$, $6$, and $12$ and:

• an eigenvector corresponding to the eigenvalue $0$ is the null vector;
• an eigenvector corresponding to the eigenvalue $2$ is $\left(0,1,0,0\right)$;
• an eigenvector corresponding to the eigenvalue $6$ is $\left(-\frac13,0,1,0\right)$;
• an eigenvector corresponding to the eigenvalue $12$ is $\left(0,-\frac35,0,1\right)$.

These eigenvectors were chosen so that, in each case, the rightmost non-null entry is $1$. So, translating these vectors into polynomials, you get a basis with the property that you're after:$$\left\{0,X,X^2-\frac13,X^3-\frac35X\right\}.$$