diagonalizable linear transformation

Question

I am having difficulties solving this problem. $V$ is the vector space of all polynomials (of complex numbers) of max degree $n$. $T$ is a linear transformation from $V$ to $V$.

$$T(p)(x)=p'(x)+p(0)\cdot x^n$$ Prove that T is diagonalizable.

Many thanks.

Answer 1

If you consider the basis $\left\{P_k = \frac{X^k}{k!}\right\}_{0 \le k \le n}$ then the matrix of $T$ is a companion matrix indeed, $$T(P_k) = \left\{\begin{array}{cc} n! P_n & \text{if $k=0$} \\ P_{k-1} & \text{otherwise}\end{array}\right.$$ The characteristic polynomial of this matrix is $\chi(X) = X^{n+1} - n!$ which has distinct roots so $T$ is diagonalizable

Answer 2

A general member of $V$ will have the form

$p(x) = a_0+a_1x+a_2x^2+\dots+a_nx^n$

Then $p(0) = a_0$, so

$T(p)(x) = a_1 + 2a_2x + 3a_3x^2 + \dots + na_nx^{n-1}+a_0x^n$

If $\lambda$ is an eigenvalue of $T$ then

$T(p)(x) = \lambda(x)$

so

$a_1 = \lambda a_0$

$2a_2 = \lambda a_1 = \lambda^2a_0$

$3a_3 = \lambda a_2 \Rightarrow 6a_3 = \lambda^3a_0$

and so on, until finally:

$n!a_n = \lambda^na_0$

$a_0 = \lambda a_n \Rightarrow n!a_0 = \lambda^{n+1}a_0$

So the eigenvalues $\lambda$ are the roots of the characteristic equation

$\lambda^{n+1} = n!$

which has $n+1$ distinct roots in the complex numbers. This is a sufficient condition for $T$ to be diagonalizable.