Show that for each $λ ∈\mathbb Z$ with $λ \geqslant 0$, $λ$ is an eigenvalue of $T$, and $x_λ$ is a corresponding eigenvector

Question

Homework help

Let $$T : P(R) \to P(R)$$ be the linear map defined by $$T(p(x)) = xp'(x).$$ Show that for each $\lambda\in\mathbb{Z}$ with $\lambda\geq0$, $\lambda$ is an eigenvalue of $T$, and $x_\lambda$ is a corresponding eigenvector.

My question is can someone help me understand what the question is asking. I understand the first sentence which says $T$ is the linear operator that produces the derivative. But I'm stuck on the second sentence that asks to show that each $\lambda$ in the positive integers is an eigenvalue of $T$ corresponding to the eigenvector $x^\lambda$.

Any help would be greatly appreciated. Thanks,

Answer 1

Assuming that $P(\Bbb R)$ is the ring of polynomials with real coefficients, also denoted by $\Bbb R[x]$, then the map

$T: P(\Bbb R) \to P(\Bbb R) \tag 1$

defined by

$T(p(x)) = xp'(x) = x \dfrac{dp(x)}{dx} \tag 2$

is not the derivative, but the derivative multiplied by $x$, so that if

$p(x) = \displaystyle \sum_0^n p_i x^i, \tag 3$

then

$T(p(x)) = x \displaystyle \sum_1^n i p_i x^{i - 1} = \sum_1^n ip_i x^i. \tag 4$

For $0 \le \lambda \in \Bbb Z$, we find

$T(x^\lambda) = x \lambda x^{\lambda - 1} = \lambda x^\lambda, \tag 5$

which shows that $\lambda$ is an eigenvalue of $T$ with corresponding eigenvector $x^\lambda$.

Note: In all honesty I should point out the mild abuse of notation in (5), since technically $d(x^0)/dx = 0$, not $x^{-1}$; but since $\lambda = 0$ in this case, the equation $T(x^0) = 0 x^0$ still works, since $x^0 = 1$ and $1' = 0$. It's a stretch, hopefully a forgivable one. End of Note.