$\textbf{Exercise 9.}$ Define $T: \mathcal{P}(\mathbb{R}) \to \mathcal{P}(\mathbb{R})$ by $Tp = p'$. Find all eigenvalues and eigenvectors of $T$. $\textbf{Solution 9.}$ We seek solutions to the equation $\begin{equation*} p' = \lambda p, \end{equation*}$ where $p \in \mathcal{P}(\mathbb{R})$ and $\lambda \in \mathbb{R}$. Because the derivative of every nonconstant polynomial has degree one less than the degree of the polynomial, we see that the only solutions of the equation above are to take $p$ to be a constant polynomial and $\lambda = 0$. Thus 0 is the only eigenvalue of $T$ and the nonzero constant functions are the only eigenvectors. ---  --- Commentary: $\textbf{Exercise 9.}$ Define $T: \mathcal{P}(\mathbb{R}) \to \mathcal{P}(\mathbb{R})$ by $Tp = p'$. Find all eigenvalues and eigenvectors of $T$. $\textbf{Solution 9.}$ We seek solutions to the equation $\begin{equation*} p' = \lambda p, \end{equation*}$ where $p \in \mathcal{P}(\mathbb{R})$ and $\lambda \in \mathbb{R}$. Because the derivative of every nonconstant polynomial has degree one less than the degree of the polynomial, we see that the only solutions of the equation above are to take $p$ to be a constant polynomial and $\lambda = 0$. Thus 0 is the only eigenvalue of $T$ and the nonzero constant functions are the only eigenvectors. $\textit{Commentary:}$ This exercise considers the differentiation operator $T$ on the space of polynomials $\mathcal{P}(\mathbb{R})$. To find the eigenvalues and eigenvectors, we set up the eigenvalue-eigenvector equation $p' = \lambda p$. This is a differential equation, and its solutions depend on the value of $\lambda$. The key observation is that differentiation lowers the degree of a polynomial by 1, unless the polynomial is constant. Therefore, the only way for a polynomial to be an eigenvector is for it to be a constant function (so that its derivative is zero) and for the corresponding eigenvalue to be zero. This example illustrates that eigenvectors are not just vectors in the usual sense, but can be functions or other objects, depending on the context. It also shows that the eigenvalue-eigenvector equation can take forms other than a matrix equation. $\textit{Example:}$ Let $p(x) = 3$ and $q(x) = x^2 + 1$. Then: - $p$ is an eigenvector of $T$ with eigenvalue 0, because $Tp = p' = 0 = 0 \cdot p$. - $q$ is not an eigenvector of $T$, because $Tq = q' = 2x$, which is not a multiple of $q$. In fact, every non-zero constant function is an eigenvector of $T$ with eigenvalue 0, and these are the only eigenvectors of $T$. The eigenspace corresponding to the eigenvalue 0 is the one-dimensional space of constant functions. ---