$\textbf{Exercise 10.}$ Define $T \in \mathcal{L}(\mathcal{P}_4(\mathbb{R}))$ by $(Tp)(x) = xp'(x)$ for all $x \in \mathbb{R}$. Find all eigenvalues and eigenvectors of $T$. $\textbf{Solution 10.}$ A typical element $p$ of $\mathcal{P}_4(\mathbb{R})$ is given by expression $\begin{equation*} p(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4, \end{equation*}$ where $a_0, a_1, a_2, a_3, a_4 \in \mathbb{R}$. With that expression, the eigenvalue-eigenvector equation $Tp = \lambda p$, which in this case is $xp'(x) = \lambda p(x)$, becomes $\begin{equation*} a_1x + 2a_2x^2 + 3a_3x^3 + 4a_4x^4 = \lambda(a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4). \end{equation*}$ Comparing coefficients in the equation above shows the eigenvalue-eigenvector equation is equivalent to the system of equations $\begin{align*} 0 &= \lambda a_0 \\ a_1 &= \lambda a_1 \\ 2a_2 &= \lambda a_2 \\ 3a_3 &= \lambda a_3 \\ 4a_4 &= \lambda a_4. \end{align*}$ From the equations above, it is clear that if $j \in \{0, 1, 2, 3, 4\}$ and $a_j \neq 0$, then $\lambda = j$ and $a_k = 0$ for each $k \neq j$. Thus the eigenvalues of $T$ are $0, 1, 2, 3, 4$ and the corresponding eigenvectors are of the form $c$, $cx$, $cx^2$, $cx^3$, $cx^4$, where $c \in \mathbb{R}$ and $c \neq 0$. ---  --- Commentary: $\textbf{Exercise 10.}$ Define $T \in \mathcal{L}(\mathcal{P}_4(\mathbb{R}))$ by $(Tp)(x) = xp'(x)$ for all $x \in \mathbb{R}$. Find all eigenvalues and eigenvectors of $T$. $\textbf{Solution 10.}$ A typical element $p$ of $\mathcal{P}_4(\mathbb{R})$ is given by expression $\begin{equation*} p(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4, \end{equation*}$ where $a_0, a_1, a_2, a_3, a_4 \in \mathbb{R}$. With that expression, the eigenvalue-eigenvector equation $Tp = \lambda p$, which in this case is $xp'(x) = \lambda p(x)$, becomes $\begin{equation*} a_1x + 2a_2x^2 + 3a_3x^3 + 4a_4x^4 = \lambda(a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4). \end{equation*}$ Comparing coefficients in the equation above shows the eigenvalue-eigenvector equation is equivalent to the system of equations $\begin{align*} 0 &= \lambda a_0 \\ a_1 &= \lambda a_1 \\ 2a_2 &= \lambda a_2 \\ 3a_3 &= \lambda a_3 \\ 4a_4 &= \lambda a_4. \end{align*}$ From the equations above, it is clear that if $j \in \{0, 1, 2, 3, 4\}$ and $a_j \neq 0$, then $\lambda = j$ and $a_k = 0$ for each $k \neq j$. Thus the eigenvalues of $T$ are $0, 1, 2, 3, 4$ and the corresponding eigenvectors are of the form $c$, $cx$, $cx^2$, $cx^3$, $cx^4$, where $c \in \mathbb{R}$ and $c \neq 0$. $\textit{Commentary:}$This exercise considers an operator $T$ on the space of polynomials of degree at most 4, defined by $Tp = xp'$. This operator multiplies a polynomial by $x$ and then differentiates it. To find the eigenvalues and eigenvectors, we set up the eigenvalue-eigenvector equation $xp' = \lambda p$ and compare coefficients. This leads to a system of equations where each coefficient must be an eigenvector of the corresponding eigenvalue. The solutions show that the eigenvalues are the integers from 0 to 4, and the eigenvectors are the monomials (multiplied by a non-zero constant). This example illustrates how the eigenvalue-eigenvector problem can be solved by comparing coefficients, and how the eigenspaces are related to the standard basis of monomials for the polynomial space. $\textit{Example:}$ Let $p(x) = 1 + x + x^2 + x^3 + x^4$. Then: - $1$ is an eigenvector of $T$ with eigenvalue 0, because $T(1) = x \cdot 0 = 0 = 0 \cdot 1$. - $x$ is an eigenvector of $T$ with eigenvalue 1, because $T(x) = x \cdot 1 = x = 1 \cdot x$. - $x^2$ is an eigenvector of $T$ with eigenvalue 2, because $T(x^2) = x \cdot 2x = 2x^2 = 2 \cdot x^2$. - $x^3$ is an eigenvector of $T$ with eigenvalue 3, because $T(x^3) = x \cdot 3x^2 = 3x^3 = 3 \cdot x^3$. - $x^4$ is an eigenvector of $T$ with eigenvalue 4, because $T(x^4) = x \cdot 4x^3 = 4x^4 = 4 \cdot x^4$. However, $p$ itself is not an eigenvector of $T$, because $Tp = x(1 + 2x + 3x^2 + 4x^3) = x + 2x^2 + 3x^3 + 4x^4$, which is not a multiple of $p$. This is because $p$ is a sum of eigenvectors corresponding to different eigenvalues.