$\textbf{Exercise 12.}$ Suppose $V = U \oplus W$, where $U$ and $W$ are nonzero subspaces of $V$. Define $P \in \mathcal{L}(V)$ by $P(u + w) = u$ for each $u \in U$ and each $w \in W$. Find all eigenvalues and eigenvectors of $P$. $\textbf{Solution 12.}$ Suppose $\lambda \in \mathbb{F}$ is an eigenvalue of $P$. Then there exists a nonzero vector $v \in V$ such that $Pv = \lambda v$. Writing this equation using the representation $v = u + w$ with $u \in U$ and $w \in W$, we have $u = \lambda(u + w)$. Thus $\begin{equation*} (1 - \lambda)u - \lambda w = 0. \end{equation*}$ Because $V = U \oplus W$, if 0 is written as the sum of a vector in $U$ and a vector in $W$, then both vectors are 0. Thus the equation above implies that $(1 - \lambda)u = \lambda w = 0$. Because $u$ and $w$ are not both 0 (because $v \neq 0$), this implies that $\lambda = 1$ or $\lambda = 0$. For $v \in V$ with representation as above, the equation $Pv = 0$ is equivalent to the equation $u = 0$, which is equivalent to the equation $v = w$, which is equivalent to the statement that $v \in W$. This means that 0 is an eigenvalue of $P$ (because $W$ is a nonzero subspace of $V$) and that the set of nonzero vectors in $W$ equals the set of eigenvectors corresponding to the eigenvalue 0. For $v \in V$ with representation as above, the equation $Pv = v$ is equivalent to the equation $v = u$, which is equivalent to the statement that $v \in U$. This means that 1 is an eigenvalue of $P$ (because $U$ is a nonzero subspace of $V$) and that the set of nonzero vectors in $U$ equals the set of eigenvectors corresponding to the eigenvalue 1. --- : $\textbf{Exercise 12.}$ Suppose $V = U \oplus W$, where $U$ and $W$ are nonzero subspaces of $V$. Define $P \in \mathcal{L}(V)$ by $P(u + w) = u$ for each $u \in U$ and each $w \in W$. Find all eigenvalues and eigenvectors of $P$. $\textbf{Solution 12.}$ Suppose $\lambda \in \mathbb{F}$ is an eigenvalue of $P$. Then there exists a nonzero vector $v \in V$ such that $Pv = \lambda v$. Writing this equation using the representation $v = u + w$ with $u \in U$ and $w \in W$, we have $u = \lambda(u + w)$. Thus $\begin{equation*} (1 - \lambda)u - \lambda w = 0. \end{equation*}$ Because $V = U \oplus W$, if 0 is written as the sum of a vector in $U$ and a vector in $W$, then both vectors are 0. Thus the equation above implies that $(1 - \lambda)u = \lambda w = 0$. Because $u$ and $w$ are not both 0 (because $v \neq 0$), this implies that $\lambda = 1$ or $\lambda = 0$. For $v \in V$ with representation as above, the equation $Pv = 0$ is equivalent to the equation $u = 0$, which is equivalent to the equation $v = w$, which is equivalent to the statement that $v \in W$. This means that 0 is an eigenvalue of $P$ (because $W$ is a nonzero subspace of $V$) and that the set of nonzero vectors in $W$ equals the set of eigenvectors corresponding to the eigenvalue 0. For $v \in V$ with representation as above, the equation $Pv = v$ is equivalent to the equation $v = u$, which is equivalent to the statement that $v \in U$. This means that 1 is an eigenvalue of $P$ (because $U$ is a nonzero subspace of $V$) and that the set of nonzero vectors in $U$ equals the set of eigenvectors corresponding to the eigenvalue 1. $\textit{Commentary:}$ This exercise considers a projection operator $P$ on a vector space $V$ that is the direct sum of two subspaces $U$ and $W$. The operator $P$ is defined by $P(u + w) = u$, i.e., it projects a vector onto the subspace $U$ along the subspace $W$. The exercise asks to find the eigenvalues and eigenvectors of $P$. The solution proceeds by considering an eigenvector equation $Pv = \lambda v$ and expressing $v$ as $u + w$. This leads to the equation $(1 - \lambda)u - \lambda w = 0$, which implies that either $\lambda = 1$ (and $v = u \in U$) or $\lambda = 0$ (and $v = w \in W$). Therefore, the eigenvalues of $P$ are 0 and 1, with the eigenvectors being the non-zero vectors in $W$ and $U$ respectively. This example illustrates the eigenstructure of a projection operator and its relation to the direct sum decomposition of the space. $\textit{Example:}$ Let $V = \mathbb{R}^2$, $U = \{(x, 0) : x \in \mathbb{R}\}$ (the $x$-axis), and $W = \{(0, y) : y \in \mathbb{R}\}$ (the $y$-axis). Then $V = U \oplus W$. The projection $P$ onto $U$ along $W$ is given by $P(x, y) = (x, 0)$. - For $\lambda = 1$, the eigenvectors are all non-zero vectors in $U$, i.e., vectors of the form $(x, 0)$ with $x \neq 0$. For example, $(1, 0)$ is an eigenvector with eigenvalue 1, because $P(1, 0) = (1, 0) = 1 \cdot (1, 0)$. - For $\lambda = 0$, the eigenvectors are all non-zero vectors in $W$, i.e., vectors of the form $(0, y)$ with $y \neq 0$. For example, $(0, 1)$ is an eigenvector with eigenvalue 0, because $P(0, 1) = (0, 0) = 0 \cdot (0, 1)$. Every vector in $\mathbb{R}^2$ can be uniquely written as a sum of an eigenvector for $\lambda = 1$ (its component in $U$) and an eigenvector for $\lambda = 0$ (its component in $W$).