$\textbf{Exercise 8.}$ Suppose $P \in \mathcal{L}(V)$ is such that $P^2 = P$. Prove that if $\lambda$ is an eigenvalue of $P$, then $\lambda = 0$ or $\lambda = 1$.
$\textbf{Solution 8.}$ Suppose $\lambda$ is an eigenvalue of $P$. Thus there exists $v \in V$ such that $v \neq 0$ and
$\begin{equation*}
Pv = \lambda v.
\end{equation*}$
Thus
$\begin{align*}
\lambda v &= Pv \\
&= P^2v \\
&= P(Pv) \\
&= P(\lambda v) \\
&= \lambda Pv \\
&= \lambda^2 v.
\end{align*}$
The equation above implies that $(\lambda^2 - \lambda)v = 0$. Thus $\lambda^2 - \lambda = 0$, which implies that $\lambda = 0$ or $\lambda = 1$.
---
$ is such that $P^2 = P$. Prove that if $\lambda$ is an eigenvalue of $P$, then $\lambda = 0$ or $\lambda = 1$.
$\textbf{Solution 8.}$ Suppose $\lambda$ is an eigenvalue of $P$. Thus there exists $v \in V$ such that $v \neq 0$ and
$\begin{equation*}
Pv = \lambda v.
\end{equation*}$
Thus
$\begin{align*}
\lambda v &= Pv \\
&= P^2v \\
&= P(Pv) \\
&= P(\lambda v) \\
&= \lambda Pv \\
&= \lambda^2 v.
\end{align*}$
The equation above implies that $(\lambda^2 - \lambda)v = 0$. Thus $\lambda^2 - \lambda = 0$, which implies that $\lambda = 0$ or $\lambda = 1$.
$\textit{Commentary:}$ This exercise deals with a special type of linear operator $P$ satisfying $P^2 = P$, known as a projection operator. The condition $P^2 = P$ means that applying $P$ twice is the same as applying it once. The exercise shows that the only possible eigenvalues for such an operator are 0 and 1. The proof follows by considering an eigenvector $v$ with eigenvalue $\lambda$ and using the condition $P^2 = P$ to show that $\lambda$ must satisfy $\lambda^2 = \lambda$. This quadratic equation has only two solutions: 0 and 1. This result is fundamental in the study of projection operators and their spectral properties. In geometric terms, a projection operator "projects" vectors onto a subspace (the eigenspace corresponding to $\lambda = 1$) along another subspace (the eigenspace corresponding to $\lambda = 0$).
$\textit{Example:}$ Let $V = \mathbb{R}^2$ and define $P: \mathbb{R}^2 \to \mathbb{R}^2$ by $P(x, y) = (x, 0)$. Geometrically, $P$ projects vectors onto the $x$-axis along the $y$-axis. You can check that $P^2 = P$. The eigenvalues of $P$ are indeed 0 and 1:
- For $\lambda = 1$, the eigenvectors are all non-zero vectors of the form $(x, 0)$, which span the $x$-axis. For example, $(1, 0)$ is an eigenvector with eigenvalue 1, because $P(1, 0) = (1, 0)$.
- For $\lambda = 0$, the eigenvectors are all vectors of the form $(0, y)$, which span the $y$-axis. For example, $(0, 1)$ is an eigenvector with eigenvalue 0, because $P(0, 1) = (0, 0)$.
Every vector in $\mathbb{R}^2$ can be uniquely written as a sum of an eigenvector for $\lambda = 1$ and an eigenvector for $\lambda = 0$, corresponding to the decomposition of a vector into its projection onto the $x$-axis and its projection onto the $y$-axis.
---