How would you interpret $P_{V^{\perp}}m$ in the following extract?
Given is a vector $v\in \mathbb{R}^n$ and a symmetric tracefree $n \times n$-matrix $m$. Let $V^{\perp}$ be the orthogonal complement of the span of $v$ in $\mathbb{R}^n$ and $P_{V^{\perp}}$ be the orthogonal projection from $\mathbb{R}^n$ onto $V^{\perp}$. Since $m$ is self-adjoint, then so is the restriction of $P_{V^{\perp}}m$ to $V^{\perp}$. Hence there exists at least one eigenvector of the operator.
I try to understand the last sentence. I read that every selfadjoint transformation $T:X \rightarrow X$ ($X$ is an Euclidean space) has a real eigenvextor. But I am not sure whether this is here applicable because I thought $P_{V^{\perp}}m$ is the composition $P_{V^{\perp}} \circ f: \mathbb{R}^n \rightarrow \mathbb{R}^n \rightarrow V^{\perp}$ for a linear map $f(w) = mw$.
The restriction of $P_{V^{\perp}} \circ f$ to $V^{\perp}$ is well defined as for $x \in V^{\perp}$, $m(x) \in \mathbb R^n$ and $(P_{V^{\perp}} \circ f)(x) \in V^{\perp}$.
Then this restriction is self-adjoint as both $P_{V^\perp}$ and $f$ are self-adjoint. Therefore the restriction of $P_{V^{\perp}} \circ f$ to $V^\perp$ has an eigenvector. By the spectral theorem, it is even diagonalizable.