Let $V$ and $W$ be subspaces of $\mathbb{R}^n$ and $P_V$ be orthogonal projection on $V$. The linear operator $T:V \to V$ is the restriction of $P_V\circ P_W$ to $V$. Prove that matrix associated with $T$ is symmetric in every orthonormal basis of $V$.
My try: The matrix of orthogonal projection is symmetric but product of symmetric matrices isn't necessarily symmetric and so this not works. I think we should use direct sum here but don't know how it can be used.
Consider the symmetric matrix $S = P_VP_WP_V$. Then $V$ is an invariant subspace of $S$ and $T = S|_V$. So, $T$ is symmetric in $V$.
Or directly: If $u,v\in V$, then $$ \langle P_VP_Wu,v\rangle = \langle P_Wu,P_Vv\rangle = \langle P_Wu,v\rangle = \langle u,P_Wv\rangle = \langle P_Vu,P_Wv\rangle = \langle u,P_VP_Wv\rangle. $$ Hence, $\langle Tu,v\rangle = \langle u,Tv\rangle$.
Now, if $(e_n)$ is an orthonormal basis of $V$, then the matrix representation of $T$ with respect to this basis is $(\langle Te_j,e_i\rangle)_{i,j=1}^r$, which is clearly symmetric.