If $T$ is a self-adjoint linear operator on an inner product space $V$ such that $T\circ T=T$ then $T$ is projection on $Image(P)$.

Question

Let $T$ be a self-adjoint linear operator on an inner product space $V$ such that $T\circ T=T$. Let $W=Image\:T$ Prove that for all $v\in V$, $T(v)=proj_W(v)$ ie $T$ projects $v$ onto it's image.

Now I have some intuitions and some implications.$T$ has eigenvalue 1 since $$T(v)=\lambda v\implies \lambda v = T(v)=T\circ T(v)=\lambda^2v\implies \lambda=0,1$$ Another thing is that if our form on the space $V$ is non-degenrate, then $P=W\oplus W^{\perp}$. Intuition also tells me to use the Spectral Theorem to form an orthonormal basis of eigenvectors of $T$ to help easing computation. How do I continue from here?

Answer 1

You are correct that all eigenvalues are either $1$ or $0$. Suppose $$v=\alpha_1v_1+\cdots+\alpha_mv_m+\alpha_{m+1}v_{m+1} + \cdots+\alpha_nv_n,$$ where the first $m$ eigenvalues are $1$ and the rest are $0$ and $v_i$ are the corresponding eigenvectors, all chosen orthogonal. Then (explain why) $$Tv=\alpha_1v_1+\cdots+\alpha_mv_m.$$ On the other hand, the projection $Pv$ of $v$ is the unique vector in $W$ such that $v-Pv$ is orthogonal to all vectors in $W$. Note that $$v-Tv=\alpha_{m+1}v_{m+1} + \cdots+\alpha_nv_n$$ and is therefore orthogonal to all vectors in $W$. So $Tv$ must be $Pv$.

Answer 2

I think the main thing is that the image and the kernel are orthogonal. Let $v \in \ker T,$ $ w \in Im T,$ then $w= Ty,$ and $$ \langle v ,w \rangle = \langle v, Ty \rangle = \langle Tv, y \rangle = 0. $$ In the finite dimensional case, the rest follows from dimension considerations.