Prove there exists an inner product that makes an idempotent linear map become a projection operator

Question

Suppose $L$ is a linear map such that $L(L(v)) = L(v)$ for all $v \in V$ and suppose $W=Range(L)$ is a subspace of $V$, prove there exists an inner product such that $L={\rm proj}_{W}$. Anyone have idea on how to prove this question?

Answer 1

Since $L^2 = L$, we have that the minimal polynomial $m_L(x)$ of $L$ divides $x^2-x = x(x-1)$.

1. If $m_L(x) = x$, then $L=0$, so $W=0$ and any inner product does the job.

2. If $m_L(x) = x-1$, then $L = {\rm Id}_V$, so $W = V$ and any inner product does the job.

3. If $m_L(x)=x(x-1)$, we have that $V$ is the direct sum of the eigenspaces associated to $0$ and $1$ of $L$. They are, respectively, $\ker L$ and $W$. That is, $V = W \oplus \ker L$. Choose any inner product $\langle \cdot,\cdot\rangle_W$ on $W$, any inner product $\langle \cdot,\cdot\rangle_{\ker L}$ on $\ker L$, and declare the inner product $\langle\cdot,\cdot\rangle_V$ on $V$ to be the direct sum of those inner products. More precisely, $$\langle w_1+k_1,w_2+k_2\rangle_V \doteq \langle w_1,w_2\rangle_W + \langle k_1,k_2\rangle_{\ker L},$$for $w_1,w_2\in W$ and $k_1,k_2\in \ker L$. This automatically makes $W$ and $\ker L$ orthogonal, and $L$ becomes ${\rm pr}_W$. You have a lot of freedom to choose the inner products in $W$ and $\ker L$. The inner product on $V$ making $L = {\rm pr}_W$ is far from unique.