Let $W$ be a finite-dimensional subspace of an inner product space $V$. Show that if $T$ is the orthogonal projection of $V$ onto $W$, then $I-T$ is the orthogonal projection of $V$ onto $W^{\perp}$.
I'm pretty sure I want to be using the Spectral Theorem here. Specifically, I think I need to let $U$ be a normal (or self-adjoint) linear operator on $V$ and use either the resolution of the identity operator or the spectral decomposition of $U$, but I have no idea how to proceed.
Although reasoning about the spectral decomposition of $W$ would work, it might be overkill here. Instead, try representing an arbitrary vector $v \in V$ as the unique linear combination $v = w + u$ where $w \in W$ and $w \in W^\perp$. Then, let's consider how the linear transformation $I - T$ acts on the vector $v$.
We can get started by $$ (I - T) v = v - Tv = v - T(w + u) = v - w = (w + u) - w = u$$ I'll let you verify why these steps are correct. What does this tell us about the linear transformation $I-T$?