How to prove that $P_{w^\perp}=I-P_{w}$ for the following condition? Let $V$ be a vector space over $\Bbb{C}$ with inner product $\langle\cdot,\cdot  \rangle$. Let $W$ be a subspace of $V$ and let $P_W:V\to V$ be an orthogonal projection onto $W$ and let $P_{w^\perp}:V\to V$ be the orthogonal projection onto $W^{\perp}$. How to prove that $P_{w^\perp}=I-P_{w}$ for the following condition?
For the question I have no idea how to start. All I know that associated with projection is the calculation $P_w(x)=\sum^k_{i=1} \langle x,w_i\rangle w_i$. Could someone help?

Each vector $x\in V$ can be written in a unique manner $x=v+w$ with $v\in W$ and $w\in W^\bot$. Since $v=P_Wx$ and $w=P_{W^\bot}x$, we obtain
$$
P_{W^\bot}x=x-P_Wx=(I-P_W)x.
$$
