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?

Hint:
for any vector $v$ the orthogonal decomposition theorem say that:
$$
v=P_w(v)+P_{w^\perp}(v)
$$
