"Let $T$ be a projection operator on a finite dimensional inner product space $V$ such that $\|T(x)\| \leq \|x\|$ for all $x \in V$. Prove that $T$ is an orthogonal projection."
If I try to prove this by way of contradiction, what is the particular element in $V$ that I can take such that the given inequality doesn't hold? Any other way of solving this problem is also welcome.
Let $W = \ker(T)$ and $U = \operatorname{im}(T)$ so that $V = W \oplus U$ (because $T$ is a projection). To prove that $T$ is orthogonal, it is enough to show that $U \subseteq W^{\perp}$ because $\dim U = \dim V - \dim W = \dim W^{\perp}$ so if $U \subseteq W^{\perp}$ then $U = W^{\perp}$.
Let $u \in U$ and $w' \in W$. Decompose $u$ as $u = w + w^{\perp}$ for $w \in W$ and $w^{\perp} \in W^{\perp}$. Then
$$ \| w \|^2 + \| w^{\perp} \|^2 = \| u \|^2 = \| Tu \|^2 = \| Tw^{\perp} \|^2 \leq \| w^{\perp} \|^2 $$
which implies that $\| w \|^2 = 0$ so $w = 0$ and $u = w^{\perp} \in W^{\perp}$.
