Let T be a linear operator on a finite-dimensional inner product space V. If T is a projection such that $\|T(x)\| \leq \|x\|$ for all $x \in V$. Prove that T is an orthogonal projection.
I want to prove by contradiction, assume T is not an orthogonal projection, that means $Ker T \neq (im T)^\perp$. I am not sure how to find a vector such that $\|T(v)\|>\|v\|$. Any help is appreciated.
Suppose $\ker T$ is not orthogonal to $\operatorname{im}(T)$, there exists $u\in \ker T$, $v\in \operatorname{im}T$ with $\langle u,v\rangle\neq 0$.
Write $x_c=u+cv$, $\|x_c\|^2=\|u\|^2+2c\langle u,v\rangle+c^2\|v\|^2$
$\|T(x_c)\|^2=c^2\|v\|^2$
$\|T(x_c)\|^2>\|x\|^2$ is equivalent to $2c\langle u,v\rangle +\|u\|^2<0$ the last inequality has always a solution if $\langle u,v\rangle\neq 0$.
