Prove that $T$ is an orthogonal projection Let $T$ be a linear operator on a finite-dimensional inner product space $V$. Suppose that $T$ is a projection such that $\|T(x)\| \le \|x\|$ for $x \in V$. Prove that $T$ is an orthogonal projection.  
I can't understand well. The definition of orthogonal operator is $\|T(x)\| = \|x\|$. But why that $\|T(x)\| \le \|x\|$ for $x \in V$ means orthogonal projection?
 A: Note that $\|Tx\|=\|x\|$ is the definition of an isometry. Over a finite-dimensional real inner product space, this is equivalent to the matrix of $T$ in an orthonormal basis being "orthogonal", i.e. $A^TA=AA^T=I_n$.
Orthogonal projection means $T^2=T$ and $T^*=T$, i.e. self-adjoint idempotent. The first thing you should remark is that this is equivalent to $T$ being idempotent with $\ker T \perp \mbox{im} T$. Indeed, if $T$ is idempotent, then $T^*$ is idempotent with $\ker T^*=(\mbox{im} T)^\perp$ and $\mbox{im} T^*=(\ker T)^\perp$.
Claim: if $T$ is idempotent and $\|T\|\leq 1$ (i.e. $\|Tx\|\leq \|x\|$ for all $x\in V$), then $T$ is self-adjoint (i.e. $T$ is an orthogonal projection).
Remark: the converse is true since then the orthogonal direct sum  $V=\ker T\oplus \mbox{im} T$ yields $\|x+y\|^2=\|x\|^2+\|y\|^2\geq \|y\|^2=\|0+Ty\|^2=\|T(x+y)\|^2$ for all $x\in \ker T$ and all $y\in \mbox{im} T$. Note that all this holds on a general inner product space. No need to assume finite dimension.
Proof: we need to prove that $(x,y)=0$ for every $x\in \ker T$ and every $y\in\mbox{im} T$. Let us take two such vectors, which are characterized by $Tx=0$ and $Ty=y$. Then for every $t\in\mathbb{R}$
$$
t^2\|y\|^2=\|ty\|^2=\|T(x+ty)\|^2\leq \|x+ty\|^2=\|x\|^2+2t\,\mbox{Re}(x,y)+t^2\|y\|^2
$$
whence
$$
\mbox{Re}(x,y)\geq -\frac{\|x\|^2}{2t}\;\forall t>0\qquad\mbox{and}\qquad \mbox{Re}(x,y)\leq -\frac{\|x\|^2}{2t}\;\forall t<0
$$
which implies $\mbox{Re}(x,y)=0$ by letting $t$ tend to $\pm \infty$. In the real case, we are done since $\mbox{Re}(x,y)=(x,y)=0$. In the complex case, take $e^{i\theta}$ such that $|(x,y)|=e^{i\theta}\mbox{Re}(x,y)=\mbox{Re}(x,e^{i\theta}y)$ and apply the above to $x,e^{i\theta}y$ to conclude that $|(x,y)|=0$ whence $(x,y)=0$. QED.
