Spectral norm of projected matrix Let $M_{n,m}$ be the set of real matrices of $n\times m$, and let $T:M_{n,m}\to M_{n,m}$ be a orthogonal projection operator, i.e., $T$ is such that for any $A,B\in M_{n,m}$
$$T(A+B)=T(A)+T(B),$$
$$T(T(A))=T(A),$$
$$\langle T(A),B\rangle = \langle A,T(B)\rangle.$$
where $\langle A,B\rangle=tr(A^{\top}B)$. For $A\in M_{n,m}$ let $\|A\|$ be its spectral norm and $\|A\|_F$ its Frobenius norm. I want to prove that
$$
\|T(A)\|\leq \|A\|.
$$
I've been able to prove that $\|T(A)\|_F\leq \|A\|_F$ which is immediate since $\langle T(A),(I-T)(A)\rangle=0$ which implies that $\|A\|_F=\|T(A)+(I-T)(A)\|_F=\|T(A)\|_F+\|(I-T)(A)\|_F$.
For the spectral norm maybe I can use that $\|A\|=\sup_{\|x\|_2=1}\|Ax\|_2$ but I can't prove that $x^{\top}T(A)^{\top}(I-T)(A)x=0$. 
Any help will be appreciated. 
Edit: If it helps, the form of $T$ is $T(A)=A-P_1AP_2$, where $P_1$ and $P_2$ are some $n\times n$ and $m\times m$ projection matrices. 
This question is motivated by the paper https://arxiv.org/pdf/1011.6256.pdf, proof of Theorem 1, page 9, inequality after equation (2.15).
 A: This is not true. E.g. suppose that $n=m=2$ and $T$ is the orthogonal projection onto the linear span of $B=\operatorname{diag}(6,3)$. Let $A=5I$. Then $T(A)=B$ because $\langle A-B,B\rangle=\langle \operatorname{diag}(-1,2),\operatorname{diag}(6,3)\rangle=0$. However, $\|T(A)\|_2=\|B\|_2=6>5=\|A\|_2$.
Another counterexample: let
\begin{aligned}
&T(X)=X-\pmatrix{1\\ &0}X\pmatrix{1\\ &0},\\
&A=\pmatrix{-1&1\\ 1&1},\ B=T(A)=\pmatrix{0&1\\ 1&1},
\end{aligned}
then $\|T(A)\|_2=\|B\|_2=\frac{1+\sqrt{5}}{2}\approx1.618>1.414\approx\sqrt{2}=\|A\|_2$.

Remark. The inequality that motivated the OP's question, namely,
$$
\|T(A)\|_F\le\sqrt{\operatorname{rank}(T(A))}\,\|A\|_2
$$
where $T(A)=A-P_1AP_2$ for some two orthogonal projectors $P_1$ and $P_2$, is correct. Without loss of generality, we may assume that
$$
A=\pmatrix{X&Y\\ Z&W}
\ \text{ and }\ T(A)=\pmatrix{0&Y\\ Z&W}.
$$
Let $Y=U_y(S_y\oplus0)V_y^T$ and $Z=U_z(S_z\oplus0)V_z^T$ be two singular value decompositions, where $S_y$ and $S_z$ are two positive diagonal matrices. Define on $M_{n,m}$ a linear map
$$
F:B\mapsto\pmatrix{U_y\\ &U_z}^TB\pmatrix{V_z\\ &V_y}.
$$
Then we may write
$$
F(A)=
\left(\begin{array}{c&c|c&c}\ast&\ast&S_y&0\\ \ast&\ast&0&0\\ \hline S_z&0&E&F\\ 0&0&G&H\end{array}\right)
\ \text{ and }\ F(T(A))=
\left(\begin{array}{c&c|c&c}0&0&S_y&0\\ 0&0&0&0\\ \hline S_z&0&E&F\\ 0&0&G&H\end{array}\right).
$$
Let
$$
M_1=\pmatrix{S_y\\ 0\\ 0\\ G}\ \text{ and }\ M_2=\pmatrix{S_z&0&E&F}.
$$
Since $M_1,M_2$ and $H$ are submatrices of $F(A)$, their spectral norms are bounded above by $\|F(A)\|_2$. It follows that
\begin{aligned}
\|T(A)\|_F^2
&=\|F(T(A))\|_F^2\\
&=\|M_1\|_F^2+\|M_2\|_F^2+\|H\|_F^2\\
&\le\operatorname{rank}(M_1)\|M_1\|_2^2+\operatorname{rank}(M_2)\|M_2\|_2^2+\operatorname{rank}(H)\|H\|_2^2\\
&\le\left(\operatorname{rank}(M_1)+\operatorname{rank}(M_2)+\operatorname{rank}(H)\right)\|F(A)\|_2^2\\
&=\operatorname{rank}(F(T(A)))\|F(A)\|_2^2\\
&=\operatorname{rank}(T(A))\|A\|_2^2.
\end{aligned}
A: 
As the other poster has shown, this is not true in general but it is
  true in the special case where $T$ takes the form $T: A \mapsto PA $;
  i.e. $T$ is left multiplication by some matrix $n \times n $ matrix $P$. As
  you have pointed out there are more general tensors that satisfy the
  definition, but it is true in this special case.

The spectral norm of $A$ for a real matrix is equal to 
\begin{equation}
|A| = \sup_{|x| = 1} \|Ax\| = \sup_{x \neq 0 } \frac{\|Ax\|}{\|x\|}.
\end{equation} 
Suppose for a contradiction that 
\begin{equation}
|TA|   > |A|
\end{equation}
then using the vector norm $\| \| $ there is some $v$ such that 
\begin{equation}
\|T(Av)\| =\|(TA)v\|   > \|Av\|
\end{equation}
and therefore setting $x=Av$
we have that 
\begin{equation}
 \|Tx\|^{2}=\langle Tx,Tx\rangle =\langle Tx,x\rangle \leq \|Tx\|\cdot \|x\|
\end{equation}
and therefore we have that 
\begin{equation}
 \|Tx\| \leq  \|x\|
\end{equation}
and that 
\begin{equation}
 \|Tx\| >  \|x\|
\end{equation}
which is a contradiction. QED
