Norm of the trimming map Let $M_n$ denote the linear space of $n\times n$ matrices over $\mathbb{C}$. Consider $M_n$ as a normed space with operator norm. Is it true that the trimming map
$$
T:M_n\to M_n:
\begin{pmatrix}
a_{1,1} & a_{1,2} & \ldots & a_{1,n} \\
a_{2,1} & a_{2,2} & \ldots & a_{2,n} \\
\ldots  & \ldots  & \ldots & \ldots  \\
a_{n,1} & a_{n,2} & \ldots & a_{n,n} \\
\end{pmatrix}
\mapsto
\begin{pmatrix}
a_{1,1} & a_{1,2} & \ldots & a_{1,n} \\
0       & a_{2,2} & \ldots & a_{2,n} \\
\ldots  & \ldots  & \ldots & \ldots  \\
0       & 0       & \ldots & a_{n,n} \\
\end{pmatrix}
$$
has norm 1? I did a few numerical tests and got that $T$ is indeed contractive, but I don't know how to prove this.

*

*$T$ is a contraction for the operator $\left\|\cdot\right\|_p$ norms for $p=1$ and $p = \infty$ since those \begin{align}
\left\|A\right\|_1 &= \max_{j}\sum_{i=1}^n \left|a_{i,j}\right| &\text{and}\quad \left\|A\right\|_\infty &= \max_{i}\sum_{j=1}^n \left|a_{i,j}\right|. 
\end{align}


*However $T$ is not a contraction for the $\left\|\cdot\right\|_2$. For example take $A = \begin{pmatrix}1 & 1\\ -1 & 1\end{pmatrix}$, \begin{align}
\left\|A\right\|_2 &= \sqrt{\lambda_{\max}\left(A^*A\right)}\\ 
&=\sqrt{\lambda_{\max}\left(\begin{pmatrix}2 & 0\\ 0 & 2\end{pmatrix}\right)}\\
&= \sqrt{2}\end{align} and $T(A) = \begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}$ so
\begin{align}
\left\|T(A)\right\|_2 &= \sqrt{\lambda_{\max}\left(T(A)^*T(A)\right)} \\
&= \sqrt{\lambda_{\max}\left(\begin{pmatrix}1 & 1\\ 1 & 2\end{pmatrix}\right)} = \sqrt{\frac{3+\sqrt5}{2}} > \sqrt2\end{align}
