Norm of the trimming map

Question

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.

Answer 1

• $T$ 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}