This is not a norm!

Question

Define $\,\nu (T) = \max\limits_{1\leqslant k\leqslant n}ks_k(T)$, where $(s_k)$ are the singular values of the matrix $T\in M_n$.
I'd like to show that $\nu$ it is not a norm.

I am trying to find a counterexample, but it doesn't work.

Answer 1

Let's call your supposed norm $\nu$.

We want the triangle inequality to fail. Since $s_1(T)$ gives the operator norm, you are looking for a matrix $T$ such that $2s_2(T)>s_1(T)$. For instance $$ T=\begin{bmatrix} 4&0\\0&4\end{bmatrix}, $$ where $\nu(T)=2s_2(T)=8$. We may write $T$ as the sum of two matrices: $$ T=\begin{bmatrix} 3&0\\0&2\end{bmatrix}+\begin{bmatrix} 1&0\\0&2\end{bmatrix}. $$ If we call $T_1,T_2$ the two matrices above, we have $$ \nu(T)=8>4+2=\nu(T_1)+\nu(T_2). $$

Answer 2

You could take two non-commuting nilpotent matrices : their singular values are zero but because of the non-commutativity, the kernel of the sum might be chosen trivial.

Answer 3

Consider $A=\big(\begin{smallmatrix}0&h\\1&0\end{smallmatrix}\big)$, having singular values $\big\{|h|,1\big\}$. What are the singular values of $A+A^T$?

Can you arrange for $h$ such that subadditivity does not hold?
I.e. $\,\nu (A+A^T)\not\leq\nu(A) + \nu(A^T)$