Are there norms on $\Bbb{C}^m$ and $\Bbb{C}^n$ so that the norm $\Vert\cdot\Vert$ is a subordinate norm?

Question

Denote $$\Vert A\Vert=\sum_{1\le j,k\le m}\vert A_{j,k}\vert$$ is cleary a norm over $M_{m,n}(\Bbb{C})$ but not a subordinate norm by taking the identity matrix $I$.

So my question is: Can we make this norm a subordinate norm by changing our vector space?

If we 'remove' the identity matrix I think we lose our vector space structure, for example in $\Bbb{R}^2$ we would lose the possibility to join two points.

I hope I am clear.

EDIT (Thanks to Joonas Ilmavirta): Apparently I was unclear, the question can be written :

Are there norms on $\Bbb{C}^m$ and $\Bbb{C}^n$ so that the norm $\Vert\cdot\Vert$ is a subordinate norm?

EDIT $2$ (Thanks 2 to Joonas Ilmavirta): To make the problem more easier one can add assumption to the norm : strict convexity.

Answer 1

No, there are not.

Let $\|\cdot\|$ be a norm subordinate to an arbitrary pair of vector norms $\|\cdot\|_*$, $\|\cdot\|_{**}$: $$\|A\| = \max_{x\ne 0}\frac{\|Ax\|_*}{\|x\|_{**}}.$$ Then for rank-one matrices $$\|uv^H\| = \max_{x\ne 0}\frac{\|uv^Hx\|_*}{\|x\|_{**}} = \max_{x\ne 0}\frac{\|u\|_*\cdot\langle x, v\rangle}{\|x\|_{**}} = \|u\|_* \cdot \|v\|^D_{**},$$ where $\|\cdot\|^D_{**}$ denotes vector norm dual to $\|\cdot\|_{**}$. We have for our particular matrix norm $$\|uv^H\| = \|u\|_1 \|v\|_1,$$ thus $\|\cdot\|_*=\|\cdot\|_1$ and $\|\cdot\|_{**}=\|\cdot\|_{\infty}$ (up to rescaling). So if we assume $\|\cdot\|$ is an operator norm, it necessarily has the form $$\max_{x\ne 0}\frac{\|Ax\|_1}{\ \|x\|_{\infty}}.$$

To find a contradiction, it is now enough to consider $2\times 2$ matrices (they can be embedded into $m\times n$ matrices without change of norm). Consider $$A = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}.$$ Evidently $$ \max_{\max\{|a|,|b|\}=1} \big(|a-b| + |a+b|\big) < 4 = \|A\|, $$ thus we finally obtain a contradiction.

Update. Since posting this answer I have published a paper about distinguishing subordinate norms. According to my research, it is enough to find gradient of $\|\cdot\|$ having rank greater than 1 to prove the norm can't be subordinate. That's easy: for the matrix $A$ considered above the gradient of the norm considered is $$\begin{bmatrix}1 & -1\\ 1 & 1\end{bmatrix}$$ and has rank 2.

Answer 2

Of course, if $n=1$ or $m=1$, the linear maps are essentially vectors and the $\ell^1$ norms on $\mathbb C^n$ and $\mathbb C^m$ make the norm $\|\cdot\|$ into an operator norm.

I have the following partial answers: The answer is "no" if $\min\{n,m\}>1$, if the norm on the domain is the Euclidean one. Also, it is impossible to give a norm $|\cdot|_k$ to each $\mathbb C^k$ so that all matrices have operator norm equal to the OP's matrix norm.

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Euclidean norm on the domain: $\newcommand{\R}{\mathbb R}\newcommand{\C}{\mathbb C}$ Assume then $n,m\geq2$. For simplicity, take $n=m=2$ (this also covers the general case by a restriction argument). Suppose there were norms $\alpha:\C^m\to\R$ and $\beta:C^n\to\R$ so that for all matrices $A$ $$ \sum_{j,k}|A_{j,k}|=\max_{x\neq 0}\frac{\beta(Ax)}{\alpha(x)}. $$ Note that the maximum is always reached because of continuity.

Let $y$ and $z$ be nonzero vectors and consider matrices of the form $yz^T$. Now $\beta(yz^Tx)=|z\cdot x|\beta(y)$. Assuming $\alpha$ to be the Euclidean norm, this gives $\beta(yz^Tx)\leq\alpha(z)\alpha(x)\beta(y)$ with equality iff $x$ is a positive scalar multiple of $\bar z$. Thus $\|yz^T\|=\alpha(z)\beta(y)$, so the norm $\beta$ has to satisfy $\beta(y)=\|yz^T\|/\alpha(z)$. This in particular implies that $\|yz^T\|$ is independent of $z$ if on the sphere $\alpha(z)=1$. Let $z_1=(1,0,\dots,0)$ and $z_2=\frac1{\sqrt2}(1,1,0\dots,0)$. Now $$ \|yz_1^T\| = \sum_k|y_k| \neq \sqrt2\sum_k|y_k| = \|yz_2^T\|, $$ which is a contradiction.

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Family of norms: Suppose each $C^k$ has a norm $|\cdot|_k$ so that the OP's norm is an operator norm for any $m\times n$ complex matrix. Then in particular the identity matrix $I_n$ satisfies $k=\|I_k\|=1$ for all $k\in\mathbb N$, which is clearly impossible.

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It seems to me that the answer is negative for $m,n>1$ even without any additional assumptions, but I don't know how to prove it.