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.

  • $\begingroup$ Your question is a bit unclear. I would interpret your question this way: "Are there norms on $\mathbb C^m$" and $\mathbb C^n$ so that the norm $\|\cdot\|$ is a subordinate norm (operator norm)?" Is this what you mean? $\endgroup$ – Joonas Ilmavirta Sep 28 '14 at 9:10
  • $\begingroup$ @JoonasIlmavirta It can be viewed like that, yes :-). $\endgroup$ – Kenobi Sep 28 '14 at 9:17
  • $\begingroup$ Then it is possible if $m=1$ or $n=1$ (take the $\ell^1$ norms on the spaces). I have a feeling that it's impossible otherwise, but I don't know how to show it. Even the case $m=n=2$ seems difficult if there are no assumptions on the two norms on $\mathbb C^2$. $\endgroup$ – Joonas Ilmavirta Sep 28 '14 at 10:51
  • $\begingroup$ @JoonasIlmavirta I edited my question accordingly to your suggestion. What kind of assumptions can me made? It would be interesting too. $\endgroup$ – Kenobi Sep 28 '14 at 11:02
  • $\begingroup$ You get a simpler problem if you require one of the norms to be the Euclidean one. There are many possible assumptions you could make on the norms, like strict convexity or smoothness. If you had a specific situation in mind, the problem might be easier. For example, if you want to have a norm $|\cdot|_k$ on $\mathbb C^k$ for each $k$ so that your norm is always the operator norm, the problem seems already much easier. (As I understand, you first fix $m,n$ and then ask whether two norms exist.) $\endgroup$ – Joonas Ilmavirta Sep 28 '14 at 11:17

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.

  • $\begingroup$ Thanks, just to understand $\Vert\cdot\Vert_*$ or $**$ are two arbitrary norm? $\endgroup$ – Kenobi Oct 3 '14 at 13:24
  • $\begingroup$ @Kenobi, yes, the proof is true for arbitrary norms. $\endgroup$ – Tigran Saluev Oct 3 '14 at 14:01
  • $\begingroup$ I don't follow your proof, you are proving that $\Vert\cdot\Vert_{*}=\Vert\cdot\Vert_{1}$ for rank-one matrices not for arbitrary matrices on $M_n(\Bbb{C})$, am I right? $\endgroup$ – Krokop Oct 4 '14 at 15:19
  • 1
    $\begingroup$ @Krokop I can explain. First, I assume the norms $\|\cdot\|_*$, $\|\cdot\|_{**}$ are arbitrary. Then there are two possibilities: 1) $\|\cdot\|_* \ne \|\cdot\|_1$ or $\|\cdot\|_{**} \ne \|\cdot\|_{\infty}$, then we obtain a contradiction on rank-1 matrices. 2) $\|\cdot\|_* = \|\cdot\|_1$ and $\|\cdot\|_{**} = \|\cdot\|_{\infty}$, then we go further and obtain a contradiction on matrix with $2\times 2$ nonzero submatrix $A$. $\endgroup$ – Tigran Saluev Oct 4 '14 at 15:57
  • $\begingroup$ In fact I don't understand why $$\Vert\cdot\Vert_{\infty}=\Vert\cdot\Vert_{**}.$$ Is surely something about the dual but I don't really have knowledge on this subject.. $\endgroup$ – Kenobi Oct 9 '14 at 16:53

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.

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.

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.

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.


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