8
$\begingroup$

In spaces $V$ and $W$, define vector norms $\| \cdot \|_V $ and $\| \cdot \|_W $, respectively. Consider the operator norm induced by norms $\| \cdot \|_V $ and $ \| \cdot \|_W $

$$ \| A \| = \sup{\frac{\|Ax\|_W}{\|x\|_V}} $$

Do there exist norms $ \|\cdot\|_V $ and $ \|\cdot\|_W $ such that the Frobenius norm is induced by these norms?

It is well known that for any matrix norm induced by $ \|\cdot\| $, $ \|I\| = 1 $. But it is not truth for two different norms in $V$ and $W$. Can anybody help?

$\endgroup$
0

1 Answer 1

5
$\begingroup$

Since the argument on the norm of the identity does not work here, let's try the unitary invariance. The Frobenius norm is unitarily invariant ($\|A\|_F=\|PAQ^*\|_F$ for any unitary $P$ and $Q$ of suitable dimensions), so if there were such vector norms inducing the Frobenius norm, they would need to be unitarily invariant as well. The only unitarily invariant vector norm is the Euclidean norm (or its positive multiple), which induces the spectral norm.

$\endgroup$
4
  • 1
    $\begingroup$ Could you explain, how from matrix unitary invariant norm we can get that vector norm is unitary invariant too? If $||A||_F = ||UAV||_F$ for any $A$ and unitary $U$, $V$. Hence $\sup \frac{||Ax||_W}{||x||_V} =\sup \frac{||UAVx||_W}{||x||_V} $ for any unitary $V$ and $W$. How we can get from that $||Ux||_V = ||x||_V$ and $||Ux||_W = ||x||_W$ for any $x$ and unitary $U$? $\endgroup$ Commented Apr 22, 2015 at 13:43
  • $\begingroup$ @StanislavMorozov One would need $\|A\|_F=\|PAQ^*\|_F=\sup\frac{\|Ax\|_W}{\|x\|_V}=\sup\frac{\|PAx\|_W}{\|Qx\|_V}$. I don't say that this is very precise but could give an idea (I've changed the notation for unitary matrices to do not confuse with the norms). $\endgroup$ Commented Apr 22, 2015 at 15:04
  • $\begingroup$ And that is not true that only unitarily invariant norm - this is the Euclidean norm or its positive multiple. Any norm generated by a scalar product is unitarily invariant. $ ||Ux|| = \sqrt{(Ux, Ux)} = \sqrt{(x, U^*Ux)} = \sqrt{(x,x)} = ||x|| $ $\endgroup$ Commented Apr 27, 2015 at 22:04
  • 1
    $\begingroup$ @StanislavMorozov Obviously, any scalar product defines its own orthogonality and is invariant (or its associated norm) w.r.t. its own meaning of orthogonality. $\endgroup$ Commented Apr 27, 2015 at 22:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .