Building a norm. I've been told to build a norm that is NOT a matrix norm. I need to show that the built norm is indeed a norm on the space of $n*n$ matrices but that is not a matrix norm induced by some vector norm.
Any help?
Thanks :)
 A: The Frobenius norm (or more generally the Hilbert–Schmidt norm) of a matrix is not induced by a vector norm. Frobenius norm is for $A \in \mathbb{C}^{m \times n}$ is defined as
$$\Vert A \Vert_F = \sqrt{\sum_{i=1}^m \sum_{j=1}^n \vert A_{i,j} \vert^2}$$
This norm is popular especially in numerical linear algebra since it is much easier to compute than other norms and is invariant under rotations.
The generalization of the Frobenius norm is called the Schatten norm. The Schatten norm is also not induced by a vector norm. The Frobenius norm presented above can be shown to be equal to $$\Vert A \Vert_F = \sqrt{\sum_{k=1}^{\min(m,n)} \sigma_k^2}$$ where $\sigma_k$ are the singular values of the matrix $A$. The Schatten norm generalizes this as follows.
$$\Vert A \Vert_{\text{Schatten }p \text{ norm}} = \left({\sum_{k=1}^{\min(m,n)} \sigma_k^p} \right)^{1/p}$$ where as before $\sigma_k$ are the singular values of the matrix $A$.
Taking $p=2$ gives the Frobenius norm.
A: If you're looking for a norm on $M_n(\mathbb{R})$ that is not induced by a norm on $\mathbb{R}^n$, then you couldn't possibly ask for a better answer than Marvis's. If you're looking for even more, namely, a norm on $M_n(\mathbb{R})$ that fails to be submultiplicative (i.e., fails the condition $\|A B\| \leq \|A\|\|B\|$ for some $A$, $B$,), then this MathOverflow post suggests you can take the maximum norm
$$
 \|A\| := \max_{i,j} \left|A_{ij}\right|.
$$
Then indeed (as pointed out in the comments to that post), for $n \geq 2$, if $A \in M_n(\mathbb{R})$ is the matrix whose entries are all $1$, then
$$
 \|A^2\| = n > 1 = \|A\|^2.
$$
A: Hint. Consider one half of the $\ell_\infty$-norm defined on $M_2(\mathbb{R})$. That is,
$$
\left\|\begin{pmatrix}a&b\\ c&d\end{pmatrix}\right\|=\frac12\max\{\,|a|,\,|b|,\,|c|,\,|d|\,\} = \frac12\|(a,b,c,d)\|_\infty.
$$
As the $\ell_\infty$-norm is a norm, so is one half of it. If it is an induced norm, we should have $\|Ax\|\le\|A\|\|x\|$ for every $x\in\mathbb{R}^2$. Now consider $A=I$ and $x\neq0$.
