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In order to prove a larger assumption, I need to find a vector norm over $M_n$ such that $\|I\| < 1$. None of the standard $p$-norms, nor the infinity norm work. I know that for matrix norms, this is impossible, but I think it should be possible with a vector norm since I don't have to worry about submultiplicativity.

Any hints would be greatly appreciated. Proving an example is in fact a norm is within my grasp, I just need some idea about how to construct an example.

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Are you looking for any norm on $M_n$? If so, you could start with some norm $||\cdot||$ and show that for any positive $\lambda$, $$\frac{1}{\lambda}||\cdot||$$ is still a norm, and then choose $\lambda$ to your liking. – Olivier Bégassat Nov 1 '11 at 15:50
@OlivierBégassat: thanks for pointing it. I apparently gave a full solution which was not required, you hint works anyway. – Ilya Nov 1 '11 at 15:55
@OlivierBégassat Well I feel only slightly more silly for not seeing this prior, thanks for the hint. – brc Nov 1 '11 at 16:05
@Gortaur I saw your answer, thank you for the effort! – brc Nov 1 '11 at 16:05
@brc: you're welcome – Ilya Nov 1 '11 at 16:07
up vote 1 down vote accepted

If I do understand you correct, you're trying to find a norm $\|\cdot\|$on $n$-dim vector space $V_n$ such that the corespondent matrix norm is given by $$ \|A\|' := \sup\limits_{v\in V_n\setminus\{0\}}\frac{\|Av\|}{\|v\|}. $$ If $A = I$ (which is identity matrix I assume) then $\|I\|' = 1$ since the fraction is always $1$ because$Iv = v$ for any $v\in V_n$.

If you're looking for a norm on the space of $n$-dim matrices $M_n$ then you can just put a norm $$\|A\| = \frac1{2n}\sum\limits_{ij}|a_{ij}|.$$

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I think the OP's statement "I know that for matrix norms, this is impossible" shows the OP understands this, and he or she isn't looking for such norms. – Olivier Bégassat Nov 1 '11 at 15:53

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