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Say that we have the following matrices: $A\in M_{n,n}$ which is unknown, $Y\in M_{n,m}/\{0_{n,m}\}$ and $X\in M_{n,m}/\{0_{n,m}\}$. I want to show that the following set is convex $\Omega=\{S\in M_{m,m},\space S=(Y-A\times X)^T(Y-A\times X)\space \space |\space \space\|A\|_2\leq \frac{\epsilon}{n}\}$.

From the norm equivalence we have $\|A\|_2\leq \|A\|_1 \leq n\|A\|_2$ we have only n this is because A is a square matrix of dimension n and so $n=m$. What I would like to have as a constraint is $\|A\|_1\leq \epsilon$ but this will make the set a polygon having parabolas for each side which not convex.. so if we assume that $\|A\|_2\leq \frac{\epsilon}{n}$, then $\|A\|_1\leq \epsilon$. I'm considering $\|A\|_2 \leq \frac{\epsilon}{n}$ because the set would be convex and smaller than considering $\|A\|_1\leq \epsilon$ which is not convex.

So how can we prove that $\Omega$ is a convex set?

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When writing $A\times X$ and $\times AX$, do you simply mean the product $AX$? – 1015 Feb 13 '13 at 0:42
I think that I did a typo here – user2987 Feb 13 '13 at 0:43
I have only the expression of $A\times X$ which is the regular matrix product AX. – user2987 Feb 13 '13 at 0:44
Note that $A$ should not be given at the beginning. It ranges over all matrices which satisfy the constraint. Also, if you know the value of $\epsilon$ you should give it. Otherwise, you might as well replace $\epsilon/n$ by $\epsilon$. – 1015 Feb 13 '13 at 0:54
Your claim that "without the constraint, the set is not convex" is not true if $X=0$. In this case, your set is a singleton, hence convex. – 1015 Feb 13 '13 at 2:25

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