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Let $x \in \mathbb{R}^n$, and $M \in \mathbb{R}^{n\times n}$ be a full rank square matrix. If it is known that $ \| M x\|_2 \leq c$, then what can be said about the upper bound of $\| x\|_2$, i.e., $\|x\|_2 \leq b$, where $b$ is a function of $c$ and $M$? I tried SVD of $M$, but it didn't take me very far: $\| U\Sigma V^T x\|_2 = \| \Sigma V^T x\|_2$. Could $b$ be expressed in terms of the singular values of $M$?

Edit: When $M$ is a non-zero scalar, the above question is clearly, meaningful. I'm looking for its extension when $M$ is a matrix. Some constraints on $M$ might be necessary.

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In general nothing can be said. Suppose $x$ is in the null space of $M$, then $Mx = 0$ and we could have $c = 0$. However, we can scale the norm of $x$ to be as large as we would like.

If $M$ is full rank, then we have that $\|Mx\|_2^2 \ge \sigma_n \|x\|_2$ (so depends on the smallest singular value). Then $\|x\|_2 \le c/\sigma_n$.

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  • $\begingroup$ I guess some additional constraints on $M$ would be necessary for the question to make sense. Based on your example, here is one: $M$ is full rank. $\endgroup$
    – ChargeShivers
    Commented Oct 19, 2021 at 16:14
  • $\begingroup$ $\|Mx\|_2^2 \ge \sigma_n \|x\|_2$ looks good. Is it easy to prove? $\endgroup$
    – ChargeShivers
    Commented Oct 19, 2021 at 17:59
  • $\begingroup$ So think of the right singular vectors as a basis for the space and then write $x$ in terms of this basis. Then the multiplication $Mx$ is easier and you should be able to get the bound from that $\endgroup$
    – Rishi Sonthalia
    Commented Oct 19, 2021 at 20:41

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