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I need to find an upper bound for a matrix norm in terms of bounds of its columns. I have a vector $\varepsilon_i(x) \in R^{n\times1} $ such that $||\varepsilon_i(x)||_2\le\gamma_0$. I also have a matrix $Z=[\varepsilon_1, \varepsilon_2, \varepsilon_3, … ,\varepsilon_N] \in R^{n\times N}$.

Using the information $||\varepsilon_i(x)||_2\le\gamma_0$, can I find an upper bound for $||Z||_2$?

If this were to be a frobenius norm question, it would be quite easy to show. However, I couldn’t find an inequality for L2 norm case.

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Thank you in advance for your help.

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If you use the following definition for $\|Z\|_2$, where $Z\in\mathbb R^{n\times N}$

$$\|Z\|_2:=\sup\limits_{x\in \mathbb R^N}{\frac{\|Zx\|_2}{\|x\|_2}}$$

then you have $Zx=\sum\limits_{i=1}^{N}{x_i\epsilon_i}\Rightarrow \|Zx\|_2=\|\sum\limits_{i=1}^{N}{x_i\epsilon_i}\|_2\leq \sum\limits_{i=1}^{N}{|x_1|\|\epsilon_i\|_2}\leq \gamma_0\sum\limits_{i=1}^{N}{|x_1|}$

Also $\|x\|_2=\sqrt{x_1^2+…+x_N^2}\ge \frac{1}{\sqrt N}\sum\limits_{i=1}^{N}{|x_1|}$. So you get $$\frac{\|Zx\|_2}{\|x\|_2}\leq \gamma_0\frac{\sum\limits_{i=1}^{N}{|x_1|}}{\frac{1}{\sqrt N}\sum\limits_{i=1}^{N}{|x_1|}}=\gamma_0\sqrt N$$

Finally, $$\|Z\|_2=\sup\limits_{x\in \mathbb R^N}{\frac{\|Zx\|_2}{\|x\|_2}}\leq \gamma_0\sqrt N$$

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