- #1

mathmari

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Let $A=(a_{i,j})$ a real matrix with $m$ rows and $n$ columns, $x\in \mathbb{R}^n$ and \begin{equation*}\|A\|:=\sup_{\|x\|_2\leq 1}\|Ax\|_2, \ \ \|A\|_{\text{Eucl}}:=\sqrt{\sum_{i=1}^m\sum_{j=1}^n|a_{i,j}|^2}\end{equation*}

I want to show that $$\|A\|\leq \|A\|_{\text{Eucl}} \leq \sqrt{n}\|A\|$$ I have already shown the first inequality:

Since $\displaystyle{(Ax)_i=\sum_{j=1}^na_{i,j}x_j}$, we get $\displaystyle{\|Ax\|_2^2=\sum_{i=1}^m\left (\sum_{j=1}^n|a_{i,j}x_j|\right )^2=\sum_{i=1}^m\left (\sum_{j=1}^n|a_{i,j}||x_j|\right )^2}$.

From the Cauchy–Schwarz inequality we get \begin{equation*}\left (\sum_{j=1}^n|a_{i,j}||x_j|\right )^2\leq \left (\sum_{j=1}^n{|a_{i,j}|^2}\right )\left (\sum_{j=1}^n|x_j|^2\right )=\left (\sum_{j=1}^n{|a_{i,j}|^2}\right )\|x\|_2^2\end{equation*}

Sp we get\begin{equation*}\|Ax\|_2^2=\sum_{i=1}^n\left (\sum_{j=1}^n|a_{i,j}||x_j|\right )^2\leq \sum_{i=1}^m\sum_{j=1}^n|a_{i,j}|^2\|x\|_2^2=\|A\|_{\text{Eucl}}^2\,\|x\|_2^2 \end{equation*}

Therefore we have that \begin{equation*}\|Ax\|_2\leq \|A\|_{\text{Eucl}}\,\|x\|_2\Rightarrow \sup_{\|x\|_2\leq 1}\|Ax\|_2\leq \sup_{\|x\|_2\leq 1}\|A\|_{\text{Eucl}}\,\|x\|_2=\|A\|_{\text{Eucl}}\end{equation*}

From that it implies that $\|A\|\leq \|A\|_{\text{Eucl}}$. Is everything correct? (Wondering)

Could you give me a hint for the second inequality? (Wondering)