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In Wikipedia about matrix norm, two induced matrix norms, $\|A\|_2=\sigma_{\operatorname{max}}\left(A\right)$ and $\|A\|_{1,2}=\underset{1\le j \le n}{\max}\|A_{\cdot,j}\|_2$, have been introduced.

Now I am interested in the equivalence of these two matrix norms. How could I get a tight bound when I prove the equivalence?

my attempt: $$ \|Ax\|_2^2=\sum_{i=1}^n\left(\sum_{j=1}^na_{ij}x_j\right)^2 $$ now apply the Cauchy inequality for each term $\sum_{j=1}^n a_{ij}x_j$ as follows: $$ \left(\sum_{j=1}^na_{ij}x_j\right)^2\le\left(\sum_{j=1}^na_{ij}^2\right)\left(\sum_{j=1}^nx_j^2\right). $$ Since $\|x\|_2=1$, we have: $$ \|Ax\|_2^2\le \sum_{i=1}^n\sum_{j=1}^na_{ij}^2 $$

mathhahaha
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1 Answers1

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$\|A\|_{1,2} = \max_j \|Ae_j\|_2$, $\|A\|_2 = \max_{\|x\|_2 \le 1} \|Ax\|_2$, hence $\|A\|_2 \ge \|A\|_{1,2}$.

The Frobenius norm $\|A\|_F = \sqrt{ \sum_{i,j} |A_{ij}|^2}$ satisfies $\|A\|_F = \sqrt{\sum_j \|Av_j\|^2}$ for any orthonormal $v_k$. In particular, this shows that $\|A\|_F \ge \|A\|_2$.

For the other direction, let $y_j = \|Ae_j\|_2$ and note that $\|A\|_{1,2} = \|y\|_\infty \ge {1 \over \sqrt{n}} \|y\|_2 = {1 \over \sqrt{n}}\sqrt{\sum_j \|Ae_j\|_2^2} = {1 \over \sqrt{n}}\|A\|_F \ge {1 \over \sqrt{n}}\|A\|_2$.

To see that this is tight, let $Q$ be an orthogonal matrix such that $Q e_1 = {1 \over \sqrt{n}} (1,...,1)^T$. Then $\|Q\|_{1,2} = {1 \over \sqrt{n}} = {1 \over \sqrt{n}} \|Q\|_2$.

To summarise, $\|A\|_2 \ge \|A\|_{1,2} \ge {1 \over \sqrt{n}}\|A\|_2$.

copper.hat
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