# inequality between matrix norms [duplicate]

A is a $n\times k$ matrix.

I have to show that $\|A\|_2\leq \sqrt{\|A\|_1\cdot \|A\|_\infty}$.

I know that $\|A\|_2^2 = \rho(A^H\cdot A)\leq \|A^H \cdot A\|$ for every $\| \cdot \|$ submultiplicative matrix norm, but I don't know how to conclude.

Any idea?

## marked as duplicate by Surb, Algebraic Pavel, user147263, Jyrki LahtonenSep 13 '15 at 17:00

• What do you mean by $\rho(.)$? – Rajat Sep 12 '15 at 10:47
• tha maximum of the eigenvalues – user269663 Sep 12 '15 at 10:48
• I think $\lambda(.)$ would be a better notation in that case. – Rajat Sep 12 '15 at 10:50
• Usually we use $\rho(A)$ to denote the maximum of the eigenvalues (taking moduli) of $A$, and we call it the spectral radius of $A$, so I think this is the case. That is to say, let $\lambda _i$ be the eigenvalues of $A$, then $\rho (A) = \max \left\{ \left| \lambda _i \right| \right\}$. – Miguel Mars Sep 12 '15 at 10:55
• here $\rho$ and $\lambda_\max$ agree as $A^H A$ is positive semidefinite... – user251257 Sep 12 '15 at 11:10

As you noted that you now the inequality $\|A\|_2^2=\rho(A^HA)$ then it is assumed that you work with the standard induced matrix norms.

The definition is as follows:

Let $B\in \mathbb R^{n\times k}$ and let $\|.\|$ is some vector norm in $\mathbb R^k$. Then the induced matrix norm, denoted again with $\|.\|$ is: $$\|B\|=\sup\limits_{\vec x\in \mathbb R^k}{\frac{\|B\vec x\|}{\|\vec x\|}}$$ So we have: $$\|A\|_2^2=\rho(A^HA)\leq \|A^HA\|_\infty\leq \|A^H\|_\infty\cdot \|A\|_\infty =\|A\|_1\cdot \|A\|_\infty$$.

The first inequality is known for you, as you said, and it is shown like this:

Let $\vec x$ be an eigenvector of $B$ for the eigenvalue $\lambda$, where $\max\limits_{i}|\lambda_i|=|\lambda|$. Then for arbitrary matrix norm $\|.\|$, subordinate to the vector norm $\|.\|$, we have $\|B\|=\max\limits_{\vec y\neq \vec 0} \frac{\|B\vec y\|}{\|\vec y\|}\ge \frac{\|B\vec x\|}{\|\vec x\|}=\frac{\|\lambda \vec x\|}{\|\vec x\|}=|\lambda|$

The second inequality is because the $\|.\|_\infty$ is induced norm from the vector $\infty$ - norm, so it is sub-multiplicative.

The last equality is because $\|A^H\|_\infty$ is the maximum absolute row sum of the matrix $A^H$, which is equal to the maximum absolute column sum of $A$ which is equal to the $\|A\|_1$ .

or use $\|A\|_2^2 = \rho(A^H\cdot A)\leq \|A^H \cdot A\| _\infty\leq\|A^H\|_\infty \cdot \|A\|_\infty \leq \|A\|_1.\|A\|_\infty$

Because $\|A^H\|_\infty=\|A\|_1$

• he has mentioned it in the question that he know it. – R.N Sep 12 '15 at 11:27
• oh sorry. I missed that – user251257 Sep 12 '15 at 11:28
• @Razieh Noori Isn't it $\|A^H\|_\infty=\|A\|_1$ ? – Svetoslav Sep 12 '15 at 11:47
• @Svetoslav No, dear. $\|A\|_\infty$ is operator norm, which is equal to biggest singular value, but $\|A\|_1$ is sum of singular values. see the reference i have said in first answer. those are interesting norms on matrices (called schatten p-norm) – R.N Sep 12 '15 at 11:53
• @Razieh Noori , see en.wikipedia.org/wiki/Matrix_norm – Svetoslav Sep 12 '15 at 12:24

$\|A\|_{2}^2\leq trac(A^H\cdot A)\leq{\Vert A^H\cdot A\Vert_1\leq \| A\|_1\|A\| _\infty}$.

for more information see $D.46$ and $D.52$ of abstract harmonic analysis hewitt&ross pages 706 and 709

• here $\|\cdot\|_2$ looks like the operator norm than the Frobenius norm. So it should be $\le trace(A^H A)$. – user251257 Sep 12 '15 at 11:13
• then what is $\|A\|_\infty$ ? – R.N Sep 12 '15 at 11:15
• the operator norm to the vector norm $\|\cdot\|_\infty$ – user251257 Sep 12 '15 at 11:17
• thus i think my below answer is better than this one – R.N Sep 12 '15 at 11:20
• probably ${}{}{}$ – user251257 Sep 12 '15 at 11:20