Proving result on matrix norms How do I prove that, letting, for $A\in\mathbb{C}^{n\times n}$:
$$(a)\quad\|A\|_1=\max\limits_{i=1,\cdots,n}\left(\sum\limits_{j=1}^n|a_{ij}|\right),$$
$$(b)\quad\|A\|_2=\rho^{\frac12}(A^HA),$$ with $\rho(A)$ and $A^H$ defined as in this question, and
$$(c)\quad\|A\|_\infty=\max\limits_{j=1,\cdots,n}\left(\sum\limits_{i=1}^n|a_{ij}|\right),$$
then the following hold?
$$(1)\quad\dfrac{1}{\sqrt{n}}\|A\|_\infty\leq\|A\|_2\leq\sqrt{n}\|A\|_\infty;$$
$$(2)\quad\dfrac{1}{\sqrt{n}}\|A\|_1\leq\|A\|_2\leq\sqrt{n}\|A\|_1;$$
$$(3)\quad\max\limits_{i,j}|a_{ij}|\leq\|A\|_2\leq n\max\limits_{i,j}|a_{ij}|;$$
$$(4)\quad\|A\|_2\leq\sqrt{\|A\|_1\|A\|_\infty}?$$ 
 A: All mentioned matrix norms are special cases of the matrix $p$-norm:
$$\tag{a}
\|A\|_p:=\max_{x\neq 0}\frac{\|Ax\|_p}{\|x\|_p}.
$$
If, for two vector $p$-norms $\|\cdot\|_p$ and $\|\cdot\|_q$ ($p,q\in[1,\infty]$), there are positive constants $\alpha$ and $\beta$ such that
$$\tag{b}
\alpha\|x\|_q\leq\|x\|_p\leq\beta\|x\|_q
$$
holds for all $x$, then we have from (a) that
$$
\|A\|_p=\max_{x\neq 0}\frac{\|Ax\|_p}{\|x\|_p}\begin{cases}\leq\max\limits_{x\neq 0}\frac{\beta\|Ax\|_q}{\alpha\|x\|_q}&=\frac{\beta}{\alpha}\|A\|_q\\
\geq\max_{x\neq 0}\frac{\alpha\|Ax\|_q}{\beta\|x\|_q}&=\frac{\alpha}{\beta}\|A\|_q.
\end{cases}.
$$
Hence,
$$\tag{c}
\frac{\alpha}{\beta}\|A\|_q\leq\|A\|_p\leq\frac{\beta}{\alpha}\|A\|_q.
$$
To get (1) and (2), use in (c) and (b) the well known relations
$$\tag{d}
\|x\|_{\infty}\leq\|x\|_2\leq\sqrt{n}\|x\|_{\infty},
\quad
\frac{1}{\sqrt{n}}\|x\|_1\leq\|x\|_2\leq\|x\|_1.
$$

For (4), note that the 2-norm of $A$ is the square root of the largest eigenvalue of $A^*A$, that is, there is a $\lambda=\|A\|_2^2$ and a nonzero $y$ such that $A^*Ay=\lambda y$. Let $\|\cdot\|$ be any vector norm and $\|\cdot\|$ its associated operator norm. The eigenvector $y$ can be chosen such that $\|y\|=1$. Then 
$$\|A\|_2^2=\lambda=\|A^*Ay\|\leq\|A^*\|\|A\|\|y\|=\|A\|\|A^*\|.$$
Hence for any operator matrix norm
$$
\|A\|_2\leq\sqrt{\|A\|\|A^*\|}.
$$
In particular,
$$\tag{e}
\|A\|_2\leq\sqrt{\|A\|_1\|A^*\|_1}=\sqrt{\|A\|_1\|A\|_{\infty}}.
$$

To show (3), let $e_i$ be the $i$th column of the $n\times n$ identity matrix.
Since $\|e_i\|_2=1$, (a) and (d) imply that
$$
\|A\|_2\geq\|Ae_i\|_2\geq\|Ae_i\|_{\infty}=\max_{j}|a_{ij}|.
$$
Since the above is valid for any $i$, $\|A\|_2\geq\max_{i,j}|a_{ij}|.$
The second inequality can be easily obtained from (e) and the fact that 
$$
\left.\begin{array}{l}
\|A\|_1\\\|A\|_{\infty}
\end{array}\right\}\leq n\max_{i,j}|a_{ij}|.
$$
