I assume that the author tries to derive the matrix norms $\|A\|_1$ and $\|A\|_\infty$ induced by vector norms $\|x\|_1$ and $\|x\|_\infty$.
Let's take, for example, $\|\cdot\|_\infty$. We write
$$\|Ax\|_\infty=\sup_i \left|\sum_j A_{ij}x_j\right|\le \sup_i \sum_j |A_{ij}||x_j|\le \sup_j |x_j| \sup_i \sum_j |A_{ij}|.$$
Hence, a good candidate for $\|A\|_\infty$ is $\sup_i \sum_j |A_{ij}|$. We need to prove that this boundary is indeed achieved; it is true, since we can take $i$ where that supremum is achieved and impose $x_j=\mathrm{sign}(A_{ij})$. With such $x$ all our inequalities degenerate to equalities and we conclude that $\|A\|_\infty = \sup_i \sum_j |A_{ij}|$ is a matrix norm induced by $\|x\|_\infty = \sup_j |x_j|$.
The case $\|\cdot\|_1$ is done likewise.
More generally, consider $n\times n$ real matrices $A,X$ and the function $f:X\rightarrow ||AX||^2=trace(AXX^TA^T)$. Then its derivative is $D_Xf:H\rightarrow 2trace(X^TA^TAH)$. moreover its Hessian is $Hess_X(f):(H,K)\rightarrow 2trace(K^TA^TAH)$, a symmetric bilinear form that does not depend on $X$. Let $(E_{i,j})$ be the canonical basis of $M_n$. Then, if $j\not=l$, then $Hess_X(f)(E_{i,j},E_{k,l})=0$ and $Hess_X(f)(E_{i,j},E_{k,j})=2u_{i,k}$ where $u_{i,j}$ is the $(i,j)$ entry of $A^TA$.
Let $vec(U)$ denotes the vectorization of a $n\times n$ matrix $U$, formed by stacking the rows (resp. the columns) of $U$ into a single column vector. Then the $n^2\times n^2$ symmetric matrix associated to $Hess_X(f)$ is $2A^TA\otimes I_n$ (resp. $2I_n\otimes A^TA$). cf.:
http://en.wikipedia.org/wiki/Kronecker_product
The eigenvalues $(\lambda_i)_i$ of $A^TA$ are non-negative. The eigenvalues of $2A^TA\otimes I_n$ and $2I_n\otimes A^TA$ are $(2\lambda_i)_i$, $n$ times each ; hopefully, $Hess_X(f)$ is sym. non-negative (semidefinite). In particular, if $A=I_n$, then $Hess_X(f)$ is associated to the matrix $2I_n\otimes I_n=2I_{n^2}$ as 127 wrote above.
EDIT: Charles, that you write is the gradient $\nabla_X(f)=2A^TAX$. The gradient is defined from the matrix scalar product $(U,V)=trace(U^TV)$ by the formula $(\nabla_X(f),H)=D_Xf(H)$. We conclude by identification.
Best Answer
(not doing all, but here's a few)
The key here is to notice that $$ \|A\|_F^2=\text{Tr}_n(A^TA). $$ Then $$ \|A\|_2^2=\lambda_\max(A^TA)\leq\text{Tr}_n(A^TA)\leq n\lambda_\max(A^TA), $$ so $\||A\|_2\leq\|A\|_F\leq\sqrt n\|A\|_2$.
Here, writing $e=(1,\ldots,1)^T\in\mathbb R^n$, $$ \|A\|_\infty=\max_i\sum_j|A_{ij}|=\|Ae\|_\infty\leq\|Ae\|_2\leq\|A\|_2\|e\|_2=\sqrt n\|A\|_2 $$