estimation of condition number for column equilibration I have trouble with the following problem:

Let $A$ be an invertible square matrix. Let $D$ be the diagonal matrix with entries $d_j=\dfrac{||A||_1}{\sum_i |a_{i,j}|}$. Show that $||D||^{-1}_\infty cond_1(A)\le cond_1(AD)\le cond_1(A)$ and for any other diagonal matrix $C$: $cond_1(AD)\le cond_1 (AC)$

For the first set of inequalities I first observed that $||AD||_1=||A||_1$, thus we have $cond_1(AD)=||AD||_1||(AD)^{-1}||_1=||A||_1||D^{-1}A^{-1}||_1\le ||A||_1||D^{-1}||_1||A^{-1}||_1\le cond_1(A)$ since the entries of $D^{-1}$ are $\ge 1$. For the the other two inequalities i have no idea where to start.
 A: The proofs use just easy matrix manipulations and the norm submultiplicativity. More general results can be found in the work by van der Sluis.
Let $A=[a_1,\ldots,a_n]$. We have $d_i=\|A\|_1/\|a_i\|_1$. $\newcommand{\cond}{\mathrm{cond}_1}$
To show the second inequality, we have (denoting $[\cdot]_i$ the $i$th column of $\cdot$) as you already got:
$$\tag{1}
\|AD\|_1=\max_{1\leq i\leq n}\|[AD]_i\|_1=\max_{1\leq i\leq n}\|a_i\|_1\frac{\|A\|_1}{\|a_i\|_1}=\|A\|_1.
$$
For an arbitrary nonsingular and diagonal $C$, we have
$$
\begin{split}
\|(AD)^{-1}\|_1&=\|D^{-1}CC^{-1}A^{-1}\|_1\leq\|D^{-1}C\|_1\|(AC)^{-1}\|_1
\\&=\frac{1}{\|A\|_1}\max_{1\leq i\leq n}(\|a_i\|_1|c_{ii}|)\|(AC)^{-1}\|_1=\frac{1}{\|A\|_1}\|AC\|_1\|(AC)^{-1}\|_1\\&=\frac{1}{\|A\|_1}\cond(AC).
\end{split}
$$
Hence
$$\tag{2}
\cond(AD)=\|AD\|_1\|(AD)^{-1}\|_1\leq\cond(AC).
$$
Note that (2) implies that
$$
\cond(AD)=\min_{\substack{\text{$C$ diagonal}\\\text{nonsingular}}}\cond(AC).
$$
We can show the second part of the first set of inequalities also using (2) with $C=I$:
$$
\cond(AD)\leq \cond(AI)=\cond(A).
$$
For the other, we have from (1)
$$
\begin{split}
\cond(A)&=\|A\|_1\|A^{-1}\|_1=\|DA\|_1\|DD^{-1}A^{-1}\|_1\\&\leq\|DA\|_1\|D\|_{\infty}\|(AD)^{-1}\|_1=\|D\|_{\infty}\cond(AD).
\end{split}
$$
