Let $A=(a_{ij})_{1 \leq i,j \leq n} \in {R^{n\times n}}$ be a regular matrix and regard the 'classical' condition number $k_\infty(A):=\|A\|_\infty\|A^{-1}\|_\infty$ of the matrix $A$.
Now some basic calculations show that for
$$D=diag(\alpha_1, \ldots, \alpha_n ), \quad\text{where}\quad \alpha_j =\sum_{\ell =1}^n |a_{j\ell }|,$$
we have $k_\infty(DA) \leq k_\infty(A)$ and $k_\infty(DA)$ is optimal with respect to the chosen norms.
Moreover, let $k_S(A):=\||A^{-1}||A|\|_\infty$, with $|A|=(|a_{ij}|)_{1 \leq i,j \leq n}$, the Skeel condition number of $A$ (compare https://dl.acm.org/citation.cfm?id=322148).
My question is: For some basic examples we have that the condition number of the equlibrated matrix equals the Skeel condition number, i.e. $k_\infty(DA)=k_S(A)$ (see example below), where $D$ is a diagonal Matrix, constructed as above. But is this true in general? Or is it possible to construct a matrix $A$ such that $k_\infty(DA)>k_S(A)$?
Here is my basic example:
Let
$$B=\begin{bmatrix}1 & 100 \\
-1 &10
\end{bmatrix}\quad \text{and}\quad D=diag(\frac{1}{101},\frac{1}{11}).
$$
Then $k(B)=101 $ and $k(DB)=k_S(B) \approx 19.18$.
Thank you very much in advance!
Best Answer
Indeed, the Skeel condition number can be characterized as $$ k_S(A)=\min\{k_\infty(DA): D\text{ diagonal}\}. $$
Three observations are useful here:
We can express the matrix $\infty$-norm as $$ \|X\|_{\infty}=\||X|e\|_{\infty}, $$ where $e=[1,\ldots,1]^T$ of suitable size. Note that there's a vector norm on the right-hand side.
If $DA$ is the row-equilibrated matrix, we have $|DA|e=e$, where $e=[1,\ldots,1]^T$.
The Skeel condition number is independent of row scaling: $k_S(DA)=k_S(A)$ for any nonsingular diagonal $D$.
This gives us $$ \begin{split} k_\infty(DA) &=\||(DA)^{-1}|e\|_\infty\||DA|e\|_\infty =\||(DA)^{-1}||DA|e\|_\infty =k_S(DA)=k_S(A). \end{split} $$