Let $W$ denote a full rank $d \times d$ matrix, the condition number of $W$ is very bad, say $\kappa(W) = 2^d$. Is there existing some diagonal matrix $D$ with $D_{ii} >0$, such that $\kappa(DW) = \text{poly}(d)$?.
What's the best condition number for a full rank matrix?
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linear-algebra
eigenvalues-eigenvectors
numerical-linear-algebra
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1A nice approach might be to look at the Gershgorin disks of $$ (DW)(DW)^T = D[WW^T]D^T $$ – 2017-02-03
2 Answers
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This is not an answer but you might be interested in looking on this paper.
Summarizing the related statements, if $A\in\mathbb{R}^{n\times n}$ is nonsingular, $p\geq 1$, and $$ D=\mathrm{diag}(\|A(1,:)\|_p^{-1},\ldots,\|A(n,:)\|_p^{-1}), $$ then $$ \kappa_p(DA)\leq n^{\frac{1}{p}}\min_{\tilde{D}\in\mathcal{D}_n}\kappa_p(\tilde{D}A), $$ where $\mathcal{D}_n$ is a set of $n\times n$ diagonal nonsingular matrices. In particular for $p=\infty$, the row scaling gives the optimal condition number.
Personally, I don't think it is possible to find what you are looking for. Using the other answer, consider $$ A=\begin{bmatrix}\alpha&1\\0&1\end{bmatrix} $$ where $\alpha\in(0,1)$. For $p=\infty$, we get $$ \kappa_\infty(DA)=1+\frac{2}{\alpha}. $$ You can make the condition number arbitrarily large by choosing a small enough $\alpha$, but for a fixed $\alpha$ you cannot make it smaller anymore by any row scaling.
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I don't think this is possible in general. Consider a matrix whose row vectors have length 1, and the matrix is poorly conditioned because the row vectors are very nearly aligned. Scaling the row vectors to be longer or shorter (IE, multiplying by a diagonal matrix) doesn't help this.