Let $A=LDL^T$ be a symmetric positive definite matrix, where $L$ is a unit lower triangular matrix, and $D=\textrm{diag}(d_{ii}).$ Show that $$\textrm{Cond}_2(A) \geq \frac{\max (d_{ii})}{\min (d_{ii})}.$$
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I think the assumption "L is a unit lower triangular matrix" plays a critical role. – David Mayar Jun 02 '15 at 15:46
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Hint: note the following:
- $\textrm{Cond}_2(A) = \|A\| \|A^{-1}\|$
- Because $A$ is positive semidefinite, $\|A\| = \max_{\|x\| = 1} x^TAx$
- Because $A$ is positive semidefinite, $\|A^{-1}\| = \max_{\|x\| = 1} \frac{1}{x^TAx}$
For an lower bound of each of these maxima, plug in the $j$th standard basis vector for $x$.
Let $L_j$ denote the $j$th row of $L$. We have $$ e_j^T(LDL^T)e_j = d_{jj}\cdot \|L_j\|^2 \geq d_{jj} $$
Ben Grossmann
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