Show that $\frac{||L|| \ ||U||}{||A||} < \min(κ(L),κ(U))$, when $A=LU$ andv$ A$, $L$, $U$ both non-singular

Question

1. $κ(L)=||L||\ ||L^{-1}||$ and $κ(U)=||U|| \ ||U^{-1}||$

2. Now I can only get that they are both greater than 1 by $$\frac{||L||\ ||U||}{||A||} \geq \frac{||LU||}{||A||} = \frac{||A||}{||A||} = 1$$

   and

   $κ(L)=||L||\ ||L^{-1}|| \geq ||LL^{-1}|| = ||I|| = 1$; $κ(U)=||U||\ ||U^{-1}|| \geq ||UU^{-1}|| = ||I|| = 1$

Can anyone help me with the inequality: $$\frac{||L||\ ||U||}{||A||} \leq \min(κ(L),κ(U))$$ when $A=LU$ and $A$, $L$, $U$ both non-singular?

Answer 1

Based on your assumptions, you can write

$$U = L^{-1}A.$$

Hence, $$ \|U\| \leq \|L^{-1}\|\|A\|.$$

Multiply on both sides by $\|L\|,$ and dividing by $|A\|$, you get$$ \frac{\|U\| \|L\|}{\|A\|} \leq \|L^{1}\|\|L\| = \kappa(L).$$

Repeat for $U$ to get the results.