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Let A, L, U be $n \times n$ non-singular matrices such that A=LU.

$\gamma_1 = \frac{\||L|\|_2\||U|\|_2}{\|A\|_2}$

$\gamma_2 = \frac{\|L\|_2\|U\|_2}{\|A\|_2}$

Show that in 2-norm case, if L is unitary,

$$ \gamma_2 = 1, \gamma_1 \leq n$$

for $ \gamma_2 = 1 $, this is my idea:

$\gamma_2 = \frac{\|L\|_2\|U\|_2}{\|A\|_2} \leq \frac{\|L\|_2\|L^{-1}A\|_2}{\|A\|_2} \leq \frac{\|L\|_2\|L^{-1}\|_2\|A\|_2}{\|A\|_2} = \|L\|_2\|L^{-1}\|_2 = 1$
$\gamma_2 = \frac{\|L\|_2\|U\|_2}{\|A\|_2} \geq \frac{\|LU\|_2}{\|A\|_2} = \frac{\|A\|_2}{\|A\|_2} = 1$

So, $\gamma_2$ = 1

But I have no idea for $\gamma_1 \leq n$

By the way, there are some facts:

(1) $\gamma_2 \leq min(\kappa(L),\kappa(U))$

(2) $\gamma_1 \leq \gamma_2$ when 1-norm or $\infty$-norm is used

Can any one help me?

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  • $\begingroup$ I took out $\gamma$ since it was a bit superfluous and also if you want $n \times n$, use \times, not x. $\endgroup$ Nov 1, 2019 at 15:34
  • $\begingroup$ Thank you ! \times looks much prettier than x ! $\endgroup$
    – gccat
    Nov 2, 2019 at 0:55

1 Answer 1

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L is unitary so the 2 norm is 1 (as well as it's absolute value.)

Next |U| = (U*)U = ((L*)A)* ((L*)A) = (A*)(LL*)A=A*A = |A|

So you want 1 = |||A|||{2}/||A||{2} = \gamma_{1}.

Well, this is the two norm, so they're the same... So \gamma_{1} = 1.

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