# In 2-norm case, if L is unitary, $\frac{\||L|\|_2\||U|\|_2}{\||A|\|_2} \leq n$

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?

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