In LU factorization of a square matrix , show that
$\|L\|_F \le \frac{n(n+1)}{2}$
$\|L\|_{\infty}\le n $
$\|L\|_1\le n$
$\|L\|_2 \le n$
Now here in $L$ all diagonal elements are $1$.Now Frobenius norm is $\sqrt{trace(L^tL)}$ . Therefore we have to prove that $\sqrt{trace(L^tL)} \le \frac{n(n+1)}{2}$ .
For (2) $ \|L\|_{\infty} =max_i\sum_{j=1}^n|a_{ij}| \le n$ then sum of all row sum will be $\le$ n but how if some elements is bigger than n??