Operator norm vs Square Frobenius Norm

Question

I got this problem. Be a linear transformation $L=\mathbb{R}^n \to \mathbb{R}^m $ with Matrix $A = (a_{i,j})$. $$ ||L||_{tr} = \sum_{i=1}^{n}\sum_{j=1}^{m}(a_{i,j})^2$$ and $$||L||_{op}=sup \{|L(x)| : |x|=1 \}$$ Demonstrate that: $$ ||L||_{op} \leq ||L||_{tr} \leq \sqrt{n}||L||_{op}$$ where $ ||L||_{op}$ is the operator norm. The left part I have done using Cauchy-Schwartz, but I have a problem with the second part, and i am not sure if I have found a counterexample: If we take n=m=2, lets suppose L= Identity, $$||L||_{op} = \sqrt{(a_{11}x_1 + a_{12}x_2)^2+(a_{21}x_1+a_{22}x_2)^2} $$ if we use the second inequality in the identity matrix $2x2$, we have that $2 \leq \sqrt{2} \sqrt{x_1^2+x_2^2}$
If you can tell me where is the mistake in this or help me to demonstrate could be very helpful. Thanks

Answer 1

For the identity map $I$

$$\|I\|_{op}=\sup \{|L(x)| : |x|=1 \}=\sup \{|x| : |x|=1 \}=1.$$

$$ \|I\|_{tr} = \sum_{i=1}^{n}\sum_{j=1}^{n}(a_{i,j})^2= \sum_{i=1}^{n}\sum_{j=1}^{n}(\delta_{i,j})^2=n$$

(I recall that $\delta_{i,j}=1$ if $i=j$ and $0$, otherwise).

I guess that there should be taken a root from $\|L\|_{tr}$ in order to make it norm, that is to assure that $\|\lambda L\|_{tr}=|\lambda|\|L\|_{tr}$ for each real $\lambda$, because in the present state we have $\|\lambda L\|_{tr}= |\lambda|^2\|L\|_{tr}$. In particular, $\|\lambda I\|_{op}=|\lambda|$, whereas $\|\lambda I\|_{tr}=|\lambda|^2$, so $0<|\lambda|<1$ the left part inequality fails too. But for corrected norm $\|L\|_{tr}=\sqrt{\sum_{i=1}^{n}\sum_{j=1}^{m}(a_{i,j})^2}$ the inequality $ \|L\|_{op} \leq \|L\|_{tr}$ indeed follows from Cauchy-Schwartz inequality.