Calculating operator (matrix) norms using eigenvalues? A remark that went unproven in class. 
It was said that the operator norm of a real linear transformation (real matrix) is the square root of the abs value of the max eigenvalue of $A^T*A$ (or maybe $A*A^T$?). 
If $A$ is symmetric (and therefore orthonormally diagonalizable) everything is straightforward... 
But the same calculation for $A^T*A$ only yields the norm of $A^T*A$....
So I want to prove the following lemmas, but they might not be true:
1. The norm of $A$ equals the norm of its transpose.
2. Multiplicative property of the norm when restricting to $A^T*A$ 
Are any of these true, and why?
If not, how can I prove the original claim?
 A: The fact you need is that, for the operator norm, $\|AB\|\leq\|A\|\,\|B\|$. With this,:
\begin{align}
\|A\|^2&=\sup\{\|Ax\|^2:\ \|x\|=1\}=\sup\{(Ax)^TAx:\ \|x\|=1\}
=\sup\{x^TA^TAx:\ \|x\|=1\}\\
&\leq\sup\{\|A^TAx\|\,\|x\|:\ \|x\|=1\}=\sup\{\|A^TAx\|:\ \|x\|=1\}\\
&\leq\|A^TA\|,
\end{align}
where the first inequality is Cauchy-Schwarz and the second one is just by definition of the supremum norm. 
Then
$$
\|A\|^2\leq\|A^TA\|\leq\|A^T\|\,\|A\|,
$$
from where we obtain $\|A^T\|\leq\|A\|$. But we can apply this last inequality to $A^T$, and so $\|A^T\|=\|A\|$, and $\|A\|^2=\|A^TA\|$. 
A: *

*The norm you talking about is a matrix $L_2$ norm  which is defined as following
$$\|A\|_2=\sqrt{\max_n\lambda_n(AA^t)}
$$
where $\lambda_n(B)$ is a n'th eigenvalue of matrix $B$.


This norm is induced from a vector $L_2$ norm given by 
$$\|v\|_2=\sqrt{\sum_i |v_i|^2}$$


*A matrix norm induced from a vector norm is defined by
$$\|A\|=\max_{v}\frac{\|Av\|}{\|v\|}$$
however this is not the only way to get matrix norm (there are norms which aren't induced from a vector norm, e.g. Frobineous norm).

*The property $\|A\|=\|A^T\|$ is not always true, for example for
$L_1$ and $L_\infty$ for matrices the following is true
$\|A\|_1=\|A^T\|_\infty$

*For a quadratic matrices we have $\|AB\|\le \|A\|\|B\|$

