Norm of a Positive definite matrix is the largest eigenvalue Let $A$ be a positive definite symmetric matrix. I need to show that
$$\lambda_n=\max \{\frac{\|Ax\|}{\|x\|}: x\ne 0\}$$ is the largest eigen value of $A$. My try is $\frac{\|Ax\|}{\|x\|}\le \lambda,\forall \lambda$ as a not eigen value of $A$, and the equality occurs when $\lambda$ is an eigen value. So $\lambda$ is maximum? May be I am not even understanding the question.

It is all of the answer
first you can prove that for each eigenvalue of positive definite matrix $M$ you have
$$ \operatorname{eig}(M^2)=\operatorname{eig}(M)^2.$$
In the other words if $s$ is the eigenvalue of $M$ then $s^2$ is the eigenvalue of $M^2$ since $s$ is the eigenvalue of $M$: $|s\cdot I-M|=0$
now by means of determinant properties we can write
|s^2*I-M^2|=|(sI-M)(sI+M)|=|sI-M||sI+M|=0
we proved that s^2 is the eigenvalue of M^2
we now that 2 norm of a matrix M is derived from this equation
||M||=sqrt(max(eig(M'*M)))
you should now that your claim  is only true for symmetric matrix M
so we can write
||M||=sqrt(max(eig(M^2)))
according to our proof we can write
||M||=sqrt(max(s^2)))
since M is positive definite s is positive now we can write
||M||=max(s)=max(eig(M))
