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.
By the spectral theorem, there is an orthonormal basis of eigenvectors of $A$, say  $x_1,\dots,x_n$. WLOG, assume $0 &lt; \lambda_1 \leq \dots \leq \lambda_n$, where $Ax_i = \lambda_i x_i$. For any non-zero $x$, write $x = c_1x_1 + \cdots + c_nx_n$. Then
$$ \frac{\lVert{Ax\rVert}^2}{\lVert{x\rVert}^2} = \frac{\lVert{c_1\lambda_1x_1 + \cdots + c_n\lambda_nx_n\rVert}^2}{c_1^2 + \cdots + c_n^2} = \frac{c_1^2\lambda_1^2 + \cdots + c_n^2\lambda_n^2}{c_1^2 + \cdots + c_n^2} \leq \lambda_n^2 \frac{c_1^2 + \cdots + c_n^2}{c_1^2 + \cdots + c_n^2} = \lambda_n^2,$$
with equality achieved when $x = x_n$ (whence $\lVert{Ax\rVert}=\lVert{\lambda_n x_n\rVert}=\lambda_n$).
You may wish to read about Rayleigh quotients or see Problem 37 in Brezis.
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))