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.
 A: By the spectral theorem, there is an orthonormal basis of eigenvectors of $A$, say  $x_1,\dots,x_n$. WLOG, assume $0 < \lambda_1 \leq \dots \leq \lambda_n$, 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.
A: 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))
