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 $0 < \lambda_1 \leq \dots \leq \lambda_n$, 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.
