I am taking a first course on numerical methods and wanted to prove the below result on spectral radius. I would like to ask if my proof is correct.
We are given that, if $A$ is an $n \times n$ matrix, then $\vert \vert A \vert \vert_2 = [\rho(A^TA)]^{1/2}$.
Prove that if $A$ is a symmetric matrix, $\vert \vert A \vert \vert_2 = \rho(A)$.
Proof.
(1) Let $\lambda$ be the eigenvalue of the matrix $A$ corresponding to the eigenvector $x$.
$\begin{align} Ax &= \lambda x \\ (A^T A)x &= \lambda A^T x \\ &= \lambda Ax \space \text{...($A$ is symmetric)} \\ &= \lambda^2 x \end{align}$
So, $A^T A$ has the eigenvalues $\lambda^2$ corresponding to the eigenvector $x$.
(2) By definition, the spectral radius of $(A^T A)$ is $\rho(A^T A) = \max \vert \lambda \vert^2$.
(3) Let $x$ be an arbitrary vector of length $1$. $\vert \vert x \vert \vert_2 = 1$.
$\begin{align} \vert\vert A \vert\vert_2 &= \max_{\vert\vert x \vert\vert_2=1} \vert\vert Ax\vert\vert_2 \\ \vert\vert Ax\vert\vert_2 &=\left(\sum_i (Ax)_i^2\right)^{1/2}\\ &= \left( (Ax)^{T} (Ax) \right)^{1/2}\\ &= \left( x^T A^T A x \right)^{1/2}\\ &= \left( \lambda^2 x^T x \right)^{1/2}\\ &= \left( \lambda^2 \vert \vert x \vert \vert_2^2 \right)^{1/2}\\ &= \lambda \end{align}$
From (2) and (3), we have :
$\vert\vert A \vert\vert_2 = [\rho(A^T A)]^{1/2}=\max \vert \lambda \vert$ = $\rho(A)$
and
$\vert\vert A \vert\vert_2 =\max \vert\vert Ax \vert\vert_2$ = $\max \vert \lambda \vert$