For a symmetric matrix $A = Q\Lambda A^T$ where $Q$ is an orthogonal matrix. How can one prove that
$$\frac{||Ax||}{||x||} = \frac{||\Lambda x||}{||x||}?$$ I understand that intuitively it happens because $Q$ does not change the norm of $x$, but how can one prove this?
I assume you mean $A=Q \Lambda Q^{T}$ with $\Lambda$ (real-valued) diagonal and $Q$ orthogonal ($Q^T Q = Q Q^T = I$). $$A^T A = Q \Lambda Q^T Q \Lambda Q^T = Q \Lambda \Lambda Q^T$$ So $$||Ax||^2=x^T A^T A x = x^T Q \Lambda \Lambda Q^T x = ||\Lambda Q^T x||^2$$ and $$||Q^T x||^2 = x^T Q Q^T x = x^T x = ||x||^2$$ So $$\frac{||Ax||}{||x||} = \frac{||\Lambda Q^T x||}{||x||} = \frac{||\Lambda Q^T x||}{||Q^Tx||}$$ Letting $x$ range over all non-zero vectors, $y = Q^T x$ ranges over all non-zero vectors so $$||A||_{o} = \text{sup}_{x \ne 0} \frac{||Ax||}{||x||} = \text{sup}_{y \ne 0} \frac{||\Lambda y||}{||y||} = ||\Lambda||_{o}$$ where $||||_{o}$ denotes the operator norm.
For the Hilbert-Schmidt (Frobenius) norm, $$||A||^2_{\text{hs}} = \text{trace}(A^T A) = \text{trace}(Q \Lambda^2 Q^T) = \text{trace}(\Lambda^2 Q^T Q) = \text{trace}(\Lambda^2) = ||\Lambda||^2_{\text{hs}}$$