Calculating $\|A\|_2$ in terms of eigenvalues of $A^\ast A$ Let $A$ be a real matrix. I'm supposed to calculate $\|A\|_2$ in terms of the eigenvalues of $A^t A$. I thought to just diagonalize $A^t A$ as $UD^2U^t$ but then I have $\|Ax\|_2 =x^tUD^2U^tx$ instead of $xD^2x=\sum x_i^2\lambda_i ^2$ which I think shows $|\lambda_{\max}|$ should be $\|A\|_2$. What am I missing?
I'm thinking maybe a basis transition is the key here. On the other hand, I'm asked to calculate the norm of a matrix and I don't see how this is only dependent on similarity classes... Help appreciated!
Update: Okay, the calculation below, taken from here, shows this holds (for the Euclidean norm):
$$||A||^2=\sup_{x\neq 0}\frac{||Ax||^2}{||x||^2}=\sup_{x\neq 0}\frac{||J^*DJx||^2}{||x||^2}=\sup_{x\neq 0}\frac{||DJx||^2}{||Jx||^2}\frac{||Jx||^2}{||x||^2}=\sup_{x\neq 0}\frac{||Dx||^2}{||x||^2}.$$

As the answer you looked at indicates, the key here is that a unitary transition of basis preserves the matrix norm.  In particular, if $A$ has SVD $A= UDV^T$, then we'll have $\|A\|=\|D\|$.  So indeed, $\|A\|=\sqrt{\lambda_{max}(A^TA)}$.
