Spectral radius, eigenvalues and singular values

Question

We know $\lambda_{max} = \rho(A) \le \lVert A \rVert$, and $\lVert A \rVert_2 = \sigma_{max}$, and so $$\lambda_{max} = \rho(A) \le \lVert A \rVert_2 = \sigma_{max} = \sqrt{\lambda_{max}} \implies \lambda_{max} \le \sqrt{\lambda_{max}}$$

But that can't be true, especially if $\lambda_{max} > 1$. What am I missing?

Answer 1

The line $\sigma_\text{max} = \sqrt{\lambda_{\text{max}}}$ isn't true. Also, strictly speaking, you should have $\rho(A) = |\lambda_{\text{max}}|$. The singular values of a matrix are the square roots of the eigenvalues of $A^\dagger A$, which are all $\geq 0$. So if $A = A^\dagger$ then $\sigma_{\text{max}} = |\lambda_{\text{max}}| = \rho(A)$ but if $A \neq A^\dagger$ then $\sigma_{\text{max}}$ could, for example, be larger than $|\lambda_{\text{max}}|$.