How can I derive this bound for the spectral radius of $A$?

Question

Suppose that $A \in \mathbb C^{n \times n}$, such that $\rho(A)$ is the spectral radius of $A$ (its largest eigenvalue in magnitude) and $\bar\sigma(A)$ is the largest singular value of $A$. How can I show that $$\rho(A) \leq \inf_{\det(D) \neq 0} \bar\sigma(DAD^{-1})$$ for invertible matrices $D \in \mathbb C^{n \times n}$? I'm aware that $A$ and $DAD^{-1}$ have the same spectral radius because they are similar, but I'm not sure if this helps.

Answer 1

Lemma 1: For any matrix $A$ we have $\rho(A)\leq \overline{\sigma}(A)$.

Proof: see Prove that the largest singular value of a matrix is greater than the largest eigenvalue.

Lemma 2: If $D$ is invertible, then $\rho(DAD^{-1}) = \rho(A)$.

Proof: Similar matrices have the same spectral value.

We deduce that for any invertible $D$ we have $$\rho(A)=\rho(DAD^{-1}) \leq \overline{\sigma}(DAD^{-1})$$ It follows that $$\rho(A)\leq \inf_{det(D)\not=0} \overline{\sigma}(DAD^{-1}),$$ as desired.