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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.

mhdadk
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  • Seems like you just need to prove that $\rho(A)\leq \overline{\sigma}(A)$. Do you know how to do that? – Yanko Nov 30 '22 at 23:51
  • @Yanko yes, but how does this help? – mhdadk Dec 01 '22 at 00:03
  • Because then you get that $\rho(DAD^{-1})\leq \overline{\sigma}(DAD^{-1})$, but you just noted that $\rho(A)=\rho(DAD^{-1})$. So here you get that $$\rho(A)\leq \overline{\sigma}(DAD^{-1})$$ for all invertible $D$ which should give the inequality you desire. – Yanko Dec 01 '22 at 00:05
  • @Yanko just to check my understanding, the inequalities $$\rho(A)\leq \overline{\sigma}(DAD^{-1})$$ and $$\rho(A) \leq \inf_{\det(D) \neq 0} \bar\sigma(DAD^{-1})$$ are equivalent, right? – mhdadk Dec 01 '22 at 00:08
  • Yes, exactly, that is if you know the first inequality for all invertible $D$. – Yanko Dec 01 '22 at 00:10
  • @Yanko I see. Actually, this must mean that (by definition of $\inf$) $$\rho(A) \leq \inf_{\det(D) \neq 0} \bar\sigma(DAD^{-1}) \leq \bar\sigma(DAD^{-1})$$ right? – mhdadk Dec 01 '22 at 00:12
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    Yes, but this alone is not useful. What you need is that if $\rho(A)\leq \overline{\sigma}(DAD^{-1})$ for all invertible $D$, then $\rho(A)\leq \inf_{D} \overline{\sigma}(DAD^{-1})$. – Yanko Dec 01 '22 at 00:13
  • @Yanko got it. Thanks! Feel free to post an answer and I can accept it. – mhdadk Dec 01 '22 at 00:14

1 Answers1

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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.

Yanko
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