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I've read in several places now that $\rho(AB) \leq \rho(A)\rho(B)$ holds if the matrices are normal. I've scoured the web for a proof since I can't seem to get there myself. How would one go about to prove this statement? Links to proof is also welcome.

Václav Mordvinov
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CupinaCoffee
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1 Answers1

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Note that for any matrix, the induced 2-norm of that matrix is defined as $$\|A\|_2:=\sup_{x\neq0}\frac{\|Ax\|_2}{\|x\|_2},$$ and thus by the spectral theorem for normal matrices, there exists a basis of eigenvectors of $A$ for $\mathbb{C}^{n\times n}$, and thus $\|A\|_2=\rho(A)$. Similarly $\|B\|_2=\rho(B)$. For $AB$, which is not necessarily normal, we hace $\|AB\|_2=\sigma_{\mathrm{max}}(AB)$, the maximal singular value of $AB$, see also this post.

Now using this, we can see directly that $$\rho(AB)\leq\sigma_{\mathrm{max}}(AB)=\|AB\|_2\leq\|A\|_2\|B\|_2=\rho(A)\rho(B),$$ where the second inequality holds because the induced $2$-norm $\|\cdot\|$ is indeed a norm.

For the first inequality, observe that for any induced matrix $p$-norm, we have for any $v\in\mathbb{C}^n$, and therefore also for $v$ the eigenvector with norm $1$ corresponding to $|\lambda|=\rho(A)$: $$\|A\|_p:=\sup_{x\neq0}\frac{\|Ax\|_p}{\|x\|_p}=\sup_{\|x\|_p=1}\|Ax\|_p\geq\|Av\|_p=\|\lambda v\|_p=\rho(A)\|v\|_p=\rho(A).$$

Václav Mordvinov
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  • Thank you for your comment. This seem to be the particular step I'm having trouble with. I simply don't understand when the inequality can be deduced. All I'm getting is that $\rho({AB}) = \rho({BA})$. – CupinaCoffee May 23 '18 at 12:58
  • Give me some time to make an edit. My answer is not clear as I wrote it. – Václav Mordvinov May 23 '18 at 13:00
  • I thought $\rho(AB) = \sigma_{max}(AB)$. Isn't the spectral radius defined as the largest absolute eigenvalue $\vert{\lambda}{max}\vert$, and the singular value as $\sqrt{\lambda{max}^2}$? – CupinaCoffee May 23 '18 at 13:48
  • Singular values are defined to be the square roots of the eigenvalues of $A^*A$. In case the matrix is not normal (we do not know whether $AB$ is normal), we cannot apply the spectral theorem to conclude that the singular values are the absolute values of the eigenvalues. – Václav Mordvinov May 23 '18 at 13:51
  • Ah. Right. Thank you so much Václav. It makes sense now. – CupinaCoffee May 23 '18 at 13:55