For any square matrix $C$ with real entries, denote by $\rho(C)$ its spectral radius, i.e. the maximum magnitude of its eigenvalues. For symmetric matrices $A$ and $B$ with $AB=BA$ show that $$\rho(AB)\le \rho(A)\rho(B)$$

I think simultaneous diagonalization of $A$ and $B$ is to be used here, but couldn't find my way out.

Also will the proposition hold if the condition of symmetry is dropped?

  • 3
    $\begingroup$ The simultaneous diagonalization means you may assume WLOG that $A$ and $B$ are both diagonal. Can you solve this problem in that case? $\endgroup$
    – Arthur
    Apr 28, 2018 at 9:18
  • $\begingroup$ Oh thanks Arthur. It is easy to show for diagonal matrices. But what if the symmetry condition is dropped $\endgroup$
    – QED
    Apr 28, 2018 at 9:20
  • $\begingroup$ Perhaps this could help with the $AB=BA$ case. $\endgroup$
    – hypernova
    Apr 28, 2018 at 9:45
  • $\begingroup$ Even without the symmetry condition, commuting matrices are simultaneously triangulable over $\mathbb C$. So, this isn't really different from the diagonalisable case. $\endgroup$
    – user1551
    Apr 28, 2018 at 14:57

3 Answers 3


Symmetry is not required. Spectral radius formula says $\rho (A)= \lim \|A^{n}\|^{1/n}$. If $AB=BA$ then $\|(AB)^{n}\|=\|A^nB^n\| \leq \|A^n\|\|B^n\|$. Take $n$-th roots and take the limit.


Since both matricea are simultaneously diagonalizable you can also simultaneously diagonlize $AB$ and find a link between its eigenvalues and the eigenvalues of $A$ and $B$.

A further hint: Calculate $P^{-1}AP\cdot P^{-1}BP$


If $A$ and $B$ are symmetric with $AB = BA$, you have the following even simpler proof : $$\|AB\|_2 = \sqrt{\rho((AB)^TAB)} = \sqrt{\rho((AB)^2)} = \rho(AB)$$ so that by sub-multiplicativity of the spectral norm, $\rho(AB) \leq \|A\|_2 \cdot \|B\|_2$ and the result follows by symmetry of $A$ and $B$ for which $\|\cdot\|_2 = \rho$.


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