Suppose that $A$ and $B$ are symmetric and non-negative matrices.

Let $\rho(A)$ denote the spectral radius of $A$ and $\rho(B)$ denote the spectral radius of $B$.

Does the following result hold?

$\rho(A+B)\le \rho(A)+\rho(B)$

I tried several examples and found the result true.

But is it true in general?


marked as duplicate by Martin R, Jendrik Stelzner, Yanior Weg, John B, Hayk May 21 at 12:25

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It is true if $A$ and $B$ are real. For symmetric real matrices, the spectral radius agrees with the operator norm $$ \|A\|=\max\{\|Ax\|_2:\ \|x\|_2=1\}. $$ And the sum of real symmetric matrices is real symmetric. Thus $$ \rho(A+B)=\|A+B\|\leq\|A\|+\|B\|=\rho(A)+\rho(B). $$ Non-negative is not necessary for all the above.

The same result holds for complex matrices if we replace "symmetric" with "selfadjoint".

Another way to prove it is to use the Min-Max Theorem (Courant-Fischer-Weyl): if $\lambda_1\leq\cdots\leq\lambda_n$ are the eigenvalues of $A$ (and $A$ is Hermitian, so real and symmetric qualifies) $$ \lambda_k=\min_{\dim K=k}\max_{x\in K, \|x\|=1} \langle Ax,x\rangle. $$ When $k=n$, \begin{align} \lambda_n(A+B)&=\max_{x\in K, \|x\|=1} \langle (A+B)x,x\rangle\\ &\leq \max_{x\in K, \|x\|=1} \langle Ax,x\rangle+\langle Bx,x\rangle\\ &\leq \max_{x\in K, \|x\|=1} \langle Ax,x\rangle+\max_{x\in K, \|x\|=1}\langle Bx,x\rangle\\ &=\lambda_n(A)+\lambda_n(B)\\ &\leq\rho(A)+\rho(B). \end{align} Since $\rho(A+B)$ is either $|\lambda_1(A+B)|$ or $|\lambda_n(A+B)|$, we get $$ \rho(A+B)\leq \rho(A)+\rho(B). $$ Again, "non-negative" is not needed.

  • $\begingroup$ Yes they are real.But can you prove this fact "For symmetric real matrices, the spectral radius agrees with the operator norm" $\endgroup$ – Math_Freak May 21 at 3:52
  • $\begingroup$ Is is possible to prove this fact that $\rho(A+B)\le \rho(A)+\rho(B)$ using linear algebra without using any functional analysis techniques.Please comment $\endgroup$ – Math_Freak May 21 at 3:56
  • $\begingroup$ I included another argument. You would have to say what "functional analysis techniques" are. $\endgroup$ – Martin Argerami May 21 at 4:21
  • $\begingroup$ Its clear now,thank you very much for your time & effort $\endgroup$ – Math_Freak May 21 at 4:35

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