Does the following result hold? [duplicate]

Question

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?

Answer 1

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".

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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.