Does the following result hold? [duplicate]

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, HaykMay 21 at 12:25

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.
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.
• 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 – Math_Freak May 21 at 3:56