# On the Rayleigh quotient of two matrices

For a given real symmetric matrix $$M$$ and nonzero vector $$x$$, the Rayleigh quotient of $$M$$ is defined as $$R(M,x)=\frac{x'Mx}{x'x}$$. I know that for a given matrix, the Rayleigh quotient reaches its minimum value $$\lambda_{\min}$$ (the smallest eigenvalue of $$M$$) when $$x$$ is $$v_\min$$ (the corresponding eigenvector). Similarly, $$R ( M , x )\leq \lambda _{\max }$$ and $$R(M, v_\max) = \lambda_\max$$.

Let $$A$$ be a non-negative irreducible matrix and $$B$$ is a matrix obtained from $$A$$ by adding some ones to it.

how can I conculde that $$\lambda^A_{\max}<\lambda_{\max}^B$$?

I suppose that you mean $$\rho(A)<\rho(B)$$, otherwise your question does not make sense, as the spectra of $$A$$ and $$B$$ may contain non-real eigenvalues. If so, the inequality is true and Rayleigh quotient has nothing to do with it. The inequality is a simple consequence of Perron-Frobenius theorem, as discussed here. In general, if $$A$$ and $$B$$ are two different nonnegative matrices such that $$A+B$$ is irreducible, then $$\rho(A)<\rho(A+B)$$.