Showing matrix C has spectral radius less than 1

Question

We are given that $A$ is symmetric and positive definite, and $B$ is such that $A-B-B^T$ is also symmetric and positive definite. We are asked to show that $C=-(A-B)^{-1} B$ has spectral radius less than 1, i.e. its eigenvalues are all between -1 and 1. I know that positive definite implies all eigenvalues are positive and that the matrix is therefore invertible.

I started with the eigenvalue problem, $Cv=\lambda v$ where I will call $\lambda$ the eigenvalue of $C$, and $v$ eigenvector. $$ -(A-B)^{-1} Bv=\lambda v \\ -Bv=\lambda(A-B)v $$ From here I attempted numerous things, like multiplying both sides by v^T, and trying to find a contradiction if $|\lambda|>1$, however I have not had any success. The key is probably to use $A-B-B^T$, but I do not know how. Any help would be much appreciate, thank you. EDIT: ALL matrices above are real n*n.

Answer 1

First, consider in detail the case where $A = I$. Note that $$ I - B - B^T = \frac 12[(I - 2B) + (I - 2B)^T] $$ So, the fact that $I - B - B^T$ is positive definite implies that $x^T(I - 2B)x > 0$ for all $x \neq 0$. From the result here, we can conclude that the eigenvalues of $I - 2B$ have positive real part. It follows that the eigenvalues of $B$ have real part strictly less than $1/2$.

Now, note that the eigenvalues of $-(I - B)^{-1}B$ have the form $$ -(1 - \lambda)^{-1}\lambda = - \frac{\lambda }{1 - \lambda} $$ for any eigenvalue $\lambda$ of $B$. If we write $\lambda = a + bi$, then the square absolute value of this eigenvalue satisfies $$ \left|- \frac{\lambda }{1 - \lambda}\right|^2 = \frac{|\lambda|^2}{|1 - \lambda|^2} = \frac{a^2 + b^2}{(1-a)^2 + b^2}. $$ Now, because $a < 1/2$, we have $$ (1 - a)^2 = a^2 + (1 - 2a) > a^2 + 0 = a^2. $$ Thus, $$ \left|- \frac{\lambda }{1 - \lambda}\right|^2 = \frac{a^2 + b^2}{(1-a)^2 + b^2} < \frac{a^2 + b^2}{a^2 + b^2} = 1. $$ For the general case, note that we can write $$ A - B - B^T = A^{1/2}(I - A^{-1/2}BA^{-1/2} - [A^{-1/2}BA^{-1/2}]^T)A^{1/2}. $$ Because $A - B - B^T$ is positive semidefinite, we can conclude that $I - M - M^T$ is positive semidefinite, where $M = A^{-1/2}BA^{-1/2}$. From our earlier work, conclude that the matrix $-(I - M)^{-1}M$ has spectral radius less than $1$. On the other hand, note that $$ -(I - M)^{-1}M = \\ -(I - A^{-1/2}BA^{-1/2})^{-1}A^{-1/2}BA^{-1/2} = \\ -(A^{-1/2}[A - B]A^{-1/2})^{-1}A^{-1/2}BA^{-1/2} = \\ -A^{1/2}[A - B]^{-1}A^{1/2}A^{-1/2}BA^{-1/2} = \\ -A^{1/2}[A - B]^{-1}A^{1/2}A^{-1/2}BA^{-1/2} = \\ -A^{1/2}[A - B]^{-1}BA^{-1/2}. $$ This matrix is similar to $[A - B]^{-1}B$ and thus has the same eigenvalues. So, the eigenvalues of $[A - B]^{-1}B$ are all less than $1$ in magnitude, which is what we wanted.

Answer 2

This answer is similar to Ben Grossmann’s, but involves fewer calculations. Note that $$ C=(A-B)^{-1}(-B)=(A-B)^{-1}[(A-B)-A]=I-(A-B)^{-1}A=I-(I-A^{-1}B)^{-1} $$ is similar to $Z=A^{1/2}CA^{-1/2}=I-(I-A^{-1/2}BA^{-1/2})^{-1}$. Since $A\succ B+B^T=B+B^\ast$, we obtain $$ (I-Z)^{-1}+(I-Z^\ast)^{-1}=2I-A^{-1/2}(B+B^\ast)A^{-1/2}\succ2I-A^{-1/2}(A)A^{-1/2}=I. $$ Let $(\lambda,v)$ be a possibly complex eigenpair of $Z$ with $|\lambda|=\rho(Z)=\rho(C)$. By the previous inequality, $$ 1<v^\ast\big[(I-Z)^{-1}+(I-Z^\ast)^{-1}\big]v =\frac{1}{1-\lambda}+\frac{1}{1-\overline{\lambda}} =\frac{2-(\lambda+\overline{\lambda})}{|1-\lambda|^2}. $$ Hence $1-(\lambda+\overline{\lambda})+|\lambda|^2=|1-\lambda|^2<2-(\lambda+\overline{\lambda})$, meaning that $\rho(C)=\rho(Z)=|\lambda|<1$.