Showing matrix C has spectral radius less than 1

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.

• I would suggest that you begin by focusing on the case that $A$ is the identity matrix Nov 29, 2023 at 17:44
• Also, it's incorrect to say that $C$ has eigenvalues between $-1$ and $1$ in the absence of further information; it is possible that $C$ has complex eigenvalues (with absolute value less than $1$). Nov 29, 2023 at 17:57
• @Ben Grossmann, thank you very much for your help and for editing my post, I could not figure out to how to the notation looks nice. I made 1 important clarification, the matrices are all real. Nov 29, 2023 at 18:23
• The fact that the matrices are real (which I indeed took as given) does not change the fact that $C$ might have complex eigenvalues. As an example, consider $$A = \pmatrix{1&0\\0&1}, \quad B = \pmatrix{0&-1\\1&0}.$$ Nov 29, 2023 at 18:31
• You are right, I did not consider that. Then it makes sense that I had no success with manipulating Cv=λv. Nov 29, 2023 at 18:35

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.
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, $$1 Hence $$1-(\lambda+\overline{\lambda})+|\lambda|^2=|1-\lambda|^2<2-(\lambda+\overline{\lambda})$$, meaning that $$\rho(C)=\rho(Z)=|\lambda|<1$$.