# Upper bound of eigenvalues of a matrix $(A + B)^{-1} B$

Consider a positive-definite matrix $$A \in \mathbb{R}^{n \times n}$$ and a positive semi-definite matrix $$B \in \mathbb{R}^{n \times n}$$, and consider the $$i$$-th largest eigenvalue of $$(A + B)^{-1} B$$. I know all eigenvalues of this matrix are less than 1, and is it possible to obtain a non-trivial upper bound on $$\lambda_i(A + B)^{-1} B)$$ that is less than 1?

I've tried Weyl's inequality and a conclusion from Bathias' "Matrix Analysis" that says, for any two operators $$A, B$$ on Hilbert space $$H$$ with dimension $$n$$, for all $$i, j$$ such that $$i+j \leq n+1, \lambda_{i+j-1}(A B) \leq \lambda_i(A) \lambda_j(B)$$, where $$\lambda_i(A)$$ is the $$i$$-th largest eigenvalue of $$A$$. But the bound I obtain in terms of the eigenvalues of $$A$$, $$B$$ is trivial - the bound I obtain may be larger than 1.

Any help is greatly appreiciated.

Let $$B=\pmatrix{B_1&0\\ 0&0}$$ and $$A=\pmatrix{X&Y^T\\ Y&Z}$$, where $$B_1$$ is an $$r\times r$$ positive definite matrix. Then $$(A+B)^{-1}B =\pmatrix{(B_1+S)^{-1}&\ast\\ \ast&\ast}\pmatrix{B_1&0\\ 0&0} =\pmatrix{(I+SB_1^{-1})^{-1}&0\\ \ast&0},$$ where $$S=X-Y^TZ^{-1}Y$$ the Schur complement of $$Z$$ in $$A$$. Therefore, when $$i,j,k\in\{1,2,\ldots,r\}$$ and $$i+j=k+1$$, \begin{align} \lambda^{\downarrow}_k\left((A+B)^{-1}B\right) &=\frac{1}{1+\lambda^{\uparrow}_k(SB_1^{-1})} =\dfrac{1}{1+\dfrac{1}{\lambda^{\downarrow}_k(S^{-1}B_1)}}\\ &\le\dfrac{1}{1+\dfrac{1}{\lambda^{\downarrow}_i(S^{-1})\lambda^{\downarrow}_j(B_1)}}\\ &\le\dfrac{1}{1+\dfrac{1}{\lambda^{\downarrow}_i(A^{-1})\lambda^{\downarrow}_j(B_1)}} =\dfrac{1}{1+\dfrac{\lambda^{\uparrow}_i(A)}{\lambda^{\downarrow}_j(B)}} \end{align} where the first inequality is the one mentioned in your question and the second one is due to the fact that $$S^{-1}$$ is a principal submatrix of $$A^{-1}$$.

• Thank you! One question: why can we assume B has the form $\begin{bmatrix} B_1 & 0 \\ 0 & 0 \end{bmatrix}$? I understand we can have a matrix of this form (let $B' = \begin{bmatrix} B_1 & 0 \\ 0 & 0 \end{bmatrix}$) with the same eigenvalues of a positive definite $B$ and but I don't know why $(A + B')^{-1} B'$ would necessarily have the same eigenvalues as $(A + B)^{-1} B$. Could you clarify?
– Yang
Feb 7, 2020 at 15:54
• @Yang You can orthogonally diagonalise $B$ and apply the same similarity transform to $A$. Feb 7, 2020 at 16:05
• Yes I understand diagonalization could give us the diagonal form, but I wonder whether the sum of $\begin{bmatrix} B_1 & 0 \\ 0 & 0 \end{bmatrix}$ and $A$ would still have the same eigenvalues as $A+B$
– Yang
Feb 7, 2020 at 16:14
• @Yang If you apply the same similarity transform to $A$, the eigenvalues certainly remain unchanged. This is just a change of basis for both linear operators. Feb 7, 2020 at 16:16
• I am sorry that I still do not get it. Say I have $B = P D P^{\top}$, then are you saying $\lambda(A+B) = \lambda(PAP^{\top} + P D P^{\top})$? (And I understand $\lambda(PAP^{\top} + P D P^{\top}) = \lambda(A+D)$)
– Yang
Feb 7, 2020 at 16:27