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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.

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1 Answer 1

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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 $A^{-1}$.

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  • $\begingroup$ 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? $\endgroup$
    – Yang
    Feb 7, 2020 at 15:54
  • $\begingroup$ @Yang You can orthogonally diagonalise $B$ and apply the same similarity transform to $A$. $\endgroup$
    – user1551
    Feb 7, 2020 at 16:05
  • $\begingroup$ 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 $A+B$ $\endgroup$
    – Yang
    Feb 7, 2020 at 16:14
  • $\begingroup$ @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. $\endgroup$
    – user1551
    Feb 7, 2020 at 16:16
  • $\begingroup$ 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 $\lambda(PAP^{\top} + P D P^{\top}) = \lambda(A+D)$) $\endgroup$
    – Yang
    Feb 7, 2020 at 16:27

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