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Assume $Q \in \mathbb R^{n\times n}$ is positive semidefinite and $I$ is the identity matrix. How can I prove that the matrix $$M = (I + \alpha Q)^{-1} Q $$ is positive semidefinite for all $\alpha \geq 0$? For all examples I check with Matlab, matrix $M$ is positive semidefinite.

If $Q \in \mathbb R^{n\times n}$ is positive definite, I can proof positive definiteness of the matrix $M$ by including $Q^{-1}$ inside the inversion, obtaining $$ M = (Q^{-1} + \alpha I)^{-1}, $$ which is positive definite, but this is not possible when $Q$ is only semidefinite.

Trb2
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    Hint: $x/(1+\alpha x)\ge 0$ if $x\ge 0$ – Exodd May 26 '24 at 13:46
  • I thought about computing $x^\top M x$ and inspecting whether it is $\geq 0$, but I didn't achieve anything. Do you have a hint on how to continue? – Trb2 May 26 '24 at 13:55

2 Answers2

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Suppose $\lambda$ is an eigenvalue of $M$. Then $$Mv = \lambda v\implies Qv = \lambda(I+\alpha Q)v \implies (1-\lambda \alpha)Qv = \lambda v\implies Qv = \frac{\lambda}{1-\lambda \alpha} v $$ so $\mu = \lambda /(1-\lambda\alpha)$ is an eigenvalue $\mu$ of $Q$ and thus nonnegative. In particular, this shows that $\lambda\ge 0$, otherwise $\mu$ would be negative.

Exodd
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  • with that, in fact you prooved only that the eigenvalues are nonnegative, but not that $M$ is symmetric, and thus also not that $M$ is positive semidefinite, is that right? – Trb2 May 31 '24 at 09:57
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    $M$ is symmetric because $M = (I + \alpha Q)^{-1} Q = (I + \alpha Q)^{-1} Q (I + \alpha Q)(I + \alpha Q)^{-1} = (I + \alpha Q)^{-1} (I + \alpha Q) Q(I + \alpha Q)^{-1} =Q(I + \alpha Q)^{-1} = M^T$ – Exodd May 31 '24 at 11:28
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We notice that the matrix $I+aQ$ is positive-semidefinite, since $x^T(I+aQ)x = x^TIx + x^T(aQ)x$, a sum of two non-negative numbers. The inverse is also a positive semi-definite, since the eigenvalues will just be reciprocals of positive eigenvalues.

One property of positive-semidefinite matrices is that they have "square roots", meaning $Q$ is positive-semidefinite iff there exists positive-semidefinite matrix $R$ such that $Q = RR$. Then $R$ is the square root of $Q$, we can also denote this as $R =Q^{\frac{1}{2}}$.

Let's say $B = (I+aQ)^{-1}$. Then we have $M = BQ = B^{\frac{1}{2}}(B^{\frac{1}{2}}Q).$ We know that $B^\frac{1}{2}QB^\frac{1}{2}$ is positive semi-definite, because $x^TB^\frac{1}{2}QB^\frac{1}{2}x =(B^{\frac{1}{2}}x)^TQ(B^{\frac{1}{2}}x)$ is positive-semidefinite and the equality between the two is allowed by symmetry of $B^\frac{1}{2}$. You can confirm this by taking the transpose of the LHS.

Finally let $S, T$ be matrices, then we have $ST$ and $TS$ with same eigenvalues. Let $u$ be an eigenvector of $ST$, then $Tu$ is an eigenvector of $TS$ with the same eigenvalue: $STu = \lambda_{0}u$, then $TSTu = T(STu) = T\lambda_0u = \lambda_0Tu$.

From this we get that $B^\frac{1}{2}QB^\frac{1}{2}$ is positive semi-definite with the same eigenvalues as $M = B^{\frac{1}{2}}(B^{\frac{1}{2}}Q)$, which is symmetric. Therefore $M$ is positive-semidefinite.

Eemil Wallin
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    This answer is heavily inspired by an answer by Robert Israel: link – Eemil Wallin May 26 '24 at 14:54
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    Addendum: $M$ has the same positive real eigenvalues and $M$ is real, this is equivalent to the positive-semidefinite conditions as long as $M$ is real (which it is). This is why $M$ is symmetric. If $M$ wasn't real, it would be harder to deduce if it is symmetric. – Eemil Wallin May 26 '24 at 16:27
  • I do not understand the last comment, why do we know that $M$ is symmetric? We know that it has nonnegative eigenvalues and that the entries are real, right? – Trb2 May 26 '24 at 16:41