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I have an empirical result (meaning it is always true by simple simulation e.g. in R) which I cannot prove to myself:

Let $A$ be a $n \times n$ covariance matrix (i.e. it is symmetric PSD), let $I_n$ be the identity matrix, $\theta_1$ and $\theta_2$ some scalars (in my case they are always positive but it does not matter). Let:

$V = (\theta_1 A + \theta_2I_n)^{-1}A$

It seems that $V$ is always symmetric! Can we prove it?

E.g. in R:

A <- cov(rbind(c(1,2.1,3), c(3,4,5.3), c(3,4.2,0)))
isSymmetric(solve(2 * A + 3 * diag(3)) %*% A)
[1] TRUE

To anyone interested: it is important to me mainly because this means I have two symmetric matrices $A, B$ which multiply to a symmetric matrix $AB$, in which case its eigenvalues are in fact multiplications of the eigenvalues of $A$ and $B$ according to this, which also simplifies its trace.

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  • $\begingroup$ Related: math.stackexchange.com/questions/126306/… $\endgroup$ Sep 14, 2021 at 7:16
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    $\begingroup$ Note that $A$ can be decomposed into $P\Lambda P^{-1}$. Furthermore, $I = PIP^{-1}$. So, $(\theta_1 A + \theta_2I)^{-1}A = (\theta_1P\Lambda P^{-1} + \theta_2PIP^{-1})^{-1}P\Lambda P^{-1} = P(\theta_1\Lambda + \theta_2I)^{-1}P^{-1}P\Lambda P^{-1}$, which means that the resulting matrix is of the form $P\Xi P^{-1}$ with $\Xi$ being diagonal. The product is hence symmetric (as $P$ is orthonormal, ie $P'=P^{-1}$) $\endgroup$
    – lmaosome
    Sep 14, 2021 at 7:17
  • $\begingroup$ @lmaosome why is $\Xi$ diagonal? $\endgroup$ Sep 14, 2021 at 7:33
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    $\begingroup$ because $P(\theta_1\Lambda + \theta_2 I)^{-1}P^{-1}P\Lambda P^{-1} = P(\theta_1\Lambda + \theta_2I)^{-1}\Lambda P^{-1}$. Now put $\Xi = (\theta_1\Lambda + \theta_2I)^{-1}\Lambda$. It is obvious that $\theta_2I$ is diagonal, $\Lambda$ is diagonal by construction, and the product of diagonal matrices is also diagonal. So $\Xi$ is diagonal $\endgroup$
    – lmaosome
    Sep 14, 2021 at 8:11
  • $\begingroup$ Right, got confused with $A$. That's actually a great proof! $\endgroup$ Sep 14, 2021 at 9:18

2 Answers 2

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$$ V = (\theta_1 A + \theta_2I_n)^{-1}A \implies (\theta_1 A + \theta_2I_n)V = A\\ \implies V^*(\theta_1 A + \theta_2I_n)^* = A^* \implies V^*(\theta_1 A + \theta_2I_n) = A\\ \implies V^*(\theta_1 A + \theta_2I_n)A^{-1} = I_n \implies V^*(\theta_1 I_n + \theta_2A^{-1}) = I_n\\ \implies V^*A^{-1}(\theta_1 A + \theta_2I_n) = I_n \implies V^*A^{-1} = (\theta_1 A + \theta_2I_n)^{-1}\\ \implies V^* = (\theta_1 A + \theta_2I_n)^{-1}A $$

Notice: it is way shorter if you know that $()^*$ and $()^{-1}$ commute


To state a general theorem, given $p(x,y),q(x,y)$ two polynomials (or even any function, if you know how to apply those to matrices), then $p(A,A^{-1})$ and $q(A,A^{-1})$ commute if $A$ is symmetric (Hermitian if complex).

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  • $\begingroup$ Accepting this answer but see also lmaosome's comment on why those matrices commute. $\endgroup$ Sep 14, 2021 at 11:24
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If $$XY=YX$$ then by multiplying by left and right by $X^{-1}$: $$YX^{-1}=X^{-1}Y$$ Thus since $[(\theta_1 I+\theta_2 A),A]=0$ also $[(\theta_1 I+\theta_2 A)^{-1},A]=0$, thus the product is symmetric (all this assuming all inverses exist etc).

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  • $\begingroup$ That is what I don't get, why they commute? Thanks. $\endgroup$ Sep 14, 2021 at 7:33
  • $\begingroup$ my comment contains the proof; Note that $(\theta_1\Lambda + \theta_2I)^{-1}$ and $\Lambda$ commute. After interchanging their positions, factor back in $P$ and $P^{-1}$ $\endgroup$
    – lmaosome
    Sep 14, 2021 at 8:15
  • $\begingroup$ @GioraSimchoni is this edit better? $\endgroup$
    – user619894
    Sep 14, 2021 at 9:04
  • $\begingroup$ @user619894 I got why they commute with lmaosome's comment. This $[A, B]$ notation is unclear to me but I'm guessing I'm the problem. Thanks. $\endgroup$ Sep 14, 2021 at 9:22
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    $\begingroup$ $[A,B]=AB-BA$ and is called the commutator $\endgroup$
    – user619894
    Sep 14, 2021 at 10:01

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