# Why the multiplication of a covariance matrix with the inverse of its sum with the identity matrix is symmetric?

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)

 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.

• Sep 14, 2021 at 7:16
• 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 $P'=P^{-1}$) Sep 14, 2021 at 7:17
• @lmaosome why is $\Xi$ diagonal? Sep 14, 2021 at 7:33
• 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 Sep 14, 2021 at 8:11
• Right, got confused with $A$. That's actually a great proof! Sep 14, 2021 at 9:18

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

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).

• Accepting this answer but see also lmaosome's comment on why those matrices commute. Sep 14, 2021 at 11:24

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).

• That is what I don't get, why they commute? Thanks. Sep 14, 2021 at 7:33
• 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 $P^{-1}$ Sep 14, 2021 at 8:15
• @GioraSimchoni is this edit better? Sep 14, 2021 at 9:04
• @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. Sep 14, 2021 at 9:22
• $[A,B]=AB-BA$ and is called the commutator Sep 14, 2021 at 10:01