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Given are real symmetric invertible matrices $A_1, A_2, ..., A_n$, a real symmetric invertible matrix $B_0$, the identity matrix $I$ and real numbers $w_1,w_2,...,w_n$. Define the linear combinations $$ R_1 = \sum_{i=1}^n w_i (I+B_0A_i)^{-1} $$ and $$ R_2 = \sum_{j=1}^n w_j A_j(I+B_0A_j)^{-1} $$

Question: Is the matrix $R = R_1^T R_2$ symmetric?

I have quickly tested a Mathematica code, which gives me a true statement for all random cases I have tried.

d = 4;
n = 5;
nd = 10^3;
s = {};
Do[
  A = Symmetrize /@ RandomInteger[{-10, 10}, {n, d, d}];
  B0 = Symmetrize@RandomInteger[{-10, 10}, {d, d}];
  Id = IdentityMatrix@d;
  w = RandomInteger[{-10, 10}, n];
  R1 = Sum[w[[i]]*Inverse[Id + B0.(A[[i]])], {i, n}];
  R2 = Sum[w[[j]]*A[[j]].Inverse[Id + B0.(A[[j]])], {j, n}];
  R = Transpose[R1].R2;
  AppendTo[s, R == Transpose[R]]
  , {di, nd}
  ];
And @@ s

True

Naturally, this only slightly motivates the possibility of the property. I can reformulate $R_1$ and $R_2$ as

$$ R_1 = \sum_{i=1}^n w_i A_i^{-1}(A_i^{-1}+B_0)^{-1} \ , \quad R_2 = \sum_{j=1}^n w_i (A_i^{-1}+B_0)^{-1} $$ such that I can see that $R_2$ is symmetric, but $R_1$ not in general. I am not able to check if $0 = R - R^T = R_1^TR_2 - R_2^TR_1$ holds. I have tried several perspectives the whole day but now I am blind to obvious ones. Can you help me? Thank you!

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  • $\begingroup$ @Crostul what do you mean by multiplication? A product of two symmetric matrices need not be symmetric ! $\endgroup$ – Arnaud D. Jun 28 '17 at 9:18
  • $\begingroup$ @Crostul sorry, I dont know what you mean there, could you elaborate your comment more explicitly? The matrix product of 2 symmetric matrices is not symmetric in general, as far as I remember. $R_1$ is not symmetric. $\endgroup$ – Mauricio Fernández Jun 28 '17 at 9:19
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Yes, it is true.

$R_2 = \sum_j w_j B_0^{-1}(-I+I+B_0A_j)(I+B_0A_j)^{-1}$

Hence

$R_2 = w B_0^{-1} - B_0^{-1} \sum_j w_j(I + B_0A_j)^{-1}$

where $w = \sum_j w$. Thus

$R_2 = w B_0^{-1} - B_0^{-1}R_1$

So $R_1^T R_2 = w R_1^T B_0^{-1} - R_1^T B_0^{-1} R_1$

The second term is clearly symmetric. Let us consider the first term, while dropping the scalar factor $w$ to simplify the notation:

$R_1^T B_0^{-1} = \sum_i w_i [(I+B_0A_i)^T]^{-1} B_0^{-1} = \sum_i w_i (B_0 + B_0A_i B_0)^{-1}$

which is symmetric (I thank the OP, it was kind of fun).

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  • $\begingroup$ Thank you very very much! This was driving me nuts! $\endgroup$ – Mauricio Fernández Jun 28 '17 at 11:38
  • $\begingroup$ You're welcome. Yes, I did try several things out before figuring it out. It wasn't that obvious in a way, though once understood, it "looks" easy. $\endgroup$ – Malkoun Jun 28 '17 at 11:41

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