Is it somehow possible to reformulate the following exuation into something easier to calculate:
$$(A^{-1}+ B^{-1})^{-1}$$
A and B are both square real matrices: $A, B \in \mathbb{R}^{n \times n}$, and are positive definite and therefore invertible.
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1What’s a quadratic matrix? Is that the same as symmetric? – Joe Jul 15 '20 at 10:55
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2I don't think so as the sum is not neccessarily invertible (take A = Id = A^-1 and B = -Id = B^-1) – Stockfish Jul 15 '20 at 10:58
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2By quadratic matrix did you mean square matrix? – J. W. Tanner Jul 15 '20 at 10:59
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Oh, I get it: quadratic = second power = square. I thought quadratic was a reference to quadratic form, meaning a symmetric matrix. – Joe Jul 15 '20 at 11:06
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1@Joe More likely it's that "quadrata" is the word for square in Italian – Ben Grossmann Jul 15 '20 at 11:52
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1Related post – Ben Grossmann Jul 15 '20 at 12:02
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@joe Sorry for the confusion, yes i meant square matrix when writing quadratic. Its adapted now – Manumerous Jul 15 '20 at 12:18
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@Stockfish, thanks for pointing out, i adapted the problem to positive definite matrices. – Manumerous Jul 15 '20 at 12:20
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Note that $$ A^{-1}(A + B)B^{-1} = A^{-1}AB^{-1} + A^{-1}BB^{-1} = B^{-1} + A^{-1}. $$ That is, we have $$ A^{-1} + B^{-1} = A^{-1}(A + B)B^{-1} \implies (A^{-1} + B^{-1})^{-1} = B(A + B)^{-1}A. $$ If you prefer, this can also equal to $A(A + B)^{-1}B$.
Note that because $A,B$ are positive definite, $A + B$ is also positive definite and therefore invertible.
Kai
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Ben Grossmann
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5Furthermore, we can't expect to get any other expressions, because the special case for $1\times1$ matrices has to be $AB(A+B)^{-1}$, so the general case must use these three factors. +1. – J.G. Jul 15 '20 at 11:57
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1The first line should read $A^{-1}(A+B)B^{-1} = A^{-1}AB^{-1} + A^{-1}BB^{-1} = B^{-1} + A^{-1}$ – Kai Jul 15 '20 at 21:24