How can I proof this result? $$ X(X+Y)^{-1}Y=(X^{-1} +Y^{-1})^{-1} $$ where $X$, $ Y$, $(X+Y)$, and $(X^{-1} +Y^{-1})$ are symmetric and invertible matrices,each one with dimensions $p\times p$.
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2Have you tried multiplying the left hand side by $(X^{-1}+Y^{-1})$ ? – hardmath Sep 17 '14 at 11:40
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You of course need to assume $X,Y$ are invertible. Symmetry is not important here. – LinAlgMan Sep 17 '14 at 11:50
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1@LinAlgMan And also $X+Y$ and $X^{-1}+Y^{-1}$ ;-) – Jean-Claude Arbaut Sep 17 '14 at 11:51
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We have
$$(X(X+Y)^{-1}Y)^{-1}=Y^{-1}(X+Y)X^{-1}=Y^{-1}XX^{-1}+Y^{-1}YX^{-1}\\=Y^{-1}+X^{-1} $$
and the result follows.
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For number you have $$ \frac{1}{x^{-1} + y^{-1}} = \frac{xy}{y+x}$$
For matrices, let $Z = X^{-1} + Y^{-1}$, I want to show that $X(X+Y)^{-1}Y Z = I$.
$$ X(X+Y)^{-1}Y ( X^{-1} + Y^{-1} ) = X(X+Y)^{-1}YX^{-1} + X(X+Y)^{-1}YY^{-1} = $$ $$ = X(X+Y)^{-1}(YX^{-1} + I) = X(X+Y)^{-1}(YX^{-1} + I)XX^{-1} = $$ $$ = X(X+Y)^{-1}(Y + X)X^{-1}) = XX^{-1} = I$$
LinAlgMan
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