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I have constant matrices $A$,$S$ & $B$

$$(A'SA)^{-1}A'SB$$

Can I simplify to $(A'A)^{-1}S^{-1}SA'B$, where $S^{-1}S = I$ and therefore drops out?

snoram
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  • $(A'SA)^{-1} A'S B= A^{-1} (A'S)^{-1} (A'S) B=A^{-1}B$. Assuming all inverses exit. –  Feb 16 '14 at 17:00

2 Answers2

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No you cannot, at least not directly.

Hint: for invertible (and compatible) matrices $A,B$, we have $(AB)^{-1} = B^{-1}A^{-1}$. Likewise, we can show $(ABC)^{-1} = C^{-1}B^{-1}A^{-1}$. That is, ${}^{-1}$ reverses the order of multiplication.

In general, $AB \neq BA$, and thus $(AB)^{-1} \neq A^{-1}B^{-1}$.

Ben Grossmann
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Not sure what your driving at here, but if you note that

$(A'SA)^{-1} = A^{-1}S^{-1}A'^{-1}, \tag{1}$

and substitute this back into $(A'SA)^{-1}A'SB$ you obtain

$(A'SA)^{-1}A'SB = A^{-1}S^{-1}A'^{-1}A'SB= A^{-1}S^{-1}SB = A^{-1}B, \tag{2}$

and that's about as far as I can take it. Not that $S$ still drops out, as does $A'$. I assume $A$ and $S$ are invertible, hence is $A'$, which I take to be the transpose of $A$.

Robert Lewis
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