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How can I prove that

$$(C_{N}^{-1} + W_{N})^{-1} = C_{N}(I + W_{N}C_{N})^{-1}$$

I tried to use the Woodbury identity:

$$(A + BD^{-1}C)^{-1} = A^{-1}-A^{-1}B(D+CA^{-1}B)^{-1}CA^{-1}$$

which seems to be useful in this case, but I can't simplify enough to get the result.

Suggestions?

r_31415
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3 Answers3

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Think about what happens if they were just numbers (but don't forget that it's not commutative):

Let's call $A:=C_N$ and $B:=W_N$ for simplicity. If these were numbers, you are looking for $\displaystyle\frac1{1/A\,+B}$. Now we want to pull out $\displaystyle\frac1{1/A}$ (alias $A$) from the left, so that, because of the inverse, we have to pull out $1/A$ from the right: $$1/A+B=(1+BA)\cdot 1/A\\ \frac1{1/A\,+B}=\frac1{1/A}\cdot\frac1{1+BA}\,.$$ Writing it with inverses, we get the same line:

As $A^{-1}+B=(I+BA)A^{-1}$, we have $$(A^{-1}+B)^{-1}= A(I+BA)^{-1}\,.$$

Berci
  • 90,745
  • That's ingenious. How far can I push that analogy? (assuming I take care of the commutative property) – r_31415 May 15 '13 at 22:35
  • Multiplication is multiplication. – Berci May 15 '13 at 22:37
  • Well, the directions are not that trivial. When I first wrote up, it wanted to be $1+AB$, but since the inverse changes the direction, actually $A^{-1}$ was pulled out from the right, so it became $1+BA$. – Berci May 15 '13 at 22:39
  • Right. That's why I asked. I can imagine that as long as an identity is obtained only by left or right multiplications, then I can apply it as a matrix. – r_31415 May 15 '13 at 22:52
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What happens when you just multiply the RHS by $C_N^{-1} + W_N$?

Zach L.
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Another simpler trick : $$(C_{N}^{-1}+W_{N}^{})^{-1}_{}=[(I+W_{N}^{}C_{N}^{})C_{N}^{-1}]^{-1}_{}=(C_{N}^{-1})^{-1}_{}(I+W_{N}^{}C_{N}^{})^{-1}_{}=C_{N}^{}(I+W_{N}^{}C_{N}^{})^{-1}_{}$$

Sunyam
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