Matrix identity proof

cogsci33 · **Joined:** Sat, 23 Oct 2010 19:53:28 UTC **Posts:** 5

Hi all,
Searle (1982) poses the following problem:
Assuming the stated Inverses exist, prove the following Identity:

$\mathbf{A} - \mathbf{A} \left( \mathbf{A+B} \right)^{-1}\mathbf{A}= \mathbf{B} - \mathbf{B} \left(\mathbf{A+B}^\right)^{-1}\mathbf{B}$

Does anybody know the solution or could give me a hint?

Many thanks, Alex

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Searle (1982), Matrix Algebra useful for Statistics

outermeasure · **Posted:** Sat, 23 Oct 2010 20:21:05 UTC

cogsci33 wrote:

Hi all,
Searle (1982) poses the following problem:
Assuming the stated Inverses exist, prove the following Identity:

$\mathbf{A} - \mathbf{A} \left( \mathbf{A+B} \right)^{-1}\mathbf{A}= \mathbf{B} - \mathbf{B} \left(\mathbf{A+B}^\right)^{-1}\mathbf{B}$

Does anybody know the solution or could give me a hint?

Many thanks, Alex

----------
Searle (1982), Matrix Algebra useful for Statistics

One-liner:
$\begin{aligned} &B(A+B)^{-1}B-A(A+B)^{-1}A\\ &=B(A+B)^{-1}B+A(A+B)^{-1}B-A(A+B)^{-1}B-A(A+B)^{-1}A \end{aligned}$
Group, and rearrange.

cogsci33 · **Joined:** Sat, 23 Oct 2010 19:53:28 UTC **Posts:** 5

Thank you for the quick reply. Sorry but I fail to appreciate your logic. I would appreciate if you could explain a bit more. It worked btw. Thank you a lot!
Thank you

Shadow · **Posted:** Sat, 23 Oct 2010 20:40:52 UTC

cogsci33 wrote:

Thank you for the quick reply. Sorry but I fail to appreciate your logic. I would appreciate if you could explain a bit more.
Thank you

There's no logic involved, he just wrote down a formula. Work out the formula.

Sam Snyder · **Posted:** Sun, 24 Oct 2010 14:45:29 UTC

outermeasure wrote:

cogsci33 wrote:

Hi all,
Searle (1982) poses the following problem:
Assuming the stated Inverses exist, prove the following Identity:

$\mathbf{A} - \mathbf{A} \left( \mathbf{A+B} \right)^{-1}\mathbf{A}= \mathbf{B} - \mathbf{B} \left(\mathbf{A+B}^\right)^{-1}\mathbf{B}$

Does anybody know the solution or could give me a hint?

Many thanks, Alex

----------
Searle (1982), Matrix Algebra useful for Statistics

One-liner:
$\begin{aligned} &B(A+B)^{-1}B-A(A+B)^{-1}A\\ &=B(A+B)^{-1}B+A(A+B)^{-1}B-A(A+B)^{-1}B-A(A+B)^{-1}A \end{aligned}$
Group, and rearrange.

Hi Alex (cogsci33).

You asked for a hint (or a solution) and outermeasure gave you a hint that is so good that it is essentially a solution. Don't dismiss it lightly.

Ask yourself, "How is
$\begin{aligned} &B(A+B)^{-1}B-A(A+B)^{-1}A\end{aligned}$
related to the identity I'm trying to prove?" ... repeat until you have answered the question.

Now, examine
$\begin{aligned} &B(A+B)^{-1}B+A(A+B)^{-1}B-A(A+B)^{-1}B-A(A+B)^{-1}A \end{aligned}$
and notice that it's equivalent to the previous expression.

Factor $$(A+B)^{-1}B$ from $$B(A+B)^{-1}B+A(A+B)^{-1}B$ and simplify. Do similarly for the next two terms.

Now, can you see (after a little bit of rearranging) that outermeasure has given you exactly what you need to prove your identity?

alfiery · **Joined:** Tue, 26 Oct 2010 10:30:33 UTC **Posts:** 1

Good information..thank you.

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