Linear Algebra Proof Help

**Danneedshelp** · June 25th 2009, 12:52 PM

Hello, I need help with a proof again.

Q: prove that if $A^{2}=A$ , then $I-2A=(I-2A)^{-1}$ .

So, from that I gather...

Givens: $A^{2}=A \rightarrow A$ is square.

Goal: $I-2A=(I-2A)^{-1}$

So, I don't have very much ha.

Should I just approach this multiplying the the inverse quantity to both sides of the quantity $I-2A$ and hope to get the Identity $I$ back?

Not sure how to use the first fact $A^{2}=A$ .

Thanks,

I'll keep plugging away at it.

**Plato** · June 25th 2009, 01:06 PM

$\left( {I - 2A} \right)\left( {I - 2A} \right) = I^2 - 2IA - 2AI + 4A^2 = I - 4A + 4A = I$

**Danneedshelp** · June 25th 2009, 01:22 PM

Originally Posted by Plato

$\left( {I - 2A} \right)\left( {I - 2A} \right) = I^2 - 2IA - 2AI + 4A^2 = I - 4A + 4A = I$

Is that all that has to be shown?

**Plato** · June 25th 2009, 01:34 PM

Originally Posted by Danneedshelp

Is that all that has to be shown?

If each of $A~\&~B$ is a matrix, how does one show that $A=B^{-1}?$

**Danneedshelp** · June 25th 2009, 01:43 PM

oh, or is it
$AB=I=BA$

**Plato** · June 25th 2009, 01:46 PM

Originally Posted by Danneedshelp

$AB^{-1}=I=B^{-1}A$
correct?

No it is not correct? Do you have a textbook?

$AB=BA=I$

**Danneedshelp** · June 25th 2009, 01:52 PM

Originally Posted by Plato

No it is not correct? Do you have a textbook?

$AB=BA=I$

Yeah, I see that now. I just forgot to delete first part of my last post when I made the edit. I put $AB=BA=I$ [/quote] at the bottom. My bad.

Thanks for the help.

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