# Linear Algebra Proof Help

• Jun 25th 2009, 12:52 PM
Danneedshelp
Linear Algebra Proof Help
Hello, I need help with a proof again.

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

So, from that I gather...

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

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

So, I don't have very much ha.

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

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

Thanks,

I'll keep plugging away at it.
• Jun 25th 2009, 01:06 PM
Plato
$\displaystyle \left( {I - 2A} \right)\left( {I - 2A} \right) = I^2 - 2IA - 2AI + 4A^2 = I - 4A + 4A = I$
• Jun 25th 2009, 01:22 PM
Danneedshelp
Quote:

Originally Posted by Plato
$\displaystyle \left( {I - 2A} \right)\left( {I - 2A} \right) = I^2 - 2IA - 2AI + 4A^2 = I - 4A + 4A = I$

Is that all that has to be shown?
• Jun 25th 2009, 01:34 PM
Plato
Quote:

Originally Posted by Danneedshelp
Is that all that has to be shown?

If each of $\displaystyle A~\&~B$ is a matrix, how does one show that $\displaystyle A=B^{-1}?$
• Jun 25th 2009, 01:43 PM
Danneedshelp
oh, or is it
$\displaystyle AB=I=BA$
• Jun 25th 2009, 01:46 PM
Plato
Quote:

Originally Posted by Danneedshelp
$\displaystyle AB^{-1}=I=B^{-1}A$
correct?

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

$\displaystyle AB=BA=I$
• Jun 25th 2009, 01:52 PM
Danneedshelp
Quote:

Originally Posted by Plato
No it is not correct? Do you have a textbook?

$\displaystyle AB=BA=I$

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

Thanks for the help.