Question 87883
Ok, I'll go over it in more detail.<br />

For part a we are told that A is idempotent, this means that A=A*A. We are told to show that I-A is idempotent. To do this we need to calculate (I-A)*(I-A) and show it is equal to (I-A). Well<br />

(I-A)*(I-A) = I*I - I*A -A*I + A*A<br />

But we know that the identity times anything is itsself, so I*A = A*I = A. Also I*I=I. So<br />

(I-A)*(I-A) = I - A - A + A*A<br />

If we now use the fact that A is idempotent (which means A*A=A) we get<br />

(I-A)*(I-A) = I - A - A + A = I-A<br />

Therefore I-A is idempotent.<br />

Part b says that if A is idempotent show that (2A-I) is self inverse(it is its own inverse), this means show that (2A-I)*(2A-I)=I. This is because anything times its inverse is the identity. See if you can multiply out the brackets, and use the fact that A*A=A to show this is true.<br />

Kev
<hr />

Hi,<br />

If you look at your definition of idempotent A^2=A, then you can actually solve this for A and find *all* idempotent square matrices. You should be able to find 2 of them.<br />

Since there are only 2 idempotent square matrices, you can just try them both for parts a and b.<br />

The alternative is to substitute I-A into the definition for idempotent for part a, and calculate (2A-I)(2A-I) using the definition of idempotent for part b.<br />

I can't really see where the difficult part is, so write back if you want more detailed specifc help.<br />

Kev