SOLUTION: A square matrix A is idempotent if A^2 = A. a) Show that if A is idempotent, then so is I - A. b) Show that if A is idempotent, then 2A

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Question 87883: A square matrix A is idempotent if A^2 = A.

a) Show that if A is idempotent, then so is I - A.

b) Show that if A is idempotent, then 2A - I is invertible and is its own inverse.
Answer by kev82(151) (Show Source):
You can put this solution on YOUR website!
Ok, I'll go over it in more detail.

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

(I-A)*(I-A) = I*I - I*A -A*I + A*A

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

(I-A)*(I-A) = I - A - A + A*A

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

(I-A)*(I-A) = I - A - A + A = I-A

Therefore I-A is idempotent.

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.

Kev

Hi,

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.

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

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.

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

Kev

SOLUTION: A square matrix A is idempotent if A^2 = A. a) Show that if A is idempotent, then so is I - A. b) Show that if A is idempotent, then 2A - I is invertible and is its own