Linear Algebra help

1. Feb 16, 2010

ephemeral1

1. The problem statement, all variables and given/known data

Prove that if A is an n x n matrix that is idempotent and invertible, then A=I sub n

2. Relevant equations

none

3. The attempt at a solution

I don't know how to prove this. Can anyone help me with this? Thank you

2. Feb 16, 2010

Staff: Mentor

What does it mean that A is idempotent? Invertible? Start with those.

3. Feb 16, 2010

ephemeral1

A is idempotent if A=A^2 and invertible if there exists an n x n matrix B such that AB=AB=I sub n.

4. Feb 16, 2010

Staff: Mentor

OK, then what does A2 - A equal?

Also, AB is always equal to itself, so that doesn't do you any good.

5. Feb 16, 2010

ephemeral1

I meant to write AB=BA=I sub n.
A^2 - A= 0

6. Feb 16, 2010

Staff: Mentor

Yeah, that's more like it.
OK, now what can you do with that?

7. Feb 16, 2010

ephemeral1

(a^2)- (a^-1) i=0

8. Feb 16, 2010

Staff: Mentor

What do you mean by that?

9. Feb 16, 2010

ephemeral1

Since B is the inverse of A, I can write AB=I as I=AA^-1. So A^2 - AA^-1 = I

10. Feb 16, 2010

Staff: Mentor

No, A2 - AA-1 = A - I
But so what?

What I asked was whether you could do anything with A2 - A = 0?

You said
I still don't know how you got that or how it relates to A2 - A = 0.

11. Feb 16, 2010

ephemeral1

I could factor out A^2 - A=0, which will become A(A-1)=0

12. Feb 16, 2010

Staff: Mentor

OK, now you're on the right track, with a correction.

A2 - A = 0
<==> A(A - I) = 0

Everything in the equation above is a matrix. You can't subtract a scalar (1) from a matrix (A), so that's the reason I wrote it this way. I has the same role in matrix multiplication that 1 has in scalar multiplication.

There are two obvious things you can say about the equation above, and one not-so-obvious thing. For starters, what are the obvious things you can say?