Prove A Invertible, Idempotent Matrix A = In Sub n

[ephemeral1]

Homework Statement

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

Homework Equations

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The Attempt at a Solution

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

 

[Mark44]

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

 

[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.

 

[Mark44]

OK, then what does A² - A equal?

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

 

[ephemeral1]

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

 

[Mark44]

I meant to write AB=BA=I sub n.

Yeah, that's more like it.

A^2 - A= 0

OK, now what can you do with that?

 

[ephemeral1]

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

 

[Mark44]

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

What do you mean by that?

 

[ephemeral1]

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

 

[Mark44]

No, A² - AA^-1 = A - I
But so what?

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

You said

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

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

 

[ephemeral1]

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

 

[Mark44]

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?