Linear algebra:Prove that the only invertible nxn idempotent matrix is I

[RossH]

Homework Statement

Prove that the only invertible nxn idempotent matrix is the identity matrix.

Homework Equations

idempotent: given an nxn matrix, A^2=A
invertible: There exists a matrix B such that AB=BA=I
I, or identity matrix: for all matrices A, AI=IA=A

The Attempt at a Solution

So far I haven't gotten very far:
Suppose A is an nxn matrix.
For A to be invertible there exists a matrix B such that AB=BA=I.
For A to be idempotent A^2=A.
Otherwise, I've kind of mulled it over and gotten this far:
To be
invertible it has to have a determinant != 0. So, I think that I have
to show that the determinant of everything else that is indempotent is
0, which would mean widely defining the criteria for a matrix to be
indempotent and showing that the two are mutually exclusive. The
problem is I'm not quite sure about the properties of indempotent
matrices and, since we don't have a formula for the determinant of a
given nxn matrix, I'm not quite sure what to do.
So far, I have not covered and am not allowed to use rank or eigenvalues in my class.
Thanks!

 

[vela]

You're overthinking the problem. You have one equation, $A^2=A$, and you know $A^{-1}$ exists. Think of how to solve for A.

 

[RossH]

You're overthinking the problem. You have one equation, $A^2=A$, and you know $A^{-1}$ exists. Think of how to solve for A.

I know that A^2=A
If I multiply both sides by A^(-1) I get: A=I
Is that it? Does that show that it is universally true or that that is just one possibility?

 

[vela]

Yeah, that's it. It holds for all A that are idempotent and invertible. You're not doing anything weird or wrong that opens up the possibility of additional solutions.

You're probably worried about something along the lines of solving x^2=x, right? If you divide both sides by x, you get x=1, and you lose the solution x=0. The reason you lose x=0 is because it doesn't have a multiplicative inverse, so you can't justify the step of dividing by x when x=0. If you were, however, told that x has a multiplicative inverse, you'd know that x=1 is the only allowed solution. Well, that's essentially what you're being told here. The matrix A is invertible, so its multiplicative inverse $A^{-1}$ exists and you're justified in "dividing by A" to solve the equation.

 

[RossH]

Yeah, that's it. It holds for all A that are idempotent and invertible. You're not doing anything weird or wrong that opens up the possibility of additional solutions.

You're probably worried about something along the lines of solving x^2=x, right? If you divide both sides by x, you get x=1, and you lose the solution x=0. The reason you lose x=0 is because it doesn't have a multiplicative inverse, so you can't justify the step of dividing by x when x=0. If you were, however, told that x has a multiplicative inverse, you'd know that x=1 is the only allowed solution. Well, that's essentially what you're being told here. The matrix A is invertible, so its multiplicative inverse $A^{-1}$ exists and you're justified in "dividing by A" to solve the equation.

Oh, good point. Thank you very much for the help.