Linear Algebra Proof: Invertible Idempotent Matrix Must be Identity Matrix

[Dosmascerveza]

Homework Statement

If A is an invertible idempotent matrix, then A must be the Identity matrix I_n.

Homework Equations

A^2==A ; A^2==AA; A^(-1); I==A^(-1)

The Attempt at a Solution

Suppose A is an nxn matrix =/= I_n.

s.t. A^(2)==A

so A^(2)==A ==> AA==A

==> A^(-1)AA==A^(-1)A ==> A==I==> A^(-1)A==A^(-1)I==>I==A^(-1)I==A^(-1)==A

which yeilds a contradiction because we supposed our A =/= I_n.

Therefore A==I_nIs this correct please help me understand where I have failed...

 

[vela]

Suppose A is an nxn matrix =/= I_n.

s.t. A^(2)==A

so A^(2)==A ==> AA==A
==> A^(-1)AA==A^(-1)A
==> A==I

You should have stopped right here. You should have also stated that A is invertible.

==> A^(-1)A==A^(-1)I
==> I==A^(-1)I==A^(-1)==A

This is wrong. You don't know that the inverse of A is equal to A.

 

[Dosmascerveza]

Okay so if i stated A an invertible nxn matrix =/= I_n

s.t A^(2)==A(idempotent)... truncating the last bit of foolishness. I was correct?

 

[Dosmascerveza]

another proof...
problem statement.
prove if A and B are idempotent and AB==BA then AB is idempotent.

AB==BA ==> A^(-1), B^(-1) exist

Since A and B are idempotent invertible matrices, from previously proven theorem, we know A=I and B=I. and since II==I ==> AB==I Therefore AB==BA and AB is Idempotent,

 

[vela]

AB==BA ==> A^(-1), B^(-1) exist

This isn't true.