Linear Algebra Proof: Invertible Idempotent Matrix Must be Identity Matrix

AI Thread Summary
An invertible idempotent matrix A must be the identity matrix I_n, as shown through a contradiction. Assuming A is not equal to I_n leads to the conclusion that A must equal I, which contradicts the initial assumption. The proof relies on the properties of idempotent matrices, where A^2 = A, and the fact that A is invertible. The discussion also touches on a separate proof regarding the product of two idempotent matrices, A and B, being idempotent if they commute, but this claim is challenged. Ultimately, the consensus is that the initial proof about A being the identity matrix is correct.
Dosmascerveza
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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...
 
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Dosmascerveza said:
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.
 
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?
 
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,
 
Dosmascerveza said:
AB==BA ==> A^(-1), B^(-1) exist
This isn't true.
 

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