Can a Matrix Satisfy \(A^2 = A\) and be Non-Singular?

[skoker]

i have a simple proof is this correct?

prove that if \(A^2=A\), then either A=I or A is singular.

let A be a non singular matrix. then \(A^2=A, \quad A^{-1}A^2=A^{-1}A, \quad IA=I, \quad A=I\) therefore \(A^2=A.\)

 

[Fernando Revilla]

let A be a non singular matrix. then \(A^2=A, \quad A^{-1}A^2=A^{-1}A, \quad IA=I, \quad A=I\)

Right

therefore \(A^2=A.\)

Why do you write this? $A^2=A$ just by hypothesis.

 

[skoker]

Right
Why do you write this? $A^2=A$ just by hypothesis.

i suppose that is redundant or unnecessary. i was not sure if it needs a conclusion with the 'therefore'.

 

[CaptainBlack]

i suppose that is redundant or unnecessary. i was not sure if it needs a conclusion with the 'therefore'.

The therefore should go before the \(A=I\) and you should stop at that point.

CB