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

 

[blue_raver22]

if A is singular, then it cannot have an inverse. Therefore, \(A^2=A\) does not imply \(A=I.\) This shows that if \(A^2=A,\) then either A=I or A is singular.

Yes, your proof is correct. However, it would be helpful to explain why A being singular means it cannot have an inverse. This is because a singular matrix has determinant 0, and a matrix can only have an inverse if its determinant is non-zero. Therefore, A being singular means it cannot have an inverse, and thus cannot satisfy \(A^{-1}A=I.\)