Show that Q is an idempotent matrix

[Incognitopad]

Homework Statement

A square matrix P is called an idempotent if P^2 = P,

Show that if P is an idempotent, so is Q = (P + AP - PAP) for any square matrix A (of the
same size as P).

Homework Equations

The Attempt at a Solution

basically I factor,

Q = (I + (I - P) A) P

then I square it and try to get back to the original,... and end up with

Q^2 = (I + (I - P - P + P^2) A^2) P^2
= (I + (I -2P + P) A^2) P
= (I + (I - P) A^2) P
= (I + AA - PAA) P
= P + AAP - PAAP

how do i get rid of the AA (A^2) ..?

 

[Mark44]

Homework Statement

A square matrix P is called an idempotent if P^2 = P,

Show that if P is an idempotent, so is Q = (P + AP PAP) for any square matrix A (of the
same size as P).

From your work below, you have apparently omitted a minus sign. The above should be:
Q = (P + AP -[/color] PAP)

Please confirm.

Homework Equations

The Attempt at a Solution

basically I factor,

Q = (I + (I - P) A) P

I'm not sure that it's helpful to write Q in this form. Why not just multiply P + AP - PAP by itself?

then I square it and try to get back to the original,... and end up with

Q^2 = (I + (I - P - P + P^2) A^2) P^2
= (I + (I -2P + P) A^2) P
= (I + (I - P) A^2) P
= (I + AA - PAA) P
= P + AAP - PAAP

how do i get rid of the AA (A^2) ..?

 

[haruspex]

Q = (I + (I - P) A) P
Q^2 = (I + (I - P - P + P^2) A^2) P^2

Try that again, this time bearing in mind that matrix multiplication is not, in general, commutative.

 

[Incognitopad]

Uhhh... ah..

Q^2 = ((I + (I - P) A) P) * ((I + (I - P) A) P)
= Ok, I'm just going to leave it expanded. Too complicated this way

= (P + AP - PAP)^2
= (PP + PAP - PPAP + APP + APAP - APPAP - PAPP - PAPAP + PAPPAP)
= (P + PAP - PAP + AP + APAP - APAP - PAP - PAPAP + PAPAP)
= (P + AP - PAP)
wow, that was easy. LOL.

 

[Mark44]

= (P + AP - PAP)^2
= (PP + PAP - PPAP + APP + APAP - APPAP - PAPP - PAPAP + PAPPAP)
= (P + PAP - PAP + AP + APAP - APAP - PAP - PAPAP + PAPAP)
= (P + AP - PAP)
wow, that was easy. LOL.

I rest my case.