Show that Q is an idempotent matrix

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Homework Help Overview

The discussion revolves around the properties of idempotent matrices, specifically focusing on proving that a matrix Q, defined in terms of another idempotent matrix P and an arbitrary square matrix A, is also idempotent. The participants are exploring the implications of the definition of idempotency and the algebraic manipulation of matrices.

Discussion Character

  • Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to manipulate the expression for Q and explore its square to verify idempotency. There are discussions about factoring and rewriting Q, as well as concerns about the correctness of expressions and the implications of matrix multiplication properties.

Discussion Status

There is an ongoing exploration of the algebraic steps needed to demonstrate that Q is idempotent. Some participants have provided alternative forms and suggestions for simplification, while others express uncertainty about the approach and seek clarification on specific steps.

Contextual Notes

There is a noted concern about a potential omission of a minus sign in the definition of Q, which may affect the subsequent calculations. Additionally, participants are grappling with the non-commutative nature of matrix multiplication as they work through their reasoning.

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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) ..?
 
Last edited:
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Incognitopad said:

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.
Incognitopad said:

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?
Incognitopad said:
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) ..?
 
Last edited:
Incognitopad said:
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.
 
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.
 
Incognitopad said:
= (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.:approve:
 

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