Show that if a matrix P is idempotent meaning P=P^2 then..

[Randyyy]

hey, the task is as follows:

Show that if the matrix P is idempotent, meaning that [MATH]P=P^2[/MATH] then [MATH]E-P[/MATH] and [MATH]P+AP-PAP[/MATH] is also idempotent. Note that E is the identity matrix.

I start of by squaring E-P and it simplifies nicely, [MATH](E-P)^2=P^2-2EP+E^2=P-2P+E=E-P[/MATH], my problem is with the second one. I know that [MATH]P^n=P[/MATH], [MATH](P+AP-PAP)[/MATH] factors to [MATH]P(E+A-AP)[/MATH] and now I square it: [MATH]P^2((E+A-AP)^2)=P((E+A-AP)^2)[/MATH], expanding this monster I get: [MATH]P(A^2+2AE+E^2-2A^2P-2AEP+A^2P^2)[/MATH] and simplifying I get [MATH]P(A^2+2A+E-2A^2P-2AP+A^2P)[/MATH] and here I don't see any more simplification so I multiply P into the parenthesis: [MATH]PA^2+2PA+PE+2PA^2P-2PAP+PA^2P[/MATH] and here I get stuck, I am guessing that I have to be careful about the order in which I put my matrices and so I am likely doing things I am not actually supposed to be able to do.

 

[Steven G]

P(E+A−AP) and now I square it: P²((E+A−AP)²). I suppose that you are saying that what is in bold are equal? I don't think so! Try again.

 

[Randyyy]

I feel like you have an aura of success today Jomo, second time you comment and I somehow manage to solve my task instantly.
okay, so, instead of trying to factor anything I went ahead and did this: [MATH](P+AP-PAP)(P+AP-PAP)[/MATH] I also realized that I have to pick a way to multiply, for example, if I had (P+AP-PAP)(P+AP-PAP) and I multiply from the left, I get [MATH]P^2+PAP...[/MATH] I can go from the other side and instead do [MATH]P^2+AP^2...[/MATH] but whichever side I decide to put the terms I have to be consistent, to make sure this was not just a blind theory I had I tried doing it placing my terms to the left of what I multiplied and to the right and they both end up yielding the exact same answer. The lesson here is to be careful when 1: factoring, make sure it is correct to begin with. 2: be consistent and remember that multiplying matrices is not the same as multiplying regular variables!

I can post the solution if it is of any interest.

Is it not true that if I square [MATH]P(E+A-AP)[/MATH] that it becomes [MATH]P^2((E+A-AP)^2)[/MATH]? I was thinking that since it is of the form [MATH]x\cdot y[/MATH] i square both quantities, but maybe I am still thinking too much algebra and not enough matrices..

 

[Dr.Peterson]

I also realized that I have to pick a way to multiply, for example, if I had (P+AP-PAP)(P+AP-PAP) and I multiply from the left, I get [MATH]P^2+PAP...[/MATH] I can go from the other side and instead do [MATH]P^2+AP^2...[/MATH] but whichever side I decide to put the terms I have to be consistent, to make sure this was not just a blind theory I had I tried doing it placing my terms to the left of what I multiplied and to the right and they both end up yielding the exact same answer. The lesson here is to be careful when 1: factoring, make sure it is correct to begin with. 2: be consistent and remember that multiplying matrices is not the same as multiplying regular variables!

Is it not true that if I square [MATH]P(E+A-AP)[/MATH] that it becomes [MATH]P^2((E+A-AP)^2)[/MATH]? I was thinking that since it is of the form [MATH]x\cdot y[/MATH] i square both quantities, but maybe I am still thinking too much algebra and not enough matrices..

It all comes down to what you say several times here without quite saying it: Matrix multiplication is not commutative, so you can NEVER change order without a reason. So just do that, rather than talking about it.

If you have (AB)^2, don't assume it's A^2B^2; write it as it is, ABAB, and then see if there are any rules that let you change it to something else.

(P+AP-PAP)(P+AP-PAP) = PP + PAP - PPAP + APP + APAP + APPAP - PAPP - PAPAP + PAPPAP, right? Now check what happens to each term. Don't ever let yourself act as if these are just numbers, and only think about order after the fact. Never let your guard down.

 

[Steven G]

I feel like you have an aura of success today Jomo, second time you comment and I somehow manage to solve my task instantly.

That is the sign of a good tutor but it only works with face to face tutoring.

 

[Steven G]

[P(E+A−AP)]² = P(E+A−AP)*P(E+A−AP) which is not what you wrote,