# Homework Help: How to find P1 and P2 such that P1, P2, and P1+P2 idempotent

1. Oct 5, 2009

### julie94

Hi everyone,

I am working on the following question:
I would need to find a non trivial example of two 3by3 matrices P1 and P2 such that
P1 is idempotent
P2 is idempotent
P1+P2 is idempotent

How I should go about attempting to find a solution?

First I thought I should solve three equations
P1^n= P1
P2^n= P2
(P1+P2)^n= P1+P2

But I quickly convince myself that this would not work, and that I should try a trial and error avenue. I would appreciate your ideas.

2. Oct 5, 2009

### aPhilosopher

P1n= P1, P2n= P2 and (P1+P2)n= P1+P2 can be replaced by P12= P1, P22= P2 and (P1+P2)2= P1+P2 because matrix multiplication is associative so you can take the power one multiplication at a time.

The second set equations are just the definition of a projection matrix applied to P1, P2 and P1 + P2. Do you know what that is?

3. Oct 5, 2009

### Dick

I really don't think that's right. If P is the 2x2 matrix [[0,1],[1,0]] then P^3=P but P^2 is not equal to P, it's I. On the other hand, I don't think the question really requires you solve for anything. You only want examples of P1 and P2 that satisfy those properties. Just try and fish around for examples, as the OP said, trial and error. Are nonzero diagonal matrices non-trivial enough?

4. Oct 6, 2009

### aPhilosopher

But that matrix isn't idempotent. Doesn't an idempotent matrix have to satisfy Pn = P for all n?

Last edited: Oct 6, 2009
5. Oct 6, 2009

### Dick

Yeah, you're right. I was just thinking about the implication 'P^n=P implies P^2=P' which is false. But, yes, idempotent means P^2=P. Which does imply P^n=P. Sorry.

6. Oct 7, 2009

### julie94

>>The second set equations are just the definition of a projection matrix applied to P1, P2 >>and P1 + P2. Do you know what that is?

No. P1, P2 and (P1+P2) are projection matrices?

7. Oct 7, 2009

### julie94

Nonzero diagonal matrices are non-trivial. Do you have one?

8. Oct 7, 2009

### aPhilosopher

cook up a couple of matrices of the that have 1's on some of the diagonals and 0's elsewhere. Play around with those. You should be able to get what you want. Those are projection operators by the by. They take a vector and return the vector that is zero except on the axes that the operator projects onto where it just returns the coordinates. The key to a projection operator is that it is the identity on its image (this allows idempotence).

9. Oct 7, 2009

### julie94

Here is the example I found
A1=
1,0,0;
0,0,0;
1,0,0;

A2=
0,0,0;
0,1,0;
0,0,0;

A1+A2=
1,0,0;
0,1,0;
1,0,0;

All three are idempotent. Is there a way to make those matrices a little less trivial.

10. Oct 7, 2009

### Dick

Idempotent matrices, by their nature, tend to look somewhat trivial. That's already more non-trivial than I was thinking of. I was thinking of just using the diagonal parts. The zero matrix I would think of as trivial.

11. Oct 7, 2009

### aPhilosopher

Change basis? The grader might hate you though.

12. Oct 7, 2009