# In order for the square of a matrix to be equal to the matrix

1. Jul 16, 2009

### megalomaniac

Idempotent Matrix Proof

1. The problem statement, all variables and given/known data

Given a matrix A where A2 = A, find the properties of A.

2. Relevant equations

detA = ai1ci1 + ai2ci2 + ... + aincin (where cij = (-1)i+j*detAij)

aij = ai1a1j + ai2a2j + ... + ainanj

3. The attempt at a solution

In order for A2 to be defined, A must be a square matrix.

I have concluded that A must either equal the identity matrix I, or A must be singular.

I am having trouble proving this in the general case.

Last edited: Jul 17, 2009
2. Jul 16, 2009

### D H

Staff Emeritus
Hint: If A is not singular, it has an inverse. Use this fact.

3. Jul 16, 2009

### megalomaniac

Well I found a counter-example that disproves that A is singular. The matrix {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}} is a singular matrix that does not satisfy A2 = A.

So A must equal I. So AA-1 = I. But I still need a good way to prove this, and proofs are my deficiency.

4. Jul 16, 2009

### D H

Staff Emeritus
No, all you have done is to prove that singularity does not guarantee that A2=A.

Hint: The zero matrix is singular.

5. Jul 17, 2009

### megalomaniac

Ok, now I'm just having problems determining the general properties of singular matrices that would make A2 = A

6. Jul 17, 2009

### megalomaniac

Thus far I have shown,

Suppose A is an idempotent nxn matrix.

In the case of a non-singular matrix, in order to be invertible, there exists an nxn matrix B such that AB = BA = In, where B = A-1

Since A = B and B = A-1, then A = A-1

AA = In

Therefore In is idempotent.

Would this show that if A = I, then A is idempotent?

Also, now how would I go about showing the only other way for A to be idempotent would be if A is singular, since not all singular matrices are idempotent?

--Thank you.

Last edited: Jul 17, 2009
7. Jul 17, 2009

### D H

Staff Emeritus
That was a big leap. You need to show this.

Isn't it kind of obvious that the identity matrix is idempotent?

Trivially, yes. What you have not done is to show that the nxn identity matrix is the only nonsingular nxn idempotent matrix.

Yet another hint: An interesting subset of the idempotent matrices are the diagonal idempotent matrices. What can you say about the diagonal elements?

8. Jul 17, 2009

### megalomaniac

Isn't the only diagonal idempotent matrix the identity matrix?

9. Jul 17, 2009

### D H

Staff Emeritus
Yes, but you have not yet shown this (yet).

There are also singular diagonal idempotent matrices. You might want to think about these for a bit, and whether they have any similarity to non-diagonal idempotent matrices.

10. Jul 17, 2009

### megalomaniac

If A is idempotent and non-singular:

If I premultiply both sides by A-1, then I get:

A = InA = A-1AA = A-1A = In.

Would this be valid?

As for the diagonal idempotent matrices, they have full rank and therefore are non-singular and symmetric and non-diagonal idempotent matrices would be singular.

EDIT: And each entry of a diagonal idempotent matrix would have to be 0 or 1.

Last edited: Jul 17, 2009
11. Jul 17, 2009

### D H

Staff Emeritus
Finally.

$$\bmatrix 1 & 0 \\ 0 & 0\endbmatrix$$

This matrix is diagonal, is idempotent, and is not full rank.

Better.

12. Jul 18, 2009

### megalomaniac

So for all diagonal idempotent matrices, A2 = A iff ai2 = ai for all i=1,...,n. This can only occur for i values of 0 and 1. Therefore, a diagonal matrix is idempotent iff each diagonal entry is 0 or 1.

So any idempotent diagonal matrix will be singular (save for I), but this still doesn't encompass all singular idempotent matrices.

I appreciate your help!

13. Jul 18, 2009

### D H

Staff Emeritus
You're welcome.

Regarding the non-diagonal idempotent matrices. I dropped a BIG clue in post #9. You haven't picked up on it yet. Reread the post. Ponder over each word.

14. Jul 18, 2009

### megalomaniac

So far it seems that the sum of the diagonal entries of an idempotent singular matrix must equal one, and the rows and/or columns must be linearly dependent.

15. Jul 18, 2009

### megalomaniac

Ok, I have determined that detA equals the product of the eigenvalues. So in order for A to be singular, at least one eigenvalue must be zero. Where would I go from here to prove a singular idempotent matrix must have at least one zero eigenvalue?

-Thanks!

16. Jul 18, 2009

### D H

Staff Emeritus
You don't need to do that. Any singular matrix has at least one zero eigenvalue.

The key word in post #9 was similarity. Forget eigenvalues. Think similarity.

17. Jul 19, 2009

### megalomaniac

So am I to show that a singular diagonal idempotent matrix is similar to a singular non-diagonal idempotent matrix? I'm sorry, I'm still unsure on how to go about this, my mind's been all over the place.

18. Jul 19, 2009

### D H

Staff Emeritus
Try the other way around first. You already know that a diagonal matrix whose diagonal elements are either zero or one is idempotent. Show that an arbitrary similarity transform yields another idempotent matrix.

19. Jul 19, 2009

### megalomaniac

I haven't spent too much time dealing with similar matrices, but I'm assuming I am to use the fact that if B=P-1AP for some invertible matrix P, then A is similar to B. And A would be an idempotent diagonal matrix in this case?

20. Jul 19, 2009

### D H

Staff Emeritus
Yes. So what is B2 if B=P-1AP and A2=A?