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

[megalomaniac]

Idempotent Matrix Proof

Homework Statement

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

Homework Equations

detA = a_i1c_i1 + a_i2c_i2 + ... + a_inc_in (where c_ij = (-1)^i+j*detA_ij)


a_ij = a_i1a_1j + a_i2a_2j + ... + a_ina_nj

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.

 

[D H]

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

 

[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.

 

[D H]

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

Hint: The zero matrix is singular.

 

[megalomaniac]

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

 

[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 = I_n, where B = A^-1

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

AA = I_n

Therefore I_n 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.

 

[D H]

Since A = B

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

Therefore I_n is idempotent.

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

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

 

[megalomaniac]

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

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

 

[D H]

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.

 

[megalomaniac]

If A is idempotent and non-singular:

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

A = I_nA = A^-1AA = A^-1A = I_n.

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.

 

[D H]

If A is idempotent and non-singular:

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

A = I_nA = A^-1AA = A^-1A = I_n.

Finally.

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.

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

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

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

Better.

 

[megalomaniac]

So for all diagonal idempotent matrices, A² = A iff a_i² = a_i 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!

 

[D H]

So for all diagonal idempotent matrices, A² = A iff a_i² = a_i 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!

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.

 

[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.

 

[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!

 

[D H]

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.

 

[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.

 

[D H]

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.

 

[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?

 

[D H]

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

 

[megalomaniac]

So B²=(P^-1)²A²P² = AI = A?

 

[D H]

No. Matrix math doesn't work that way. Try again.

 

[megalomaniac]

Ha oops,

How about: P^-1APP^-1AP = P^-1AIAP = P^-1AAP = P^-1AP. So B² = B ?

 

[D H]

There you go.

 

[megalomaniac]

But how does that prove that A must be singular?

 

[D H]

It doesn't. It shows that a matrix that is similar to an idempotent matrix is idempotent.

 

[megalomaniac]

But I'm trying to prove that if an idempotent matrix is not the identity matrix, then it must be singular. How do I make this leap?

-Thank you!

 

[HallsofIvy]

But how does that prove that A must be singular?

Obvious point: the identity matrix satisfies I²= I but is not singular.

A²= A is equivalent to A²- A= A(A-I)= 0. It does NOT follow that "A= 0 or A= I" but it does follow that either A or A- I has determinant 0.

 

[megalomaniac]

A²= A is equivalent to A²- A= A(A-I)= 0. It does NOT follow that "A= 0 or A= I" but it does follow that either A or A- I has determinant 0.

Oh wow! I never looked at it that way! How does it follow that A has determinant 0?

-Thanks!

 

[megalomaniac]

I was thinking, since I've proven the only non-singular idempotent matrix is the identity matrix, wouldn't it follow that the only other idempotent matrices must be singular?

 

[D H]

Of course.

 

[megalomaniac]

Wow I'm an imbecile. I skipped over that fact and just focused on trying to prove which singular matrices would fit the bill, but this was not required.

Thank you so much for your guidance! I learned more than what the proof expected of me.