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

 
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Jul16-09, 10:23 PM   #1
 

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


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.
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Jul16-09, 10:29 PM   #2
D H
 
Mentor
Hint: If A is not singular, it has an inverse. Use this fact.
Jul16-09, 10:48 PM   #3
 
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.
Jul16-09, 10:54 PM   #4
D H
 
Mentor

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


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

Hint: The zero matrix is singular.
Jul17-09, 02:19 AM   #5
 
Ok, now I'm just having problems determining the general properties of singular matrices that would make A2 = A
Jul17-09, 02:18 PM   #6
 
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.
Jul17-09, 03:23 PM   #7
D H
 
Mentor
Quote by megalomaniac View Post
Since A = B
That was a big leap. You need to show this.

Therefore In 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?
Jul17-09, 03:37 PM   #8
 
Quote by D H View Post
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?
Jul17-09, 06:27 PM   #9
D H
 
Mentor
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.
Jul17-09, 10:58 PM   #10
 
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.
Jul17-09, 11:56 PM   #11
D H
 
Mentor
Quote by megalomaniac View Post
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.
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.
[tex]\bmatrix 1 & 0 \\ 0 & 0\endbmatrix[/tex]

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.
Jul18-09, 12:10 AM   #12
 
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!
Jul18-09, 12:28 AM   #13
D H
 
Mentor
Quote by megalomaniac View Post
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!
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.
Jul18-09, 01:04 AM   #14
 
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.
Jul18-09, 02:31 PM   #15
 
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!
Jul18-09, 02:51 PM   #16
D H
 
Mentor
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.
Jul19-09, 01:16 AM   #17
 
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.
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