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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 |
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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. |
| Jul16-09, 10:29 PM | #2 |
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Hint: If A is not singular, it has an inverse. Use this fact.
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| Jul16-09, 10:48 PM | #3 |
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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 |
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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 |
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Ok, now I'm just having problems determining the general properties of singular matrices that would make A2 = A
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| Jul17-09, 02:18 PM | #6 |
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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 |
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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 |
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| Jul17-09, 06:27 PM | #9 |
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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 |
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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 |
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This matrix is diagonal, is idempotent, and is not full rank. |
| Jul18-09, 12:10 AM | #12 |
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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 |
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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 |
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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.
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| Jul18-09, 02:31 PM | #15 |
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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 |
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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 |
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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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