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[Elementary Linear Algebra]Idempotant Matrix

  1. Jul 4, 2009 #1
    I finished this problem but I'm not sure if what I did was mathematically legal. This is a homework problem but I'm hesitant to turn it in the way it is since it's worth a lot of points. If someone can confirm what I did is correct or incorrect, I'd really appreciate it, thanks!

    1. The problem statement, all variables and given/known data
    Prove that if a matrix A is idempotent (A^2 = A), then the determinant of A is 0 (det(A) = 0).

    2. Relevant equations

    3. The attempt at a solution
    1) A^2 = A
    2) AA = A
    3) AA - A = A - A <- Here, I subtracted A from both sides
    4) AA - A = 0 <- property of the Zero matrix (A+(-A) = 0)
    5) A(I-A) = 0 <- I factored out A here, not sure if this is legal or not in matrices.

    A(I-A) = 0 implies that either I-A = 0 or A = 0; det(0) = 0, therefore, det(A) = 0.
  2. jcsd
  3. Jul 4, 2009 #2


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    Factoring is legal but you have other problems. You jump from "either I-A= 0 or A= 0" to "det(0)= 0" without saying that A= 0. What prevents I- A= 0 which gives A= I?

    Also, since not all matrices have inverses, there exist 0 divisors. That is, AB= 0 does NOT imply "A= 0 or B= 0". However, it is true that if AB= 0 then either det(A)= 0 or det(B)= 0.
  4. Jul 4, 2009 #3
    Ah, I forgot about that. Truth is, the entire question asks

    "Prove that if a matrix A is idempotent (A^2 = A), then the determinant of A is either 0 or 1."

    However, I was able to prove that the det(A) is 1 with a different method, which is why I neglect to mention it in this problem; I was thought I had to do two proofs. Perhaps I can use this method for both proofs?

    Edit: Thinking about it, if I-A=0 then A can be any random matrix, not just I. So then it wouldn't work...?
    Last edited: Jul 4, 2009
  5. Jul 4, 2009 #4


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    Have you thought about looking at the possible eigenvalues of A?
  6. Jul 5, 2009 #5
    I haven't because my class haven't gotten that far. I know eigenvalues/vectors are coming up but I haven't read that chapter myself.
  7. Jul 5, 2009 #6


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    This derivation is perfectly good. (I fixed a minor typo, colored in red) The part you were worried about is just the distributive property -- i.e. X(Y+Z) = XY+XZ and (Y+Z)X = YX+ZX -- and matrix arithmetic does have the distributive property.

    This part is not fine. This is the cancellation property -- that xy=xz implies y=z, and yx=zx implies y=z -- and matrix arithmetic does not have the cancellation property.

    Counterexamples are easy to find. (try diagonal matrices; they're simple) Understanding when a product is zero is one of the interesting problems of linear algebra.

    That said, your problem doesn't seem to be linear algebra -- your problem seems to be with proof, or more accurately, understanding logical statements. You seem to have gotten the meaning of "or" and "and" exactly backwards.

    Error 1:

    In the proof you stated above, you claimed that "A=0 or I-A=0". For the sake of argument, I'm going to assume this claim is valid. Your next step used A=0 -- but that's not correct. That wasn't what your claim was: your claim was that A=0 or I-A=0.

    If your claim was "A=0 and I-A=0", then it would be correct to continue on by using A=0, because that's (part of) what the claim said.

    However, the claim is "A=0 or I-A=0". If you want to use either of those terms, you have to split your proof into two cases: in one case you use A=0 and prove the thing you're trying to prove, and in the other case you use I-A=0 and prove the thing you're trying to prove. If you cannot do both of those, then you have failed to prove the thing you're trying to prove.


    The problem you're trying to solve asks you to show that "det(A)=0 or det(A)=1".

    Your method was to first attempt to prove det(A)=1, and then attempt to prove that det(A)=0. However, that's wrong -- that proves "det(A)=0 and det(A)=1". (And here, it leads a contradiction*, because if det(A)=0 and det(A)=1, then 0=det(A)=1, and thus 0=1)

    What you need to prove is that "det(A)=0 or det(A)=1". If you could prove det(A)=1, then you're done. ("X or Y" doesn't have to mean that X is actually possible -- if Y is always true, then "X or Y" is true, even if X is always false)

    More commonly, this means somewhere in the problem, you have to split into two or more cases. In each of those cases, you prove det(A)=0 or you prove det(A)=1. If you fail to do that in any of the cases, then you have failed to prove "det(A)=0 or det(A)=1".

    *: A contradiction doesn't necessarily mean your proof is wrong -- only that it's probably wrong. If your proof was right, then that means there's an error in the foundations of mathematics someplace, but if that were true, surely someone would have found it by now!
  8. Jul 5, 2009 #7


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    What? If I- A= 0, then I= A. It is NOT true that "any random matrix", subtracted from I, is 0.

    The main problem is, as I said and Hurkyl reiterated, "A(I-A) = 0 implies that either I-A = 0 or A = 0" is NOT true for matrices. There are plenty of non-zero matrices A, B such that AB= 0. An example is
    [tex]\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}\begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix}= \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}[/tex]

    However, it is true that if det(AB)= det(A)det(B). If A2= A, then det(A2= det(A) so (det(A))^2= det(A) or (det(A))2- det(A)= 0 and determinants are numbers.
  9. Jul 5, 2009 #8
    I understand what you mean now. I didn't quite pick up on that after the first reply, though that was what HallsofIvy was getting at. Thanks for pointing me in the right direction guys. Looks like I have some work to redo.
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