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An interesting matrix identity problem

  1. Oct 11, 2004 #1
    My linear algebra professor likes to use theorems he expects us to prove later in his proofs of theorems in class. Well it's not like that because it makes sense but they're usually more of side notes. Today he asked us to prove this:

    Given A and B are 2 by 2 matrices over the field R, prove that if A*B = I(2x2) where I(2x2) is the 2x2 identity matrix, B*A = I(2x2).

    Now I filled up a good page just to prove this and ended up turning A and B into

    [tex]A = \left( \begin{array}{cc} a1 & a2\\
    a3 & a4 \end{array} \right) , B = \left( \begin{array}{cc} b1 & b2 \\ b3 & b4 \end{array} \right)[/tex]


    [tex]A = \left( \begin{array}{cc} a1 & a2\\ a3 & a4 \end{array} \right) , B = \frac{1}{a1a4-a3a2} \left( \begin{array}{cc}a4 & -a2\\ -a3 & a1 \end{array} \right)[/tex]

    and moving on from there it was pretty easy to prove that [tex]AB = BA = \left( \begin{array}{cc} 1 & 0\\0 & 1 \end{array} \right)[/tex]

    Is there another way? I wasted so much time and my professor's proofs are usually so much more elegant. I can scan a copy of my work if you're still confused as to what I actually did (unbeautifully)

    (Please excuse this post it will take awhile to get right I am learning LaTeX)
  2. jcsd
  3. Oct 12, 2004 #2

    matt grime

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    here's one way of doing it let me use 1 for the identity

    1=AB, as given implies




    let C=BA, we want to show C=1 too.

    C satisfies C^2=C, so its characteristic polynomial divides x^2-x.

    this implies that C is one of

    [tex]\left( \begin{array}{cc} 1 & 0\\0 & 1 \end{array} \right)[/tex]

    [tex]\left( \begin{array}{cc} 1 & 0\\0 & 0 \end{array} \right)[/tex]

    [tex]\left( \begin{array}{cc} 0 & 0\\0 & 1 \end{array} \right)[/tex]

    [tex]\left( \begin{array}{cc} 0 & 0\\0 & 0 \end{array} \right)[/tex]

    but C has only 0 in its kernel, hence it must be the identity.

    In general for more dimensions, we just say C is 'idempotent' since its square is itself. idempotent matrices correspond exactly to projection onto some span of some subset of the basis vectors. C has trivial kernel, and the only such idempotent is the identity matrix.
  4. Oct 12, 2004 #3


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    how about this:
  5. Oct 12, 2004 #4

    matt grime

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    that presupposes that A has a left inverse.
  6. Oct 12, 2004 #5


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    if it has a right inverse, it is invertible, and thus has a left inverse. that is, if it has a right inverse, its columnspace spans R2, which means its determinant is nonzero, which means its transposes determinant is nonzero, which means its rowspace spans R2, which means it has a left inverse. i havent done linear algebra in a while, so i might not be thorough, but i just think theres a simpler proof of this than the one you gave.
    Last edited: Oct 12, 2004
  7. Oct 12, 2004 #6


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    I get the impression that they haven't yet proven that.
  8. Oct 13, 2004 #7

    matt grime

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    Mine is a very simple proof, if we omit the details, which the OP may not be familiar with.

    BA is clearly idempotent with trivial kernel, hence it is the identity.

    It's a one liner. Of course simplicity depends on what we assume familiarity with.

    To prove that it isn't as trivial as you might think, consider the left (L) and right (R) shift operators on l^2 with the standard basis.


    so R possesses a right inverse, but R is not invertible.

    It is in fact a standard lemma in group theory that if something has a left and right inverse then they are the same. However having a left inverse does not imply that something is invertible. We both need finite dimensionality.

    Besides, this is a proof that applies to any linear map on a finite dimensional vector space, and does not require you to pick a basis (though I did pick one to make it clearer), and as such would generally be considered better by a pure mathematician, though if you aren't one of those you're free to think something else is better.
  9. Oct 16, 2004 #8
    Sorry for being so late in responding. Yes Mattgrime my professor gave a very similar proof to that one in class. It was very set theory based as opposed to strictly linear algebra. Thank you for all your input everyone it has been enlightening :)
  10. Oct 17, 2004 #9


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    As usual I am too late. but this seems easy enough to do in an elementary way.

    I.e. AB = I implies that (BA)B = B(AB) = B, which implies that BA is the identity on the image of B, so you just need to show that multiplication by B is surjective. Since the determinant is non zero this follows, but here is another simple argument without determinants:

    If not, then every vector form Bx would have form cv for some fixed v. And then every vector of form ABx would have form A(cv) = c(Av). But not every vector x has this form, so we cannot always have ABx = x, contradiction.
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