Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Is the determinant a linear operation?

  1. Jul 26, 2011 #1
    Is the determinant a linear operation? I mean can we say that:

    [tex]\mathbb{E}[\text{det}]=\text{det}[\mathbb{E}][/tex]

    where E is the expectation operator?
     
  2. jcsd
  3. Jul 26, 2011 #2

    Char. Limit

    User Avatar
    Gold Member

    First impression says no, but I'm testing it out now, I might be wrong.
     
  4. Jul 26, 2011 #3
    No, it's not. For instance, det(aM) = andet(M). And det(M+N) doesn't have any simple relationship to det(M) and det(N). (See, for instance, http://en.wikipedia.org/wiki/Matrix_determinant_lemma" [Broken].)
     
    Last edited by a moderator: May 5, 2017
  5. Jul 26, 2011 #4
    Ok, now we have this inequality:

    [tex]\mathbb{E}\left[\log(X)\right]\leq\log\left(\mathbb{E}[X]\right)[/tex]

    Can we say in the same manner that:

    [tex]\mathbb{E}\left[\log\left(\text{det}\left\{H\right\}\right)\right]\leq\log\left(\text{det}\left\{\mathbb{E}\left[H\right]\right\}\right)[/tex]
     
  6. Jul 26, 2011 #5
    Nope. Suppose H=((1 0), (0 1)) with 50% probability and ((-1 0), (0 -1)) with 50% probability. Both of those have determinant 1, so E[log det H] = 0. But E[H] = ((0 0), (0, 0)), with determinant 0, so log det E[H] = -infinity.
     
  7. Jul 26, 2011 #6
    Ok, I see. Thanks a lot
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook