1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Trace of Matrix Product as Scalar Product

  1. Aug 2, 2013 #1
    1. The problem statement, all variables and given/known data

    Let V be the real vector space of all real symmetric n × n matrices and define the scalar product of two matrices A, B by (Tr (A) denotes the trace of A)

    Show that this indeed fulfils the requirements on a scalar product.


    2. Relevant equations

    Conditions for a scalar product:


    3. The attempt at a solution

    I'm not sure how to show the last part. Which can be summarized as:

    <A|B> = 0 if ATA = I and BTB = I

    The first 3 parts of my attempt are shown below:

  2. jcsd
  3. Aug 2, 2013 #2
    For (2), you say ##\mathrm{Tr}(AB)^T##. I don't really know what you mean with this. What is the transpose of a number? And why should

    [tex]\mathrm{Tr}(AB)^T = \mathrm{Tr}(B^T A^T)[/tex]

    For (3), you should still show that the inner product is ##\geq 0## and that it is ##=0## iff ##A=0##.
  4. Aug 3, 2013 #3
    For rule 1, I think the OP means [itex]Tr ((AB)^T)[/itex], not [itex](Tr(AB))^T[/itex]. The 1st line of the proof is unnecessary. The 2nd line looks good to me. However, you are using the fact that [itex]Tr(A) = Tr(A^T)[/itex]. This is easy to prove and you should add that proof in. For rule 2, you still need to prove that <A|A> = 0 implies A = 0. You will need to use the fact that the underlying space only includes symmetric matrices.
    Last edited: Aug 3, 2013
  5. Aug 5, 2013 #4
    Yeah [itex]Tr(A) = Tr(A^T)[/itex] because for any [itex]A_{ij}[/itex] component where i=j, switching their positions don't change anything.

    I'm more concerned about the point number 4. Which can be summarized as:

    <A|B> = 0 if ATA = I and BTB = I
  6. Aug 5, 2013 #5

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    Point number 4 says nothing of the kind. Consider A=B=I. Obviously ATA = I, as does BTB. Yet <A,B> is not zero.

    Instead think of point #4 as being a definition of what it means for two quantities to be deemed as being "orthogonal" to one another.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted