1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    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!

Linear algerba: trace of square matrix is a linear functional

  1. Apr 14, 2013 #1
    Lets define trace for each square matrix [itex]A[/itex] its trace as sum of its diagonal elements, so [itex]tr_{n}(A)=\sum_{j=1}^{n}a_{j,j}[/itex]. Now proove that trace is a linear functional for all square matrix.

    I would be happy to know what has to be true for anything to be a linear functional?

    If I understand correctly, linear functional works on a vector but returns a real or complex number. So linear functional is a scalar product. Now what?
  2. jcsd
  3. Apr 14, 2013 #2


    User Avatar
    Science Advisor
    Homework Helper

    That is correct. The trace takes a matrix and returns a number. All matrices form a vector space, which you can show by checking the properties of a vector space. Or if you want to cheat a bit, you could write down all the entries of the matrix in one big vector and consider it an element of [itex]\mathbb{R}^{n^2}[/itex].

    I'm not sure what you mean by "scalar product", are you talking about the inner product [itex]\vec a \cdot \vec b = \sum a_i b_i[/itex]?

    It has to satisfy that [itex]\operatorname{tr}_n(A + B) = \operatorname{tr}_n(A) + \operatorname{tr}_n(B)[/itex] and [itex]\operatorname{tr}_n(k A) = k \operatorname{tr}_n(A)[/itex] for all n x n matrices A, B and real numbers k. Those are the two properties that make an arbitrary function [itex]V \to \mathbb{R}[/itex] into a linear functional, see e.g. Wikipedia.
  4. Apr 14, 2013 #3
    Exactly. That is how we defined linear functional.

    So, you are trying to say that the following two rows are a complete proof that trace, defined as a sum of diagonal elements of square matrix, is a linear functional:
    [itex]tr_{n}(\lambda A)=\sum_{j=1}^{n}\lambda a_{j,j}=\lambda \sum_{j=1}^{n}a_{j,j}=\lambda tr_{n}(A)[/itex]

    Does anything change if [itex]\lambda \in \mathbb{C}[/itex]
  5. Apr 14, 2013 #4


    User Avatar
    Science Advisor
    Homework Helper

    Hmm, the scalar product is only a linear functional if you fix one of the vectors, e.g. for any fixed real vector [itex]\vec a[/itex] the functions
    [tex]f_{\vec a}(\vec v): \mathbb{R}^n \to \mathbb{R}, \vec v \mapsto \vec a \cdot \vec v[/tex]
    [tex]g_{\vec a}(\vec v): \mathbb{R}^n \to \mathbb{R}, \vec v \mapsto \vec v \cdot \vec a[/tex]
    are both linear functionals (I'll leave it to you to prove it). The inner product [itex]\mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}[/itex] is what we call bilinear, where the bi- indicates that it is linear in both arguments.

    Yes, that was what I was saying.

    Not really, except that you have to make everything complex, e.g. since [itex]\lambda A[/itex] will be an n x n matrix with complex entries you have to extend the definition of trn those matrices. Note that the example of the scalar product becomes a bit more involved, as [itex]\vec v \cdot \vec w = \sum_i v_i^* w_i[/itex] has an additional complex conjugate compared to the real case.
  6. Apr 14, 2013 #5
    THANK YOU VERY MUCH. One more question:

    How do I show that [itex]<A,B>=tr(AB^{*})[/itex] defines scalar product if A,B are both square matrix with complex elements.

    Assuming * here means complex conjugation I started like this:

    [itex]tr(AB^{*})=\sum_{i=1}^{n}[/itex][itex](\sum_{j=1}^{n}a_{i,j}[/itex][itex]\cdot {b_{j,i}^{*}}) [/itex] but how do i write [itex]<A,B>[/itex]?
  7. Apr 14, 2013 #6


    User Avatar
    Science Advisor
    Homework Helper

    A scalar product (or inner product) should satisfy three properties, e.g. one of them is [itex]\langle A, A \rangle \ge 0[/itex] with equality if and only if A = 0. Can you give me the other two properties?
  8. Apr 14, 2013 #7
    I hope I can:

    [itex]<A,B>=(<B,A>)^{*}[/itex] and [itex]<\lambda (A+B),C>=\lambda <A,C>+\lambda <B,C>[/itex]

    So, the same goes for trace: [itex]tr(AA^{*})=0[/itex], [itex]tr(AB^{*})=tr(BA^{*})[/itex] and [itex]tr(\lambda (A+B^{*}))=\lambda tr(A)+\lambda tr(B^{*} )[/itex]?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Threads - Linear algerba trace Date
2nd order non-linear pde Yesterday at 7:31 AM
Bland rule proof linear programming Mar 7, 2018
Linear algebra matrix to compute series Mar 1, 2018
Find the Fourier Series of the function Feb 24, 2018
Rearrangement and intergration of algerba Nov 26, 2009