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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

    CompuChip

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    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}(A+B)=\sum_{j=1}^{n}(a_{j,j}+b_{j,j})=\sum_{j=1}^{n}a_{j,j}+\sum_{j=1}^{n}+b_{j,j}=tr_{n}(A)+tr_{n}(B)[/itex]
    and
    [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

    CompuChip

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    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]
    and
    [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

    CompuChip

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    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]?
     
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