Linear algerba: trace of square matrix is a linear functional

[skrat]

Lets define trace for each square matrix A its trace as sum of its diagonal elements, so tr_{n}(A)=\sum_{j=1}^{n}a_{j,j}. 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?

 

[CompuChip]

If I understand correctly, linear functional works on a vector but returns a real or complex number.

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 \mathbb{R}^{n^2}.

So linear functional is a scalar product. Now what?

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

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

It has to satisfy that \operatorname{tr}_n(A + B) = \operatorname{tr}_n(A) + \operatorname{tr}_n(B) and \operatorname{tr}_n(k A) = k \operatorname{tr}_n(A) for all n x n matrices A, B and real numbers k. Those are the two properties that make an arbitrary function V \to \mathbb{R} into a linear functional, see e.g. Wikipedia.

 

[skrat]

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

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:
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)
and
tr_{n}(\lambda A)=\sum_{j=1}^{n}\lambda a_{j,j}=\lambda \sum_{j=1}^{n}a_{j,j}=\lambda tr_{n}(A)

Does anything change if \lambda \in \mathbb{C}

 

[CompuChip]

Exactly. That is how we defined linear functional.

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

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:
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)
and
tr_{n}(\lambda A)=\sum_{j=1}^{n}\lambda a_{j,j}=\lambda \sum_{j=1}^{n}a_{j,j}=\lambda tr_{n}(A)

Yes, that was what I was saying.

Does anything change if \lambda \in \mathbb{C}

Not really, except that you have to make everything complex, e.g. since \lambda A will be an n x n matrix with complex entries you have to extend the definition of tr_n those matrices. Note that the example of the scalar product becomes a bit more involved, as \vec v \cdot \vec w = \sum_i v_i^* w_i has an additional complex conjugate compared to the real case.

 

[skrat]

THANK YOU VERY MUCH. One more question:

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

Assuming * here means complex conjugation I started like this:

tr(AB^{*})=\sum_{i=1}^{n}(\sum_{j=1}^{n}a_{i,j}\cdot {b_{j,i}^{*}}) but how do i write <A,B>?

 

[CompuChip]

A scalar product (or inner product) should satisfy three properties, e.g. one of them is \langle A, A \rangle \ge 0 with equality if and only if A = 0. Can you give me the other two properties?

 

[skrat]

I hope I can:

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

So, the same goes for trace: tr(AA^{*})=0, tr(AB^{*})=tr(BA^{*}) and tr(\lambda (A+B^{*}))=\lambda tr(A)+\lambda tr(B^{*} )?