Components of the trace operationby fa2209 Tags: dual space, linear functional, matrices, trace, vector space 

#1
Dec1311, 02:31 PM

P: 21

I'm currently reading "Introduction to tensors and Group Theory for Physicists". I'm stuck on a question posed on dual spaces.
The author gives the trace as an example of a linear functional on the vector space M_n(ℝ) (n x n matrices with real entries) and then asks how one would find the components of the element of the dual space that takes an n x n matrix to a real numberthe trace. I have no idea how to solve this problem, any help would be much appreciated. 



#2
Dec1311, 05:28 PM

Emeritus
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PF Gold
P: 8,999

Start with an arbitrary member of V* acting on an arbitrary member of V, in whatever notation you prefer, e.g. ρ(x). Then try two different ideas: 1. Expand x in a basis. 2. Expand x in a basis, and ρ in the dual basis of that first basis. Then compare the results. This will give you the formula for components of dual vectors.
Then choose a basis for the vector space of matrices, and use the formula with the trace and the dual basis. Note that the components of a linear functional are always component with respect to some basis, so it's impossible to do it without choosing a basis. 



#3
Dec1311, 06:59 PM

P: 21

Thanks a lot for your reply. I'm pretty new to this mathematics so I follow what your saying (sort of) but not sure I could carry out the steps you've outlined.
A few questions: 1). I would probably choose the basis E_ij which represents the matrix with 1 in the ijth position and zero's everywhere else but I wasn't sure what you meant by "and use the formula with the trace and the dual basis". 2). I know that to find the dual basis I use [F E]_ij = δ_ij where F are the bases of the dual space, so I've written ƩF_{ik}E_{kj} = δ_{ij} but I'm not really sure where to go from here to find the dual basis. thanks 



#4
Dec1311, 07:31 PM

Emeritus
Sci Advisor
PF Gold
P: 8,999

Components of the trace operationEdit: You're new here, so you might not understand why I don't just tell you the answer. It's the forum's policy on textbookstyle questions. We're required to treat them all as homework, and only give hints, not complete answers. I think I gave you 95% of it though. 



#5
Dec1311, 08:35 PM

P: 21

Haha, thanks but I'm struggling with the next 5% because I've only been doing this kind of mathematics for about 2 days.
What I'm not sure about is how I can extend this to matrices to answer my problem. I understand how to use the orthogonality relationship with vectors but I'm not really sure about how to use it with matrices. What I mean is that if I use the fact that e^{i}e_{i}=δ_{ij} then I can instantly see that the dual basis vector can be written explicitly as a row vector (assuming I write e_{i} as a column vector) with a 1 in the same column as the 1 in the row of the vector in V and zero's elsewhere. So how do I extend this to matrices? I think I also confused myself a bit with my choice of notation, normally if I write E_{ij} I mean the ijth component of a matrix but here it actually represents the entire matrix not just one component. Finally, I'm trying to find something that maps an n x n matrix to a real number so don't I need a (1 x n) times the original n x n and then multiply on the right by an n x 1 matrix? Sorry for the ramble but writing down my thinking might help you see where I'm going wrong (apart from being a terrible mathematician). 



#6
Dec1311, 09:12 PM

Emeritus
Sci Advisor
PF Gold
P: 8,999

I think you need to start by doing what I described in the first paragraph of post #2. If you think you can't, I think you're really just assuming that it will be hard, when in fact it's very easy. I've done that myself a bunch of times. One time it took me more than an hour to prove that if a Banach algebra has an identity element, it must be unique. Banach algebras sounded scary, so I expected it to be hard. I felt really stupid when I realized that the proof looks like this: 1=1·1'=1'.



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