A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

Can someone help me prove a property of matrices?

  • This Question is Closed
  1. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Prove that id A and B are square matrices of order n, then Tr(AB)=Tr(BA)

  2. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    *if

  3. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    when you say order, is that like rank? or just the fact that they are both n x n?

  4. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I guess that they are NxN

  5. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what are traces?

  6. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The trace is the sum of the entries on the main diagonal of a matrix.

  7. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Well if AB=BA the Tra(AB)=Tra(BA)

  8. JamesJ
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    You'll want to write down explicitly what is the (i,i) term of AB; e.g., (1,1) term of AB is equal to \[ \sum_j a_{1j}b_{j1} \] and sum them up over the diagonal. Then write down the expression for a diagonal element of BA, and then sum them up. Finally, show the two sums are equal.

  9. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @rld that is only if AB = BA, but generally matrix multiplication is not commutative. James' idea is the bet one, it will be a proof that shows you can formally manipulate summations.

  10. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    best*

  11. JamesJ
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    It's not that hard and it's a very good exercise in manipulating indices and summations.

  12. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    A trace of an n-by-n square matrix A is defined to be the sum of the elements on the main diagonal (the diagonal from the upper left to the lower right).

  13. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    LOL i hate summations

  14. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You'll find that a lot of these problems, especially in deeper linear algebra and representation theory, can only be proven by frustratingly long summations that can only be solved using clever techniques and lots and lots of patience. So don't be too afraid of summations!

  15. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    LOL i guess it is time to overcome my fear

  16. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I just never learnt summations in hs so i dont have a background

  17. Zarkon
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you can also look at the eigenvalues

  18. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Didnt learn eigenvalues yet lol

  19. watchmath
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Let \(E_{i,j}\) be the matrix where all the entries are zero except 1 at the (i,j) position. I think the problem can be reduce into proving \(tr(E_{i,j}E_{k,l}=tr(E_{k,l}E_{i,j}\). What do you think?

  20. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Tr(AB)=Tr(BA) Proof: Let A=a_{ij} and B=b_{ij} with AB = c_{ij} and BA = d_{ij} Then, \[Tr(AB) = \sum_{i=1}^n c_{ii}\]\[= \sum_{i=1}^n \sum_{j=1}^n a_{ij} b_{ji}\]\[= \sum_{j=1}^n \sum_{i=1}^n b_{ji} a_{ij}\]\[=\sum_{j=1}^n d_{jj}\]\[=Tr(BA)\]

  21. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanks Guys :D U guys r awesome

  22. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.