anonymous
  • anonymous
Can someone help me prove a property of matrices?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Prove that id A and B are square matrices of order n, then Tr(AB)=Tr(BA)
anonymous
  • anonymous
*if
anonymous
  • anonymous
when you say order, is that like rank? or just the fact that they are both n x n?

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anonymous
  • anonymous
I guess that they are NxN
anonymous
  • anonymous
what are traces?
anonymous
  • anonymous
The trace is the sum of the entries on the main diagonal of a matrix.
anonymous
  • anonymous
Well if AB=BA the Tra(AB)=Tra(BA)
JamesJ
  • JamesJ
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.
anonymous
  • anonymous
@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.
anonymous
  • anonymous
best*
JamesJ
  • JamesJ
It's not that hard and it's a very good exercise in manipulating indices and summations.
anonymous
  • anonymous
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).
anonymous
  • anonymous
LOL i hate summations
anonymous
  • anonymous
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!
anonymous
  • anonymous
LOL i guess it is time to overcome my fear
anonymous
  • anonymous
I just never learnt summations in hs so i dont have a background
Zarkon
  • Zarkon
you can also look at the eigenvalues
anonymous
  • anonymous
Didnt learn eigenvalues yet lol
watchmath
  • watchmath
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?
anonymous
  • anonymous
Tr(AB)=Tr(BA) Proof: Let A=a_{ij} and B=b_{ij} with AB = c_{ij} and BA = d_{ij} Then, \[Tr(AB) = \sum_{i=1}^n c_{ii}\]\[= \sum_{i=1}^n \sum_{j=1}^n a_{ij} b_{ji}\]\[= \sum_{j=1}^n \sum_{i=1}^n b_{ji} a_{ij}\]\[=\sum_{j=1}^n d_{jj}\]\[=Tr(BA)\]
anonymous
  • anonymous
Thanks Guys :D U guys r awesome

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