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anonymous
 4 years ago
Can someone help me prove a property of matrices?
anonymous
 4 years ago
Can someone help me prove a property of matrices?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Prove that id A and B are square matrices of order n, then Tr(AB)=Tr(BA)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0when you say order, is that like rank? or just the fact that they are both n x n?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I guess that they are NxN

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The trace is the sum of the entries on the main diagonal of a matrix.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well if AB=BA the Tra(AB)=Tra(BA)

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.3You'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
 4 years ago
Best ResponseYou've already chosen the best response.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.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.3It's not that hard and it's a very good exercise in manipulating indices and summations.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0A trace of an nbyn 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
 4 years ago
Best ResponseYou've already chosen the best response.0LOL i hate summations

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You'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
 4 years ago
Best ResponseYou've already chosen the best response.0LOL i guess it is time to overcome my fear

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I just never learnt summations in hs so i dont have a background

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.0you can also look at the eigenvalues

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Didnt learn eigenvalues yet lol

watchmath
 4 years ago
Best ResponseYou've already chosen the best response.0Let \(E_{i,j}\) 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
 4 years ago
Best ResponseYou've already chosen the best response.0Tr(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
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks Guys :D U guys r awesome
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