Question

Can someone help me prove a property of matrices?

Answer 1

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

Answer 2

*if

Answer 3

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

Answer 4

I guess that they are NxN

Answer 5

what are traces?

Answer 6

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

Answer 7

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

Answer 8

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.

Answer 9

@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.

Answer 10

best*

Answer 11

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

Answer 12

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).

Answer 13

LOL i hate summations

Answer 14

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!

Answer 15

LOL i guess it is time to overcome my fear

Answer 16

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

Answer 17

you can also look at the eigenvalues

Answer 18

Didnt learn eigenvalues yet lol

Answer 19

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?

Answer 20

Tr(AB)=Tr(BA)

Proof: Let A=a_{ij} and B=b_{ij} with
AB = c_{ij} and BA = d_{ij}

Then,
\[Tr(AB) = \sum_{i=1}^n c_{ii}\]\[= \sum_{i=1}^n \sum_{j=1}^n a_{ij} b_{ji}\]\[= \sum_{j=1}^n \sum_{i=1}^n b_{ji} a_{ij}\]\[=\sum_{j=1}^n d_{jj}\]\[=Tr(BA)\]

Answer 21

Thanks Guys :D U guys r awesome