Proof the Commutativity of the Trace of Two Matrices

In this exercise we will proof that the trace of two quadratic matrices A times B is the same as the trace(B*A).
⏰ Timeline
00:00 Exercise
00:14 Trace
00:59 Proof
03:07 Conclusion
🔢 To proof
trace(A*B) = trace(B*A)
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[@imanabu5862]

this is high quality content!
I just discovered this channel and I already love it, keep up the good work :)!

[@flolu]

Thank you, I appreciate your comment!

[@drtruth6359]

@@flolu Tr(ABC)=Tr(BCA) can you please do it.

[@Trubripes]

I think the real trick is reversing the summation order of i and j, turning columns into rows and vise versa.
This allow the sum of reverse element to become the trace of matrix in reverse order.

[@minutenreis]

great explanation though if I don't misunderstand you have to swap the two sum signs before you can remove the inner sum since the rule for matrix multiplication predicates, that
sum k= 1 to n(B i,k * A k, j) = BA i,j meaning that it would transform into sum j = 1 to n ( (B*A)j,j) = trace(B*A)

[@flolu]

Yes, that would defeinitely make more sense. But I thinkg since addition is commutative, the order of sums doesn't matter.

[@minutenreis]

@@flolu it really doesnt if you swap the sums you just actually show that you use the commutativeness :)

[@whocareska1371]

Hi, nice video! I have a similar problem to solve, but in my case there is A E R(^m x n) and B E R(^n x m). How exactly would you show that trace(AB) = trace(BA) in this case?

[@bestmusic2608]

What an explanation

[@dan7nm]

Excellent video, but the music is a little distracting, thanks!

[@eyalofer5015]

The colors are not visiable. Use light color on board

[@flolu]

Ok