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)

📜 All Discrete Mathematics Exercises

📜 All Linear Algebra Exercises

🎵 Music
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[imanabu]

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

[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)

[whocareska]

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?

[rohanthomas]

Communiting only the product doesn’t work, you have to swap the indices too. Swapping only the indices as here would mean AB and BA have same diagonal entries, which isn’t always true. Take a concrete example and see how it goes.

[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.

[dannm]

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

[eyalofer]

The colors are not visiable. Use light color on board