I bet Peyam already knows this: this means two quantum mechanical observables can be measured precisely at the same time (sharing eigenvectors, sim.diag.) iff the order of observation doesn't matter (their operators commute, AB=BA)
@@drpeyam Wow I have never been pinned before! Just wanted to share some specific examples: Heisenberg uncertainty principle occurs because position and momentum don't commute and don't share eigenvectors. Knowing the particle's position means it is a position eigenvector, a linear combination of momentum eigenvectors, so it has probabilities to have one of the associated momentum eigenvalues. But for an atom you can simultaneously know the angular momentum and energy level of an electron because these commute
Fun fact: not only does this generalize for any finite set of commuting matricies {A_1,A_2,...,A_k}, but it generalized even further: for any subspace H of the n by n matricies with AB=BA for all A,B in H and any A in H diagonalizbe, then we can simultaneously diagonalize all of H, i.e. we can find a basis so that H is contained in the diagonal matricies.
At 14:30 Dr Peyam showed that ABv = kBv (k being a real number, v an eigenvector of A) which proves eigenvectors of A are eigenvectors of B. That proves AB=BA and A,B are diagonalizable => A=PDP^-1 and B=PD'P^-1 i.e. A and B are simultaneously diagonalizable. Also, since Dr Peyam had already proven A,B are simultaneously diagonalizable => AB=BA, then this is enough to prove the entire theorem, namely that AB=BA and A,B are diagonalizable A and B are simultaneously diagonalizable. What am I missing? I don't understand what work is to be done after 14:30, aren't we done at the timestamps 14:30 of the video? Hasn't Dr Peyam already proven the theorem by 14:30?
@@drpeyam Sure yes, I saw that on your math website at Cal. I worked with Prof C Evans on an approximate solution of the Hamilton Jacobi Bellman Eq. arising in optimal control theory. Your RUclips videos are superb btw.
so 4:40 isn't shocking because if PQ = I, then PAQPBQ = PABQ, and PBQPAQ = PBAQ, while all diagonal matrices commute due to their non-interacting elements, which means we know PABQ = (PAQ)(PBQ) = (PBQ)(PAQ) = PBAQ
I bet Peyam already knows this: this means two quantum mechanical observables can be measured precisely at the same time (sharing eigenvectors, sim.diag.) iff the order of observation doesn't matter (their operators commute, AB=BA)
WOW, I didn’t know this! How cool!!!
@@drpeyam Wow I have never been pinned before! Just wanted to share some specific examples:
Heisenberg uncertainty principle occurs because position and momentum don't commute and don't share eigenvectors. Knowing the particle's position means it is a position eigenvector, a linear combination of momentum eigenvectors, so it has probabilities to have one of the associated momentum eigenvalues.
But for an atom you can simultaneously know the angular momentum and energy level of an electron because these commute
I am here for this reason exactly
That's why I'm here hahaha
but doesn't that literally break down the uncertainty principle??
You are just the best! Thank you for elegantly explaining what some professors just don’t.
Thank youuuu!!! It’s such a hard topic, so I’m happy you understood it
Neta lo amooooo. Con usted no me aburro y me motiva a estar focus en el tema ♡
Gracias :)
Fun fact: not only does this generalize for any finite set of commuting matricies {A_1,A_2,...,A_k}, but it generalized even further: for any subspace H of the n by n matricies with AB=BA for all A,B in H and any A in H diagonalizbe, then we can simultaneously diagonalize all of H, i.e. we can find a basis so that H is contained in the diagonal matricies.
Nice!
Very hard for me but as always a complete video who gives good explainations. Thank you very much.
At 14:30 Dr Peyam showed that ABv = kBv (k being a real number, v an eigenvector of A) which proves eigenvectors of A are eigenvectors of B.
That proves AB=BA and A,B are diagonalizable => A=PDP^-1 and B=PD'P^-1 i.e. A and B are simultaneously diagonalizable.
Also, since Dr Peyam had already proven A,B are simultaneously diagonalizable => AB=BA, then this is enough to prove the entire theorem, namely that AB=BA and A,B are diagonalizable A and B are simultaneously diagonalizable.
What am I missing?
I don't understand what work is to be done after 14:30, aren't we done at the timestamps 14:30 of the video? Hasn't Dr Peyam already proven the theorem by 14:30?
Hi
Can you do an example on finding eigenvector of two simutaneous diagonalizable matrices?
Hello Peyam, Cal grad here; PhD in mechanical engr and MA in math supervised by C Evans.
OMG, very nice! You know he was my PhD advisor, right? :)
@@drpeyam Sure yes, I saw that on your math website at Cal. I worked with Prof C Evans on an approximate solution of the Hamilton Jacobi Bellman Eq. arising in optimal control theory. Your RUclips videos are superb btw.
Nice! He’s the reason my RUclips videos are so good, he’s my teaching inspiration :)
so 4:40 isn't shocking because if PQ = I, then PAQPBQ = PABQ, and PBQPAQ = PBAQ, while all diagonal matrices commute due to their non-interacting elements, which means we know PABQ = (PAQ)(PBQ) = (PBQ)(PAQ) = PBAQ
Thankyou sir
It helped me a alot ❤
It's like A and B are in the same conjugacy class, so they're 'close' friends that way.
I had a professor explain, not very well, that simultaneous diagonalizability is an equivalence relation. Does that have any implication here?
Conjugacy classes in group theory same matrix trace in matrix representations of group elements (=sum of eigenvals of course) same group character
Thanks man, you saved me
Awesome lecture 🔥
25:03
why do you need to prove w_i are in W.
don't we have that w_(1,2,...,k-1) are in W by induction assumption that k-1 is true
@9:22 you write what to find as WTF!
please make a video about nilpotent matrices
A^2 = 0 ruclips.net/video/yjzvnMyilYc/видео.html
Thank you ❤️
You have a method of making simple things lengthy, unintuitive and boring !
You have a method of leaving useless comments