​@@aaaa8130 to commute means to travel a specific route on a regular basis, for example most commonly from home to work. so, if a matrix were to go back and forth between its workplace and home, it would commute. finally @SchrodingerBraCat claims that matrices like to work from home (for whatever reason) and that this would be the reason why they don't commute. now i'm sure you would have understood all of this just from the first line i wrote but since i was already taking the fun out of the joke i might as well have done it properly. anyway i'm done wasting your time now, hope his helps :)

You should be ashamed of yourself 🤣

Haha I'm glad you enjoyed the joke!

@@jacquelinedoan2224 Yeah, that tickled me as well. :D

​@@steviebudden3397☠️

Maybe there is a meaningful interpretation of a complex value. Consider a complex dielectric constant or resistance/impedance. I am not saying that every component of a geometric algebra is meaningful in all physical contexts, but maybe there is a potential dialectical synthesis which broadens the understanding of the real-valued quantity.

@@polyhistorphilomath Possibly; this is actually already happening with the operators we use for quantum computing; these are non-hermitian, thus giving complex eigenvalues

@@polyhistorphilomath also, dielectric constant and impedance can indeed be complex, but this is because they are either not quantum mechanical observables (but results of many, many interactions) or they are dynamical variables instead of static ones

@@polyhistorphilomath If you're interested in complex physics, have you considered the work on PT symmetric physics (uses non-Hermitian Hamiltonians with real eigenvalues, but complex coupling coefficients)? Carl Bender gives an easy introduction in kzfaq.info/get/bejne/bJ2fqc1lyMW6gmw.html

Thanks! That application in physics is actually what inspired me to pick this subject for the video, the applications section was more of a joke (:

Thank you for your feedback! The sound quality is definitely something we have to work on, as this time we were still figuring out how to use the mic 😅 You brought up a good point about compromising the video's self-containedness for conciseness. Alex and I are still figuring this out, so we really appreciate your comment!

@@jacquelinedoan2224 so does the conjugation mean the complex conjugate then??

@@carterwoodson8818 Yup. Reflect the matrix in the leading diagonal and then take complex conjugates of the entried.

*Whips out dictionary* Time to understand

​@@thatkindcoder7510 It's not that hard to understand. Finite dimensional matrices which are symmetric can be diagonalized step by step, by first applying them to a vector again and again (this quickly produces the largest eigenvector), then considering the matrix restricted to the perpendicular space to the vector. This is an inductive procedure, and it diagonalizes any symmetric matrix step by step, by induction on the dimension. The complex analog of symmetric is Hermitian (symmetric after complex conjugation), and the same thing applies there. But the 'spectral theorem' is a result about function spaces, about INFINITE dimensional 'matrices'. These are the ones that show up in quantum mechanics, or in applications to PDEs or whatever. An infinite dimensional matrix is a linear operation on a function space. For example, consider the differentiable functions on the interval [0,1] with periodic boundary conditions. That just means periodic functions f(x+1)=f(x). This is a vector space, because you can add two such functions pointwise and they are in the same class. Any linear operation on these functions is a 'matrix' of sorts, but now with infinite dimensions. So consider the operator 'differentiation'. This has eigenvectors those functions whose derivative is proportional to themselves, i.e. exp(kx) for some k. To be periodic, k has to be imaginary and a multiple of 2\pi. The spectral theorem then tells you that this is a basis for the space, i.e. that any differentiable function can be written as a combination of these basis functions. Now consider the same operator acting on the vector space of functions on [0,1] with zero boundary conditions. The operator is still formally anti-symmetric (on those functions where it is defined and takes function in the space to other functions in the space), so i times the operator is 'hermitian', but now it has no eigenvectors in the vector space, because the eigenvalues should still be exp(ikx), and these don't obey the boundary conditions. So some infinite dimensional operators admit a basis of eigenvectors, some don't. The ones that do are a subset of the really self-adjoint ones, including the 'compact operators', which admit a limiting procedure to find the eigenvectors step by step, like in the finite dimensional case. These infinite dimension irritations, the distinction between 'symmetric' or 'hermitian' and 'self adjoint', the limit difficulties that sometimes the eigenvectors are distributional and don't even lie in the space you are analyzing, these are what make the spectral theorem interesting. The video doesn't discuss these issues, just diagonalizing finite dimensional matrices, which is easy. Physicists tend to diagnolize finite dimensional approximations, and then check the limit is sensible (which it sometimes is, and sometimes it isn't).

@@annaclarafenyo8185 “it’s not that hard to understand” *proceeds to write a short novel on the subject* All math is trivial if you study it for long enough. No one wants to see you reinforce your own comfortability/expertise in a subject via an esoteric wall of text written under the false pretense that what you’re writing is easy to understand for someone who perhaps is hearing of this stuff for the first time. The plethora of mathematical machinery you’ve assumed as prior knowledge in your reply is covered in years worth of college level math courses. What shouldn’t be hard to understand is that new math can be difficult and shouldn’t automatically be treated as easy by someone for which said math is not so new.

@@marcuslaurel5758 I haven't assumed anything, and, yes, what I wrote is not difficult, nor is it a 'short novel', it's the simplest example of infinite dimensional function spaces and the issues with diagonalizing infinite dimensional matrices.

@@marcuslaurel5758 true

While I'm here... Why the hell do I know NOTHING about matrices and they get mentioned EVERYWHERE? Is it that abnormal to not know of them? Is it supposed to be a highschool notion? Cause it certainly wasn't in my highschool.

I take full credit for that joke 😎

Lol I noticed too

It's a difference of notation between mathematical physics (where * denotes conjugates and dagger is the Hermitian adjoint) and pure mathematics (where * denotes Hermitian adjoints and \overline is for the conjugate)

Summary: [some] matrices (like derivatives) can be thought of as defining or being defined by the effect applying such objects will have on the vector or scalar operated on. So figure out the zeros of det(A-λI) for your matrix. These values are representative of the whole matrix A. Then you can conjugate and perform whatever operation you want. g h g^-1 or g^-1 h g lets you use the values on the diagonal (once you have diagonalized your matrix) as scalars. And you still have a valid result. Also the inner product generalizes the dot product. That’s with hardly any restrictions on what the operands are. If you have two abelian operations on a set of vectors you can show it (with some additional constraints, maybe?) is closed. Meaning you don’t need to worry about performing only finitely many operations on the elements. Closure. But even in this context operators like T have a 1-to-1 correspondence with matrices. So you can decompose T, an operator in a Hilbert space, the same way you can decompose a matrix.

This means you can apply the matrix exponential to T and apply that functor to a vector in the space. This is maybe two steps away from using a Fourier or Laplace transform on T. Heck, you could apply a translation operator, assuming that your differential operator D is well-defined. The world is your oyster.

I’d imagine there are many introductory linear algebra courses which talk about how to diagonalize a matrix, but fail to go into further depth beyond that. While an introductory linear algebra course can be more theory based than any previous math course, at the end of the day it can still be mostly a course in computing things involving matrices, leaving out much of the deeper theory and the more mathematically technical theorems.

Our goal was to introduce a geometric interpretation of what is happening when you diagonalize a finite-dimensional operator. Lots of people learn theorems about diagonalization as a rote process of manipulating boxes of numbers, with no intuitive understanding of the process

@@kazachekalex I think you did a great job. I was taught all of these concepts in theoretical chemistry but the video really helped deepening my understanding of it!

Thank you for your comment. Our intended target audience is students with some background in linear algebra, so we understand that it might be difficult to follow as a layman. We will improve on self-containedness in later work!

@@jacquelinedoan2224 it is not nicely done, i teach math, but but you have to be better than this

@@jacquelinedoan2224 I have a master degree in computer science and had a lot of math but still couldn't follow the video, because it was too fast and the pictures were poor.