On the fifth root of the identity matrix. 

10:08 "so, we also know that 1+1=2" finally, a part that I understood

I'm not usually a golden ratio nerd, but the fact that he didn't write them in terms of phi at the end drove me INSANE

One cool application of this is that you can use it to find functions of the form f(x)=(ax+b)/(cx+d) whose fifth power is the identity.

Quick guess: The golden ratio shows up

Those matrices are in SL(2, R). That group has interesting properties that might merit a video.

Thanks!

For once I'm not sure it was the best place to stop for the following reason. People who don't know linear algebra that well could end up thinking along the lines of "oh we were looking for fifth roots of the identity matrix and we ended up with 5 solutions so matrix are like complex numbers". I would have prefered if you had just commented on the fact that you found 5 solutions because you were looking at a very specific type of solutions and that in general, the equation A^5=I has infinitely many solutions.

Oh my god there's another channel??? With even more learning material? You are too good to us 🤩

I miss the times when the right after "that's a good place to stop" the video would cut

I really enjoy these videos, as a long-time-ago math major. But (and this is trivial, I know) I miss the trick of knocking on the chalkboard to update for the next section of calculations.

14:29

To answer Vinvic1: The general solution can be written as M=ARA^(-1) where R is a rotation of k*2*pi/5. The choice of A actually has only 2 free parameters since A can vary by a dilation. (rotation time scaling). In fact M_11= cos(mu)+a*sin(mu) ,M_22= cos(mu)-a*sin(mu) , M_12=b sin(mu) and M_21=-g sin(mu) where mu = k*2*pi/5 and 1+a^2=bg. That is the general solution of the problem M^5=Identity. You can easily compute a,b,g using M=ARA^(-1) . You can write M as cos(mu)* Identity + sin(mu) B where the matrix B=(a b// -g -a) . Identity and B form a representation of the complex numbers since B^2=-Identity. In fact, in general for any angle mu, M^n= cos(N mu)* Identity + sin(N mu) B = exp(N mu B) . Euler's formula!!!!!!!!!!

The lemma in this video is absolutely beautiful and I’m mad I didn’t encounter it in school.

Yup, I didn't learn any of this in my linear algebra class outside of finding eigenvalues. How many solutions are there for X^5=I in general? Or X^2=I for that matter?

Since the det=1, b²=1-a and it's easy to set up A³ = A^{-2} . It's pretty quick but not very insightful, however.

It becomes very simple when you realise i is really (0, 1; -1, 0). Then you just write down all 5th roots, i.e. e^(i t) where t = 2 pi k / 5.

Thanks:)!

I've never done that exercise, thanks.

it's all assuming A (the "5th root" matrix) is diagonalizable, so A = PDP^-1 and A^5 = PD^nP^-1 and in order to be equal to I, A^5 must have in particular the same eigenvalues, i.e. D^5 = I (which means λi^5 = 1) what if it's not? then this decomposition is not valid and you cannot use it. is it any lemma like "rausing non-diagonalizable matrix to a power results in non-diagonalizable matrix" or something like that? let's elaborate. p_A(x) = (a-x)(1-x) + b^2 = x^2 -(a+1) + a + b^2 if discriminant !=0 there will be 2 distinct roots (either real or complex) each of them with geometric multiplicity of 1, together 2, and the matrix will be diagonalizable. equating discriminant to 0, (a+1)^2 - 4(a+b^2) = 0 we get b^2 =set ((a+1)^2 - 4a)/4 = (a-1)^2 / 4 so b = +- |a-1| / 2 = +- (a-1) / 2 so A is [a , (a-1) / 2] [-(a-1) / 2, 1 ] in this case the only eigenvalue is x=(a+1)/2 and it's easy to see that the geometric mult. is 1 because the resulting matrix (A-xI) is [a-(a+1)/2, (a-1) / 2] [-(a-1) / 2, 1-(a+1)/2] simplified [(a-1)/2, (a-1)/2] [-(a-1) / 2, -(a-1)/2] which is clearly not the zero matrix for any a!=1. (in case a=1 the original is the identity, one of your solutions) another simple case is when a=-1 so the eigenvalue (a+1)/2 = 0, A is singular and no power will get it to be a regular matrix such as the identity. what about other cases? so A is non-singular but not diagonalizable and we can't use the A^n = PD^nP^-1 formula thus cannot calculate using the eigenvalues. maybe there are such whose 5th power is the identity?

That 1 on the main diagonal throws me off. Put an a there and suddenly it's just about complex number roots.

Shouldn't the results be verified? The implications only went to one direction. Thus we might have to show that the found numbers indeed solve the original problem. E. G. If matrix is the same, then eigenvalues are the same but it doesn't work in the reverse direction.

It remind me of the idea of the Parity operator as the square root of the identity P f(x) = f(-x) ==> P^2 f(x) = I f(x) =f(x), whose eigenvectors are the symmetric and anti-symmetric functions with +/-1 as the eigenvalues. I tried analyzing the Fourier transform as a quartic root of unity (since F^2 = P), so it has eigenvalues of 1,i,-1,-i, but I wasn't able to get a nice sense of what do the eigenvector functions mean, or what is the cubic root of unity in this family: I know there is a complicated formula for the fractional Fourier transform, so I'm now wondering "what's so special about this family of roots of unity" and if there's a sort of "generator" for them by taking a limit of an epsilon fractional transform.

Just wondering why the concept of complex conjugates was not used. If lambda(1)*lambda(2) must be a real number, this means that these are complex conjugates. Later on, when finding the fifth roots, it basically mean that only pairs with +/- the same argument come into play.

This implies that the firth root of the identity could also be: [1 b; -b a]. So how many root are there?

Love mathmajor!

Instantly, one solution must be (a,b) = 1,0 😊 Obvs there are more solutions. Like maybe another four, just at a guess...

Next show that those matrices form a representation of unit complex numbers

Nice!

Guess roots of unity

Great !

I heard you like getting viewer suggested problems, so here is one I finished solving just the other day! Given a natural number n, find m such that the line y=2nx forms an angle bisector between the lines y=nx and y=mx

So in general, the trace of the n x n identity matrix is n.

Normally we arrange eigenvalues in order of magnitude

Why was the problem set up with this antisymmetric matrix with one coefficient already known ? Is the general case not solvable ?

Can I join math major even though I’m a physics major 🥺🥺🥺

Please probability theory cours

This seems like a lot of calculation without anything to learn. What can you learn from this?

👍

Didn't you only prove the "necessity" part of the problem? The fact that one matrix has only eigenvalues 1 doesn't mean it is the identity, also the 2x2 Jordan block could work.

Eigenvector? German maths terminology? For real?