On the fifth root of the identity matrix.

[@descuddlebat]

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

[@SuperYoonHo]

me too:)

[@humbertonajera6561]

Thanks!

[@lexinwonderland5741]

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

[@candamir26]

Where exactly do you see a golden ratio? I think you are mistaken.

[@franciscomorilla9559]

@@candamir26 In the second case, a=(sqrt(5)-3)/2=φ-2, where φ is the golden ratio; also b=sqrt(3-φ). The third case is similar

[@MathFromAlphaToOmega]

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.

[@lexinwonderland5741]

Wooooo modular forms!

[@DanGRV]

I think it should be the function applied five times instead of its fifth power: f(f(f(f(f(x))))) = x

[@robertmines5577]

Is this because that function is a Mobius Transform?

[@DanGRV]

@@robertmines5577 Yes

[@davidgillies620]

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

[@marcushendriksen8415]

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

[@f5673-t1h]

Quick guess: The golden ratio shows up

[@user-nb6zu3rk4f]

Quick guess: roots of unity show up

[@Alex_Deam]

Quick guess: identity matrix shows up

[@Numbabu]

Yeah if I had to guess I would say there is probably some math

[@gon8330]

@@Numbabu don’t Jump to quick conclusions here

[@JO-fs1on]

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.

[@pietergeerkens6324]

Yes; this is an important, even if subtle, point.

[@jonasgajdosikas1125]

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

[@philstubblefield]

I was just about to type the same comment! I'm happy that he's advertising his other channels more actively, but the old, abrupt cut is nostalgic!

[@trueriver1950]

Me too. Logically it points out that "good place to stop" may not mean the best place to stop

[@pseudorandomly]

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.

[@lexinwonderland5741]

Same, haha, I love Dr. Penn's videos regardless but there was something wonderful about that little trick. Still, I love his content and the quality has been consistently wonderful overall!

[@victorcossio]

And the back flip

[@lexinwonderland5741]

@@victorcossio he did have an ab shot tho

[@goodplacetostop2973]

14:29

[@johns.8246]

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?

[@TheEternalVortex42]

In general there are infinity many solutions. For example consider that if A^5 = I then we can take any PQ s.t. PQ = QP = I and then (PAQ)^5 = I as well.

[@jceepf]

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!!!!!!!!!!

[@lexinwonderland5741]

WHOA that's a great connection!!

[@NotoriousSRG]

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

[@jimskea224]

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.

[@cmilkau]

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

[@matanah1989]

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?

[@jonathandawson3091]

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.

[@TVWJ]

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.

[@edwardlulofs444]

I've never done that exercise, thanks.

[@SuperYoonHo]

Thanks:)!

[@thomashoffmann8857]

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.

[@AlcyonEldara]

Without anything else, yes. But: 1) We have a diagonalizable matrix (the two eigenvalues are different) 2) A power of a diagonalizable matrix is a diagonalizable matrix 3) The only diagonalizable matrix with all eigenvalues equal to 1 is the identity matrix

[@thomashoffmann8857]

@@AlcyonEldara I agree in this case with the diagonal matrix solution. The other solutions have b0 and are not diagonal.

[@1089S]

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

[@eliavrad2845]

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.

[@SQRTime]

Hi Eliav. If you are into problems from math competitions, our channel can be something you enjoy. Hope it helps. One sample video Solving a nice trigonometry problem kzbin.info/www/bejne/oIbFkHyPeKiJpqM

[@wjrasmussen666]

Love mathmajor!

[@the11060]

Guess roots of unity

[@henrikr8183]

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

[@tolkienfan1972]

Normally we arrange eigenvalues in order of magnitude

[@tolkienfan1972]

Also, is Fibonacci around here somewhere?

[@krisbrandenberger544]

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

[@trueriver1950]

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

[@gibbogle]

Nice!

[@SQRTime]

Hi Gib. If you are into problems from math competitions, our channel can be something you enjoy. Hope it helps. One sample video Solving a nice trigonometry problem kzbin.info/www/bejne/oIbFkHyPeKiJpqM

[@mathadventuress]

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

[@lexinwonderland5741]

You can have a little mathmajor, as a treat 😉❣️

[@mathadventuress]

@@lexinwonderland5741 🥺🥺

[@hexeddecimals]

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

[@fix5072]

This might be a bit too easy. We know 2n=(n+m)/2, therefore m=3n

[@hexeddecimals]

@@fix5072 the average of two lines does not produce an angle bisector of those two lines. For example, the average of 3x and 1x is 2x, but the angle of 2x (63.43°) is not the average of the angles of 3x and 1x (85.28°) I used degrees just to show the disparity between the angles.

[@fix5072]

@@hexeddecimals you're right

[@fix5072]

@@hexeddecimals maybe like this: the angle of a line is arctan(m) where y=mx. Then arctan(2n)-arctan(n)=arctan(m)-arctan(2n) -> 2arctan(2n)-arctan(n)=arctan(m) Now apply tan to both sides and use tan properties to get m=4n^3+3n

[@hexeddecimals]

@@fix5072 "use tan properties" is doing a lot of work there, but yes that's correct!

[@vinvic1578]

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

[@jceepf]

In accelerator physics, we have the general solution. There is an infinite number of solutions in general. Essentially it correspounds to motion on an ellipse with a rotation angle of 2pi/5 and multiples. But the shape of the ellipse gives you a continous family of solution. By setting some coefficients here to 1, you force a discrete # of solutions. You force the ellipse.

[@paradoxica424]

even for square root of identity matrix (2 by 2), there are uncountably many solutions

[@jceepf]

@@paradoxica424 yes

[@jceepf]

See general answer above in the main thread.

[@get2113]

@@jceepf nice answer

[@egillandersson1780]

Great !

[@cmilkau]

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

[@noumanegaou3227]

Please probability theory cours

[@ojas3464]

👍

[@lorenzovittori7853]

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.

[@ConManAU]

He calculated the values of a and b explicitly, and they were 1 and 0 respectively which automatically results in the identity matrix.

[@jesusmauro3114]

Hi, I´m not sure about what you refer to, but I think it is because the statement of the problem, A=[ [ a, b ] , [ -b, 1 ] ] .

[@lorenzovittori7853]

I think my logic is wrong but I don't understand how. Maybe I can express myself with this question: how the exercise would change if, instead of the identity, there was the matrix [[1 1] [0 1]] ?

[@iang0th]

@@lorenzovittori7853 That's not allowed by the statement of the problem, since you would have both b=1 and -b=0.

[@AlcyonEldara]

Without anything else, yes. But: 1) We have a diagonalizable matrix (the two eigenvalues are different) 2) A power of a diagonalizable matrix is a diagonalizable matrix 3) The only diagonalizable matrix with all eigenvalues equal to 1 is the identity matrix

[@vladdadofganja254]

Eigenvector? German maths terminology? For real?

[@schweinmachtbree1013]

the " *Z* " for the set of integers is from the german "Zahlen" meaning "numbers", and I'm sure there are lots of other examples too :)

[@vladdadofganja254]

@@schweinmachtbree1013 quite a surprise for me)

[@henrikr8183]

@@schweinmachtbree1013 Ansatz is used in English too, albeit rarely

[@jceepf]

"proper vector" is also used but it is more British I think.

[@jimskea224]

@@jceepf I'm British and we always used eigenvector, eigenvalue and eigenfunction