start your day with some linear algebra!

Подписаться 305 тыс.

Просмотров 36 тыс.

50% 1

We look at a nice problem from the IMC.
Suggest a problem: forms.gle/ea7P...
Please Subscribe: www.youtube.co...
Merch: teespring.com/...
Personal Website: www.michael-pen...
Randolph College Math: www.randolphcol...
Randolph College Math and Science on Facebook: / randolph.science
Research Gate profile: www.researchga...
Google Scholar profile: scholar.google...
If you are going to use an ad-blocker, considering using brave and tipping me BAT!
brave.com/sdp793
Buy textbooks here and help me out: amzn.to/31Bj9ye
Buy an amazon gift card and help me out: amzn.to/2PComAf
Books I like:
Sacred Mathematics: Japanese Temple Geometry: amzn.to/2ZIadH9
Electricity and Magnetism for Mathematicians: amzn.to/2H8ePzL
Abstract Algebra:
Judson(online): abstract.ups.edu/
Judson(print): amzn.to/2Xg92wD
Dummit and Foote: amzn.to/2zYOrok
Gallian: amzn.to/2zg4YEo
Artin: amzn.to/2LQ8l7C
Differential Forms:
Bachman: amzn.to/2z9wljH
Number Theory:
Crisman(online): math.gordon.edu...
Strayer: amzn.to/3bXwLah
Andrews: amzn.to/2zWlOZ0
Analysis:
Abbot: amzn.to/3cwYtuF
How to think about Analysis: amzn.to/2AIhwVm
Calculus:
OpenStax(online): openstax.org/s...
OpenStax Vol 1: amzn.to/2zlreN8
OpenStax Vol 2: amzn.to/2TtwoxH
OpenStax Vol 3: amzn.to/3bPJ3Bn
My Filming Equipment:
Camera: amzn.to/3kx2JzE
Lense: amzn.to/2PFxPXA
Audio Recorder: amzn.to/2XLzkaZ
Microphones: amzn.to/3fJED0T
Lights: amzn.to/2XHxRT0
White Chalk: amzn.to/3ipu3Oh
Color Chalk: amzn.to/2XL6eIJ

Опубликовано:

26 сен 2024

Ссылка:

Скачать:

Готовим ссылку...

Добавить в:

Мой плейлист

Посмотреть позже

Комментарии : 93

@johongo 3 года назад

Linear algebra is the most important meal of the day

@johongo 3 года назад

@ཀཱ Don't forget the high nutritional eigenvalues

@mueezadam8438 3 года назад

Lmao

@Maxmuetze 3 года назад

Proof with less theory assumed (just definition of characteristic polynomial chi(t)=det(A-tI), multiplicativity of det, and intermediate value theorem): A^3 = A+I so taking determinants on both sides chi(0)^3=chi(-1) (*) A^3-A=I A(A+I)(A-I)=I so chi(0)*chi(-1)*chi(1)=1 (**) substitute (*) into (**): chi(0)^4*chi(1) = 1 so chi(1)>0 Now assume for contradiction that chi(0)=det(A)

@mcqueen424 3 года назад

Not sure if it’s wildly known that chi(t) is continuous so you can’t go around using IVT without showing that first

@Maxmuetze 3 года назад

@@mcqueen424 it’s a polynomial in t, which is obvious from the fact that it’s a determinant with some t entries. No matter how you attack this question, continuity is essential. If you can’t assume that polynomials are continuous, you can’t answer this question.

@mcqueen424 3 года назад

@@Maxmuetze my bad, you’re right

@aydin5131 3 года назад

Nice and neat proof!

@MathElite 3 года назад

I never took linear algebra in university so this channel is helpful in making me learn this Btw: p(2) is supposed to be 8-2-1=5 but it doesn't change the result

@MathElite 3 года назад

protecting my comment because youtube thinks it is spam

@matheusbianchesi2419 3 года назад

Really like your content man, keep it up

@MathElite 3 года назад

@@matheusbianchesi2419 thank you

@DylanNelsonSA 3 года назад

"Most popularly given in Europe" is a bit of an understatement. As far as I know, it's been held at the same university in Blagoevgrad, Bulgaria for the last 15 years or so.

@cbbuntz 3 года назад

There's a whole set of matrices involving binomial coefficients that are either inverses of themselves or are a root of themselves but I think they are usually even roots.

@ordanguric461 3 года назад

6:46 Ah yes I see he used the big point theorem, very clever!!!

@Leockard 3 года назад

More linear algebra please!

@goodplacetostop2973 3 года назад

HOMEWORK : Suppose S is a set of functions with the property that, if f(x) and g(x) are in S, then (f ◦ g)(x) = f(g(x)) is in S. Given that the functions r(x) = (x√3 + 1)/(-x + √3) and s(x) = 1/x are in S, compute the smallest possible size of S. SOURCE : General Round Tiebreaker from SMT 2020

@mykolalysynskyi8228 3 года назад

Can it be solved using group theory? :)

@Examinx 3 года назад

r(x) reminds me of the tan addition formula, looks like if x=tan(a), then r(x)=1/tan(pi/3-a)=tan(pi/6+a), and s(x)=1/tan(a)=tan(pi/2-a). Then for any a, there are 6 different values from repeatedly applying r (angle shifts from 0 to 5pi/6 before repeating), and at most another 6 different values from applying s after the r (negating a but keeping the same set of angular shifts), so there should be at least 12 functions of the form tan(k pi/6 +- arctan(x)) where k ranges from 0 to 5. Other orders and combinations of applying r and s still give one of those 12 functions.

@goodplacetostop2973 3 года назад

SOLUTION *12* Let the notation f_n(x) denote repeated composition of the same function, so f_1(x) = f(x), f_2(x) = f(f(x)), and so on. Note r_6(x) = x and s_2(x) = x (in particular, 6 and 2 are the smallest integers such that r_n(x) = x, s_n(x) = x). In addition, (s◦r◦s◦r)(x) = x.Then we can compute compositions of functions that avoid reduction of one or more terms in the composition into the identity function. This determines that S must have a minimum size of 12. The functions are id, r, r_2, r_3, r_4, r_5, s, s◦r, s◦r_2, s◦r_3, s◦r_4, s◦r_5. Alternatively, we can consider the functions geometrically as symmetries of a regular hexagon with labeled vertices. If r represents clockwise rotation of 60 degrees about the center and s represents reflection of the hexagon about a fixed line of symmetry (so that applying r six times, s twice, or r, then s, then r, then s, gives us back the vertices in their original orientation), it is easy to see that there are 12 distinct orientations of the labeled hexagon, which correspond to the minimum 12 functions in S.

@goodplacetostop2973 3 года назад

@@mykolalysynskyi8228 Can be solved with anything you want, as long as the math is correct ;)

@leandrocarg 3 года назад

This set is a subset of Möbius transformations (assuming these are complex functions) since Möbius transformations form a group under composition then it remains to check cyclic and commutativity relations of the two provided functions. If they commute and each generate n and m functions respectively by taking composition with themselves, then S has nm elements.

@purplerpenguin 3 месяца назад

Nice to see how a real mathematician actually goes about solving problems. You don't get that on most RU-vid math channels.

@noumanegaou3227 3 года назад

Marathon of linear algebra is good idea

@MichaelPennMath 3 года назад

I like that idea! Maybe at the end of the Fall semester... I think only one or two marathons per year is reasonable

@noumanegaou3227 3 года назад

@@MichaelPennMath thanks you professor for this marathon. Your free VIDEO has really helped me to understand math. In Morocco, there are no universities to study mathematics.

@daniello4038 2 года назад

det(A^3)=det(A+I) implies det(A) and det(A+I) are of the same sign. det(A^3-A)=1 implies det(A)det(A-I)det(A+I)=1 implies det(A-I) is +ve. As there is only a real root between 2 and 3, det(A) ought to be +ve, otherwise there is another root between 0 and 1.

@rbnn 3 года назад

Any real root r of the characteristic polynomial P must be positive, since r^3 = r+1. (If r were negative, r^3 is too, so r

@hakerfamily 3 года назад

Excellent!

@daniellosh8341 Год назад

@@hakerfamily Right, all real roots are positive.

@ikramefa2019 3 года назад

Hi We can say There is a polynome P with single roots and P (A) =0 Then A is diagonalisable in Mn((C) The roots of P are r witch is >0 and a and a conjugate Or A is real. So det is real and it is the product of eigan values r^p* (a)^m*(a conjugate)^m is >0

@goodplacetostop2973 3 года назад

12:50 How do you feel today, Michael? Tired?

@roberttelarket4934 3 года назад

Good Place To...: Michael is never tired he is the Energizer Bunny! Note he just finished the marathon on Sunday and still we're getting videos right after. That's your PROOF.

@MichaelPennMath 3 года назад

not to spoil the magic.... but this was recorded before the marathon... Also, I am recording several per day now to make a backlog. I have a 5 week climbing trip starting on july 9!!

@goodplacetostop2973 3 года назад

@@MichaelPennMath Nice, I hope you will enjoy it! I've also a trip planned, it's gonna be 3-4 weeks starting mid-Auguest.

@roberttelarket4934 3 года назад

@@MichaelPennMath: Where do you find time teaching at Randolph, students there, your own research and making videos one after another? You only have 24 hours/day.

@goodplacetostop2973 3 года назад

@@roberttelarket4934 Noah, one of his students, helps him for the editing.

@koenth2359 Год назад

I like the fact that the minimal polynomial is denoted mA(x).

@bilalabbad7954 2 года назад

Good explanation

@ManjotSingh-rq5qo 3 года назад

This is Brillant!

@mikesteele5935 Год назад

Damn nice !

@cicik57 3 года назад

i just want to ask, so you solved that there are only one real single - digit root, why can you extend that on matrix and say that other factors in the polynominal must be complex for any size matrix?

@Calcprof 7 месяцев назад

Cayley- Hamilton! Cayley-Hamilton! Rah Rah Rah!

@kurax9115 3 года назад

Can i somehow type with latex in the google doc fromt the description? (suggestion thingy)

@miro.s 3 года назад

What about using translation operator and laplace or other integral transform?

@NikunjKumarGupta 3 года назад

Sir it's 6:00 pm here in Delhi,India

@ourgoalisto6737 3 года назад

0:28

@harshsinghal5898 3 года назад

😳😳

@JM-us3fr 3 года назад

Hey Michael, could you do a video on how one comes up with a competition math problem? Often times I'm totally mystified as to how one would even create one of these problems.

@miro.s 3 года назад

I came up with these kinds of problems while solving something totally different. It appeared naturally on the path and it can point on something important that is generally not known on theoretical level or what stands behind.

@miro.s 3 года назад

It is part of some mathematical structures.

@krisbrandenberger544 3 года назад

The value of P(2) should be 5.

@emanuellandeholm5657 3 года назад

From A^3 = A + I, we have A^3 - A = I => A(A^2 - I) = I, ie. the inverse of A is A^2 - I. Would it be possible to prove det(A) > 0 from here? We have det(A) det(A^2 - I) = 1.

@inderjeetsingh6136 3 года назад

A is n×n matrix !!

@emanuellandeholm5657 3 года назад

@@inderjeetsingh6136 Indeed. Your point being?

@emanuellandeholm5657 3 года назад

@@dominic2446 I don't understand your third step, ie. A=(A-I)(A^2+A+I). Edit: Wait, all the intermediate powers cancel and you get A=A^3-1. But why is det(A - I) > 0? My thought process was go through all the possible signs (neg, zero, pos) for det(A^2 - 1) (= det(A - I) det(A + I)) and see if they all imply sign(det(A^2 - I)) = sign(det(A)) != 0 for some class of matrices. If so, we're golden.

@emanuellandeholm5657 3 года назад

@@dominic2446 Okay, that's nice. But that means the proposition holds for ALL matrices. I don't buy that. I'm looking for a constructive proof that produces a family of matrices that satisfy A^3 = A + I

@riadsouissi 3 года назад

@@dominic2446 what makes you think det(A^2+A+I)>0 ?

@roberttelarket4934 3 года назад

I'd like to start my day with a "donut"(torus).

@profkrinkels 3 года назад

Small correction: you only know u^2 + v^2 is positive because alpha is a complex number with non-zero imaginary component

@jorgedive5950 3 года назад

actually there's a simpler way to proof that detA > 0 just looking at the coefficients of the characteristic polynomial, since the independent term in a characteristic polynomial is, in fact, the detA. In this case, the characteristic polynomial is -x^3 + x + 1, thus prooved

@CM63_France 3 года назад

Hi, For fun: 1 "and finish it off".

@eiseks3410 2 года назад

And that's a good place to stop

@mohitkrjain9396 3 года назад

A simpler approach would be to take determinant on both sides of the eqn as det(A³)=det(A)+det(I) or, det³(A) = det(A)+1 Put det(A)=y So y³-y-1=0 Intermediate value theorem tells that there is a root between 1 and 2, and it can be easily shown that there is no other real root to the eqn as f(2)>0 and it is increasing beyond Also f(1)

@_P_a_o_l_o_ Год назад

I don’t think your argument holds, since in general det(A+I) ≠ det(A) + det(I)

@manumano3887 3 года назад

Can u make more videos on limits and make a Playlist please ??

@mueezadam8438 3 года назад

Don’t mind if I do

@cinedeautor6642 3 года назад

If you take mod p1, mod p2..and x2, 2^2,..You have a group ciclic. normally 2^^p-1 order...Theorem little Fermat....But with mod 7, mod 17, mod 37...You have two groups...Michael Penn, Why is this?...

@goodplacetostart9099 3 года назад

Good Place To Facts At 1:00

@yoav613 3 года назад

A^3-A=A(A+I)(A-I)=A^4(A-I)=I So by the rule of deteminants we get (detA)^4det(A-I)=1 so det(A-I) is bigger than 0 so detA must be also bigger than 0

@malawigw 3 года назад

how do you know that det(A-I) > 0 ?

@SanketGarg 3 года назад

how did you get A^4(A-I) from A(A+I)(A-I)??

@briemann4124 3 года назад

@@malawigw 1/det(A-I) is equal to det(A)^4 which must be positive. Or you have something to the fourth power multiplied by some unknown quantity that equals 1. The unknown quantity can’t be negative otherwise the product would be negative.

@briemann4124 3 года назад

This is essentially what I thought of but couldn’t explain the final ingredient. How does det(A-I)>0 imply det(A)>0?

@robertgerbicz 3 года назад

Yes, det(A-I)>0 from that equation, but it doesn't follow that that det(A)>0, counter-example: let A=[[1/2,0],[0,-1]], then det(A-I)=1 but det(A)=-1/2.