Matrix exponentials, determinants, and Lie algebras.

[@CreativeMathProblems]

Infact, proving it for diagonalizable matrices is enough. This is because the determinant is a polynomial (and hence continuous) and trace is a continuous functional, and that diagonalizable matrices are dense in the space of matrices

[@MichaelPennMath]

The idea is that you can write a non-diagonalizable matrix as the limit of a sequence of diagonalizable matrices. Then you apply the special case of the result to each term of the sequence. Finally continuity guarantees the result for the limit of the sequence.

[@MichaelPennMath]

@Javier Concepción Good question -- I was thinking of matching the space of nxn matrices with R^(n^2) with the usual Euclidean metric. That being said, other settings might also be ok.

[@thanhtantran3593]

@Javier Concepción We can consider the space of matrixes as a normed vector space. Because its dimension is finite, all of its norms are equivalent. So you can take any norm and use the metric induced by it. (Sorry if I make any mistake, I’m not used to doing maths in English!)

[@a5110123]

@Javier Concepción You can use operator norm, R^(n^2) p-norm, or any norm you can imagine. For matrix space, they all induce the same topology. Typically, we use operator norm for theory, and R^(n^2) 2-norm for computation. The latter one is just the square root of sum of square of each entry, which may also view as a norm induced by the inner product =A^T*B (with a fancy name : Frobenius inner product).

[@josephmathmusic]

Another way is to trigonalise on C which avoids to use density arguments.

[@Risu0chan]

The corresponding operation to the Lie group multiplication in the Lie algebra is not the Lie bracket but the addition. Indeed, we have exp(x) · exp(y) = exp(x + y) IF they commute (that is [x,y] = xy-yx = 0) .

[@howwitty]

Awesome... thanks for sharing. When I took ODE, my professor introduced me to matrix exponential and I was blown away. Sometimes I wish I could go back to that time.

[@kvazau8444]

You can. Just read Fulton & Harris' Representation Theory.

[@howwitty]

@@kvazau8444 Thanks. Copying and pasting stuff in Excel seems to pay better, but I'll look into it.

[@Boringpenguin]

Yesss a live stream on Lie algebras, I'm already hyped for that!!!

[@scottmiller2591]

13:24 Element (1,2) should be -e^3+e^6, not e^3-e^6 - signs are swapped, but corrected when the board is erased and updated a few seconds later.

[@austinmoore8041]

The fundamental group of SL(n, C) is trivial, also it's not compact. Hence, drawing it as a genus 2 torus was kind of a lie, but I do like the intuition it sparks.

[@DavidSavinainen]

To find the determinant, you don’t need to expand the multiplication: det(P M P⁻¹) = det(P) det(M) det(P⁻¹) = = det(P) det(M) det(P)⁻¹ = det(M) for all matrices M and all invertible matrices P. Edit: Maybe I shouldn’t comment before finishing the video. Now I notice that you used this exact trick for the general part. Still, it would’ve saved some effort on the example too.

[@demenion3521]

was about to write about the same :D

[@janouglaeser8049]

Was also going to write the same hahah

[@telnobynoyator_6183]

I also thought of that

[@NicholasPellegrino]

I think everyone was thinking the same 😛

[@evankalis]

I thought i knew a good amount about math when i finished calc 1 senior year. I was a newborn then. Math is so beautiful and interconnected. Thank you for sharing this with me. Ill be excited to watch the livestream :)

[@apteropith]

1:00 ah, yes, this determinant/trace exponentiation relation makes some sense, considering that in a matrix's eigenbasis the determinant is the _product_ of the diagonal entries (which are its eigenvalues, and in that basis should be its _only_ entries)

[@Handelsbilanzdefizit]

This was a good refresher about diagonalization. Thank you.

[@EntropicalNature]

Incredible video! Thanks a lot for this! @12:53 you made a small mistake in the the upper right entry (a factor of -1) This happens to me all the time when I teach. Glad to see you're human too and glad of course that it was corrected on the next blackboard.

[@pecan4434]

That last part about Lie groups and Lie algebras was really illuminating. I can't wait for the livestream!

[@MrDeath537]

I took Linear Algebra 5 years ago, I didn't study any further math (not at least for algebra) and I could follow you through the entire video, it is not only a beautiful property and a great explanation, but also brought me nice memories. The connection to Lie groups and algebras got me intrigued, I will read about it! Thanks for the amazing work, keep it up!

[@farfa2937]

It's been eons since I've diagonalized a matrix, so nice reminder! Looking forward to the stream, I learned some basics of group theory but nothing nearly as advances as Lie algebras.

[@koenth2359]

13:27 the right upper coefficient just changed sign when taking over from the previous board. A tacit correction of a previous error.

[@Celastrous]

Hey Dr. Penn. Thank you so much for these lessons. I've been out of college as an electrical engineer for about 2 years now. When I took linear algebra, I had some personal issues that led to me absorbing little from that class, barely passing. This video is such a good refresher on some of those concepts I've forgotten, like how to find null space.

[@michaelaristidou2605]

In 11:25 no need for all those multiplications. Just use the multiplicative property of the determinant.

[@oleg67664]

There is really beautiful proof of tr(A) = , it goes like follows: = Now let's recall, that the sum of roots of polynomial of n'th power, written in form x^n + a1 * x^(n - 1) + ... equals to -a1 Thus we need to find the first (or the second, depends on your notation) coefficient of the characteristic polynomial of A, or the polynomial det(A - xI) Now we just need to substitute determinant by the definition through permutations, and notice, that only one summand contributes to the coefficient in front of x^(n - 1), and that is the product of diagonal elements, i.e. (x - a_00)(x - a_11)*...*(x - a_nn), and from here it's not hard to see, that the coefficient in front of x^(n - 1) in characteristic polynomial of A equals to -tr(A), so the sum of eigenvalues of A equals to tr(A) P.S. Sorry for mistakes, I'm not really experienced in explaining math in English

[@Czeckie]

similar story explains why the product of all the eigenvalues (any contributing as many times as its multiplicity) is equal to the determinant. Just think about the absolute term of the polynomial.

[@Jack_Callcott_AU]

An observation, e^Tr(A) can never be zero or negative. So does that mean that e^A for any matrix A must always have a positive determinant > 0 , therefore e^A must always have an inverse ( must always be non-singular). Am I right?

[@Alex_Deam]

Surely the inverse of e^A is just e^(-A)

[@Jack_Callcott_AU]

@@Alex_Deam Thanks for the reply, yes , I agree, that would be right.

[@Anytus2007]

@@Jack_Callcott_AU Critically, this works because any matrix A always commutes with itself, so there is no ambiguity between e^(A)e^(-A), e^(-A)e^(A), and e^(A-A). In general, if A =/= B, then it is not necessarily true that e^(A)e^(B) = e^(A+B). If the two matrices (or more generally operators, in the infinite dimensional case) do not commute, then there are additional terms related to the commutator, commutator of the commutator, and so on.

[@Jack_Callcott_AU]

@@Anytus2007 Thanks for the reply.

[@wellingtonbalmant5965]

We use that on Quantum mechanics.

[@goodplacetostop2973]

25:44

[@kappasphere]

11:24 You don't even need to multiply all the matrices, because you can use det(XY)=det(X)det(Y) to figure out that det(PDP^-1)=det(D)=e^3 e^6=e^9

[@DeanCalhoun]

super interesting! great content as always, thanks michael

[@byronwatkins2565]

At 13:00, upper right is e^6-e^3. I suspect the second eigenvector has a sign error.

[@laverami]

This is so beautiful. I found the equation beautiful and fascinating in the beginning, and find it even more fascinating towards the end of the video. In the beginning I thought, although the symbols we use are so suggestive that this must work, it is actually a miracle how the process of exponentiation retains its structure between the realm of matrices and of scalars. Like, when a caterpillar becomes a butterfly it undergoes a phase where its body (including its brain) is a completely liquified stuff of free floating cells, and still there are experiments showing that the butterfly can remember things it learned before. Towards the end of the video I understand better how the exp, taking on different shapes, keeps its role of translating between addition and multiplication, even if these operators belong to different mathematical structures. And I am no less fascinated how this all works out.

[@Chr15T]

great video! a minor correction: as far as i remember, the trace of a product of matrices is only invariant under cyclical permutations of the matrix product inside. at about 19:15, you made a different permutation and said the trace would not change, which is true but only because your matrices were special.

[@lexinwonderland5741]

AWESOME!!! I've been waiting for more Lie algebra content, thanks for posting this!!! Can't wait for the livestream!!

[@lucachiesura5191]

just also for the jordan normal form

[@marcoottina654]

I know little to nothing about this branch of Math, indeed your explanations are so much well done that I was able to follow everything and .. this video is beautiful, it merges tons of knowledge in an armonious way. Lovely Thanks for sharing!

[@josephmartos]

This is what Feynmann used in his Path Integral right???

[@AoG2695]

Linear algebra will always be a favorite of mine. Unfortunately, I haven’t had to use the stuff

[@channalbert]

This is one of the coolest videos I've seen in YT about mathematics. Wow.

[@videolome]

Here is a way of doing it, using L’Hospital. Let A be d x d. Let p(z)=det(z I-A), the characteristic polynomial of A. Then one can argue that det(I-xA)= |I-xA|=x^d p(1/x)= 1- tr(A)x + x^2 g(x) Where g is a polynomial. Using L’Hospital, one gets that |I-xA|^(1/x) -> exp(-tr(A)), as x->0. Replacing A by -A and x by 1/n we find Det((I+1/n A)^n) -> exp(tr(A)) (The determinant is continuos and (I+1/n A)^n -> exp(A), as n-> infty. )

[@bonsairobo]

Wow. I had no idea matrix exponentiation maps from a Lie algebra to the Lie group. Very cool. I need to know what the Lie bracket does!

[@good_brake]

It captures the commutativity (or lack thereof) of the Lie group.

[@aoehler1]

we had to prove this when I was in school... one of the most satisfying math assignments I ever did. Thanks for this video!

[@Walczyk]

There's an error in here! I'll try to find it later. at 13:30 I did an auxillary step to simplify it, I wrote e^A in terms of cosh(3/2) and sinh(3/2) after factoring out an e^(9/2) and 1/3 to get e^A = 1/3 * e^[9/2] * ([3cosh[3/2] - sinh[3/2], sinh[3/2]],[2sinh[3/2], 3cosh[3/2] + sinh[3/2]]), whose trace is quite simple, factoring out a sinh(3/2) makes it easier to see: e^A = 1/3 * Exp[9/2] * Sinh[3/2] * ([3 * Coth[3/2] - 1, 1], [2, 3 * Coth[3/2] + 1]) The determinant here is 1/3 * Exp[9/2] * Sinh[3/2] * (Coth[3/2]^2 - 1) and we can use the identity cothx + 1 = cschx, which gets the det of 1/3 * Exp[9/2] * (csch(3/2) - 2*sinh(3/2))

[@mrminer071166]

Very pleasant to be able to follow the discussion at 3 different levels, one after another.

[@pedrobalodis6617]

Using the Jordan Classification Theorem (in its complex form) it is fairly easy to prove this, since as any matrix is conjugate to its Jordan form, and the determinant of the exponential of both are therefore equal. Lastly, any triangular matrix the formula its obviously true, are therefore for the Jordan form too.

[@alphalunamare]

14:20 ... by messing with the minus sign are you not correcting an earlier mistake made at 12:53 in evaluating the top right entry?

[@StanleyDevastating]

Wow, this is a very, very good video. I was familiar with determinant calculations but that's all I remembered, and it was very clear what was happening all the way through, and made Lie algebra seem like an intelligible concept too. High level math content!

[@humbledb4jesus]

ok, michael...you stating that lie algebras are some of your favorites is reason enough for me to stick through these...

[@kentgauen]

I loved the tie into Lie stuff! Thank you for posting.

[@doctorb9264]

Such an excellent problem and presentation. Your Linear Algebra students are fortunate.

[@LorenzoClemente]

the love for useless calculations obscures the beauty of the mathematical structure hinted at in this video

[@paulsalomon27]

SUPER cool. By far my favorite math channel.

[@jaeimp]

@25:28 This is phenomenal. Is the live stream posted anywhere?

[@noumanegaou3227]

We can use the same idea by triangulsable matrix And every n xn are coefficients in C Is triangulsable

[@baruchspinoza4979]

Wow. Thank you Dr. Penn. And thank you again for your excellent differential forms playlist.

[@jamiewalker329]

I'm wondering if you could do a video on the connection between conserved quantities (Noether's theorem) and Lie algebras...

[@michaelaristidou2605]

At the end, i think you need to prove the converse to in order to have that Lie(SL2C) = gl2C.

[@chrissid.3763]

Your pronunciation of "eigen-" is absolutely perfekt.

[@billconklin5003]

Lie groups and linearalgebra nostalgic memories, lots of fun

[@saptarshibhattacharya3696]

It's for proffesors like you that we students remain inspired. Appreciate your effort a lot!

[@mrminer071166]

Dr. Penn low-key thrilled to have his 9 e^9's lined up after the 1/9!

[@user-wu8yq1rb9t]

Just Great ... Please more of these (Matrices and ...) Thank you so much Professor

[@amadeepl9643]

Wow you have my attention, Sir - thank you

[@pawebielinski4903]

The same argumentation passes for Jordan normal form matrices instead of diagonal ones, and that would do the trick as every matrix has a Jnf.

[@eytansuchard8640]

It will be awesome to see a proof of Baker-Campbell-Hausdorff formula. Thank you for this professional and good lecture.

[@MasterHigure]

During the eigenvalue calculation you swap pretty suddenly from x to λ. Maybe it would be more pedagogical to stick to one of them. Also, at 14:20, instead of pulling the negative sign out, you could pull the 2 out of the other bracket and see that the two brackets are equal, letting you use the binomial theorem. Resulting in a little faster calculation.

[@jpalreis]

Nice video! It reminded me of a trick that was used during a first ODE course: matrix exponentials could be rewritten as matrix polynomials because of Cayley-Hamilton. That e^A from the example (2x2) would have become a_1*A + a_0*I, where these coefficients could be determined using C-H. I’m not sure if it holds for other types of functions, but it probably should work for smooth, analytic functions or something. Maybe that’s an idea for a video! :)

[@angeldude101]

This seems like a matrix form of the power laws of a^(x+y) = (a^x)(a^y), and when x = y, a^2x = (a^x)^2. As was mentioned, the trace is the sum of eigenvalues, and the determinant is the product of eigenvalues, so the trace living in the world of addition, and the determinant lives in the world of multiplication, and the exponential is the bridge between these two worlds.

[@nataliem4434]

This has given me so much insight, thank you!

[@PunmasterSTP]

Matrix exponentials? More like "A lot of potential for learning these videos hold!" Thanks again for recording all these lectures.

[@rodionraskolnikov6989]

Great presentation! I am not following how the SLn group is representing the double Torus (or the surface of some other "object"). Also, how does it relate to the homology groups (that count the holes in the shape)?

[@synaestheziac]

Lie algebra livestream sounds amazing, can’t wait!

[@MissPiggyM976]

Great teacher !

[@CTJ2619]

eigenvectors and eigenvales - i just had a flashback to my linear algebra class!

[@rikschaaf]

13:21 Anyone else noticing the mistake in the top right of the e^A matrix that he changed to the correct value after the next cut?

[@grchauvet]

Matrix exponentials seem quite fun. I wonder if octonion exponentials have any interesting properties.

[@speakadreamagency7518]

Very good ! Thank you so mucht o help me understand the Lie groups & Algebra

[@misterlau5246]

Ooh the big picture! @Michael Penn Have you done any video about Fourier Analysis and FFT? If not, do you have any plans for doing so? Cheers, nice job! 🖖🤓

[@nahblue]

The identity with det(Exp(A)) = Exp(tr(A)) really makes me think.. does this not mean that non-diagonal elements of A are meaningless for that determinant? And how does it work out to be that way, that only diagonal entries matter? Is there any intuition for that?

[@jeffreycloete852]

Hi Prof Penn..thanks for another wonderful exposition..looking forward to the Lie Algebra live stream...

[@rickyng1823]

This is an awesome video. I didn’t learn much about Lie group and Lie Algebra in grad school. Looking forward to your next video, or perhaps a series in the future.

[@Hope16449]

At 11:24 why did you not use that det(A^{-1}) = det(A)^{-1} even though the proof for this identity is short?

[@masonholcombe3327]

Best video yet!

[@esbenagerbo724]

Isn't det(P) = det (P^-1) = 1, as the eigenvectors form an orthogonal basis for R^n. Haven't looked at this for 30+ years

[@senhorkorracha]

Man, all I wanted was something more out of Lie algebra. I've used it to show equilibrium points of an ODE system... but I feel like matrix exponential could really provide me an upper bound for the trajectories of this ODE system as well.

[@brunocaf8656]

Seeing the calculation of the exponential of a matrix makes me wonder, have you ever heard about Functional Calculus? It's a quite interesting theory that allows you to calculate matrix functions much more easily without having to diagonalize the matrix, among several other functions, of course

[@abrahammekonnen]

Yeah I'm definitely interested in the livestream, what day is it?

[@abnereliberganzahernandez6337]

you should do more on lie theory

[@andreben6224]

Can't wait for the Lie algebra stream. Sounds really cool!

[@Bruno08931]

Your content is awesome.

[@nashaut7635]

That's brilliant indeed (no pun intended). BTW does the illustration at the end have something to do with U-V coordinate mapping of 3D objects in 3D modelling applications?

[@_emdoubleu]

Would love to see such livestream about Lie algebras

[@erds6579]

Awesome, thanks!

[@hocinemerah2453]

In case matrix of A not diagonalizable how can we calculate exp(tra(A))

[@brantonsaurus]

Fantastic! This really ties together so many loose ends for me in a concise and clear-headed fashion. Definitely looking forward to the livestream!

[@abrahammekonnen]

Great video, this is gonna take me some time to process.

[@miguelaphan58]

Great subject !!!

[@gastonsolaril.237]

Amazing subjects. After 10 years of so many maths, and still I've got so much to learn... When are you planning to do your Lie Algebra class? Tell us through an Instagram story!!

[@khaledchatah3425]

A question comes to my mind: How do we know that e^( a matrix ) can be expressed as the power series expansion of e^x for matrices . We know that the power series expansion is true on C but. However i don't know about the set of matrices. Did we define it like that or we have some proof?

[@drdca8263]

I believe that is the definition, yes.

[@tw5718]

I really want to see the livestream. I oftentimes don't see them until after they are done, however. Could you make a couple/few posts about it when it gets closer to time so there is a higher chance I will see it?

[@tw5718]

Oh I just realized this was 9 mo ago. Disregard.

[@mahiainti678]

very cool video! thank you!

[@Czeckie]

so hyped about the lie theory livestream

[@tildarusso]

This is surprisingly nice trick! Now I am wondering is there any application to this feature? Such as in the Machine Learning area?

[@raquelmelgar3531]

Does someone know where is it posible to find the stream about lie algebras?

[@pacificll8762]

Great video !

[@zhuolovesmath7483]

tr(I) ≠ 0, so the identity is not in the tangent space?

[@amiramaz]

Is it true to say that detP*detP^-1=1 so we can easily calculate det(e^A)? For all P?

[@zsomborhajdu2181]

As he mentioned the proof was valid only for diagonalizable matrices. The reason for that is they have a nonsingular P matrix which transform them into diagonal form. So for diagonalizable A matrices there are nonsingular P matrices in every case however they are not unique. We have an identity for determinants which says that det(A * B) = det(A) * det(B). Using this, det(P) * det(P^(-1)) = det(P * P^(-1)) = det(I) = 1. Moreover, det(.) is scalar, so we can use the commutative property of scalar multiplication thus det(P) * det(e^A) * det(P^(-1)) = det(P)* det(P^(-1)) * det(e^A) = det(e^A). Did I answer your question?

[@amiramaz]

@@zsomborhajdu2181 indeed, thank you!