16:12:45 Okay.
16:12:48 I don't know, we'll see what happens.
16:12:52 Okay.
16:13:04 Okay, it looks like everybody's here.
16:13:06 And
16:13:09 it's pretty good.
16:13:11 So,
16:13:15 okay, I want to continue and clean up what we did last time,
16:13:23 I'm any general questions before I begin,
16:13:31 okay. That's okay.
16:13:33 So I'm going to share the screen and
16:13:41 get started.
16:13:44 Okay.
16:14:00 Okay, so here we go.
16:14:04 So this is the file from Wednesday.
16:14:09 And so we're looking at this paper here for a moment.
16:14:22 And so we have this proposition which is fine. It just
16:14:22 tells you exactly when your evolution algebra has a unit.
16:14:27 And in that case, It's a pretty trivial case so that's called a trivial.
16:14:34 My evolution algebra in the sense that the
16:14:39 structure matrix is a diagonal matrix as lots of zeros.
16:14:44 So, okay, we have one more.
16:14:54 So, we don't need to review this.
16:15:04 Also, this.
16:15:07 The second paper that we looked at
16:15:11 was simply to get us, used to the idea of a spectrum.
16:15:18 And so it's not relevant to our.
16:15:22 What it says is the spectrum which is the eigenvalues of A matrix is something that's well known and understood.
16:15:34 And this new thing the multiple collective spectrum is a generalization of it, meaning.
16:15:43 This this is for any algebra is not necessarily associative.
16:15:50 But if the algebra is associated.
16:15:52 These two concepts coincide. So that was the point of this remark it's just an interesting remark.
16:16:04 So, now, this is where we ran into trouble last time because I wasn't careful.
16:16:12 So
16:16:15 we want to prove this formula three.
16:16:17 So, this is a revised file so the revisions are in reading.
16:16:25 Most of them anyway.
16:16:30 So if you have an algebra, and we're going to assume it does not have a unit.
16:16:35 And this is not necessarily an evolution algebra, but if it is an evolutionary algebra or even if it's just commutative, and not necessarily evolution algebra then LA and rar.
16:16:57 Of course the same, so we can we can simplify things by not worrying about our a.
16:16:59 Okay.
16:17:01 So, last time I had the misprint here which I've now corrected. So a one is the units ization of a was defined back here and it's pretty important.
16:17:17 So it's. Notice that the dimension of a one is one more than the dimension of a, of course if dimension of A is infinity infinity plus one is infinity but dimension of A is going to be finite and so dimension of a one is kind of beef, whatever it is,
16:17:35 plus one.
16:17:38 Okay, so we'll we'll use. So let's go back here and.
16:17:47 So, notice that we have.
16:17:50 Elevate here, linear transformations, on the algebra. A considered as a linear space.
16:17:58 And here we have elevate one and.
16:18:05 Okay, so here we have l Elsa Bay.
16:18:10 And here I should really write L sub A comma zero, which I do and the rest of this notes but I forgot to do it here.
16:18:19 So you, you'll see, because here a is considered is identified with a comma zero in a one. Okay.
16:18:31 So, in fact I started doing it here so.
16:18:35 So, the proof of this and, again, we're going to prove one and two and then use one on to to prove three which is the main form
16:18:46 was, I was confused last time so I want to go over it again, and make sure it's very clear and making these changes and also.
16:18:55 So step one
16:18:59 is you have an element of a in a in a nonzero complex number.
16:19:04 Or it could be real number for the reels but we got that.
16:19:10 So, la minus. Elsa A minus lambda times the identities injected if and only if now.
16:19:20 This is acting on a one so I'll call the identity of a one identity operator on the one call it by someone to just for emphasis and La really is hell of a comma zero.
16:19:40 Okay, so I think that makes it a little bit clearer.
16:19:42 In case your mind wanders or something.
16:19:45 So, anyway, we don't need to walk through the proof again but I did make a few changes in notation.
16:19:56 Some of these red marks are from before but most of them are are just clarifications this point, I did notice that this one says, if and only if, and I only have we only prove one direction.
16:20:13 And so, I included a proof of the other direction here.
16:20:18 This is Page Six. And so that's going to be on page six prime which I inserted here, and it's it's a very straightforward booth it's not necessary to go through it.
16:20:31 But if you you assume that this this fellow is
16:20:40 injected.
16:20:42 You want to prove that this is injected man so it's pretty straightforward. It's almost obvious, obvious you need to work that out.
16:20:51 So,
16:20:57 now, this is just a remark that I wanted to make.
16:21:02 We're assuming that he has no unit.
16:21:05 But even if a has a unit.
16:21:08 Let's call it, he can still consider a one.
16:21:13 And remember the unit of a one, which contains a is 01 which is not not in a Of course, it's in a one.
16:21:25 And so I just did the calculation here to show you that the unit in a is definitely not a unit of a one.
16:21:36 So if I take the unit of a which is this.
16:21:40 And I multiply it by an arbitrary element of a one.
16:21:45 Then, this is what you get.
16:21:47 And that's not equal to a land and so that's kind of obvious, but so you, it's not the case that you have two units, because this original unit of a.
16:21:59 There's not a unit available. so I just want to make them.
16:22:03 So let's go back here and see where we stand, so.
16:22:12 Okay.
16:22:12 Now step two, was
16:22:17 the equivalence.
16:22:20 La minus lambda and La zero minus number one both things are objective.
16:22:31 And we, I wrote out the proof here, but it's not really necessary if A is finite dimensional.
16:22:40 because a one would also be finite dimensional and we already proved that it injected is the same for both and so they're finite dimensions objective is the same as surgery, hope that's clear.
16:22:55 So that's number two.
16:23:01 And
16:23:01 so, if you want to read that proof again at this point here we go to page seven.
16:23:21 So that's Miranda that book so step three is kind of interesting. So, I see.
16:23:21 I should have written this L survey here is a sub zero, because we have a one here so I
16:23:30 should have should have written a sub zero there as I'm doing and the proof here, but. So this says that if you look at the spectrum of health some a sub zero, the, the eigenvalues, which are not zero, then they're the same as the eigenvalues of LA.
16:23:52 My turn on, not too.
16:23:56 So, I hope you understand this means set theoretic difference.
16:24:00 Just taking away zero, it doesn't mean that zero belongs here.
16:24:06 But if it does, then you take it away.
16:24:09 And if it doesn't, then there's no affected by this notation. Okay.
16:24:19 Well, this falls neatly from either one of the previous two statements because, you know, one of these is injected with the knowledge, the others impacted.
16:24:31 And so, so that's just basically restatement of one and two.
16:24:38 And then number four is, again, this should have been our surveys, a sub zero over here are today's ok because it's just a not a one.
16:24:51 But since, since since commutative.
16:24:56 For us, but even if it's not committed in the proof would go exactly the same way. Just change a notation,
16:25:06 putting an element on the other side.
16:25:09 So here here's the main point.
16:25:13 So, and this is where I screwed up last time.
16:25:18 And so it's a little bit messy here because I put in a lot of red, red marks.
16:25:23 But we have an element, a so what we're trying to prove is part three of this thing, way back here.
16:25:36 So, This is the multiplicative spectrum of an element, a, an announcer without a unit.
16:25:44 And it's equal to.
16:25:49 The eigenvalues of LA and the ordinary sense,
16:25:58 but it contains 00.
16:26:05 These, these fellows over here, this may or may not contain zero.
16:26:10 If it doesn't, we have to add it on here to make this formula true if it does contain zero, this union with zero does not harm.
16:26:21 Okay.
16:26:23 All right.
16:26:27 Mike transcript here.
16:26:29 But I think if you want, transparent, you have to save it so we'll take care of that at the end, remind me to say that if you have to do something, we're experimenting here.
16:26:43 Okay.
16:26:46 So, um, So this is a statement you want to prove.
16:26:52 So, this isn't a one here so we're only at the level of a one.
16:26:58 So, by definition, this multiplicative spectrum is and notice I had this was the confusion last time, this should have been written. This is the element that is not him and vertical.
16:27:17 In the algebra elevator one.
16:27:23 So
16:27:27 remember, an element is M and vertical if the left modification on the right location about project.
16:27:34 So this is not lmn vertical so at least one of these things and again, this is an estate. This is our have a zero.
16:27:46 So too bad mistakes here is constant problem last time.
16:27:50 So this is not my objective. Okay, so.
16:27:59 So let's see.
16:28:08 Okay.
16:28:10 We have to remember why we're here.
16:28:18 So, this is statement one in the proposition.
16:28:24 So, You see what it says.
16:28:28 The spectrum if your takeaway zero from each of these spectrums and then the same.
16:28:37 And
16:28:37 so, that means if I, if I throw away zero, it may or may not be in here but throw it away.
16:28:47 Then, this thing is contained in these two fellows.
16:28:54 So that's what one and two says said.
16:29:00 So, if I now.
16:29:04 Throw in zero, then to both sides.
16:29:10 It may not be necessary because it might already be there.
16:29:14 Well actually I took it away if it was there. So if I Fayyad zero here and I add zero here I still have the same input. So that's what I want to say here.
16:29:25 The spectrum multiplicative spectrum of an element, they
16:29:32 just contained here. Okay.
16:29:35 This is a little bit subtle.
16:29:38 Now, conversely, if you look at something in the right hand side, and say here,
16:29:48 then that means that this is not by objective.
16:29:54 Lambda is not zero,
16:29:57 not projecting.
16:30:04 And
16:30:04 so, it's not by objective.
16:30:10 So it's contained in this thing. So, because not by jacket.
16:30:16 And it's not.
16:30:18 And the element day.
16:30:21 Rather this this element is not M and vertical
16:30:30 will make another mistake.
16:30:33 No, the algebra here is elevate one. So these are elements in the algebra.
16:30:38 So,
16:30:42 my goodness.
16:30:47 Simon vertical means multiplication by this normal skin terrible.
16:30:54 Okay.
16:30:56 Um, may have to revise this again.
16:30:59 Anyway. Okay.
16:31:03 Let's just continue and not trying to fix it Eric short of time.
16:31:10 Um, so what this shows.
16:31:16 So,
16:31:22 yes.
16:31:29 Sorry.
16:31:39 can't be too well prepared Can you.
16:31:49 Okay, I have to revisit that. So, but this is this is true I haven't explained it well. But, and it's kind of messy to read but it says the multiplicative spectrum delay in the algebra, a salon
16:32:08 course that should be a set a comma zero
16:32:14 is given by the ordinary spectrums here.
16:32:19 So the final statement which is the third statement in the theorem is that the spectrum is equal to these two without the a one, this is the name.
16:32:32 So this is a one this is a so this kind of subtle.
16:32:37 And I'm not explaining the wall.
16:32:41 So, but life goes on. OK, so the spectrum, the multiplicative spectrum, by definition, its multiplicative spectrum and the unit ization that's the definition.
16:32:55 And that was shown. And number five, make a mistake
16:33:08 to be the union of these two sets.
16:33:18 Notice that this isn't a one, and this isn't a one and we want need to put in a there, and the price you have to pay is that you have to add zero, so that that is the point of this thing.
16:33:31 So, we now we do know that this with a one is true.
16:33:37 Number five.
16:33:39 So, that means.
16:33:43 Well, just set that aside for a minute just consider an element, a little a and then consider a zero, then if you map a one to a one.
16:34:00 It's not subjective, because if you have a zero and then an arbitrary on the day one.
16:34:08 You get zero over here, so it's it's actually back in a, and so it's not subjective.
16:34:17 And so what that means is if you look at, hell of a sub zero is a comma zero, and then minus zero.
16:34:28 I'm now introducing zero as an Eigen value,
16:34:33 then these are not by objective because not subjective.
16:34:37 And so that means that zero is in the spectrum.
16:34:43 Those la zero minus zero times the identity is not by jack.
16:34:51 Okay.
16:34:57 So,
16:35:01 notice that we have an A one here and that was shown to be if I can fix that mistake if there's no one upstairs.
16:35:09 That's Sean so this is getting very confusing but there's one more step here.
16:35:16 Mainly, I just recalled for you. if you can read this.
16:35:22 What wanting to say because I'm they use them right away.
16:35:30 So, the spectrum multiplicative spectrum of the bay and the low and the ALS ray is given my this extended thing, a one.
16:35:37 That's fine number five of them.
16:35:40 But my woman to can't, can't see these right now.
16:35:46 Let's try something.
16:35:53 Is that better.
16:35:58 Okay.
16:36:01 So, but now, if, if I take away, zero from this and from this, and we can ignore this, because we're commutative, but believe it in any way, if I take the way zero from each one, even if it's not there.
16:36:17 And then I add a zero and then I haven't done anything.
16:36:22 And but when I take away zero from this guy.
16:36:25 I can replace the one by A.
16:36:29 And then.
16:36:33 Same here.
16:36:33 And now, if I now want to.
16:36:36 So, because I have a zero here.
16:36:41 I don't need to take it away from these fellows, and unfortunately scam got off the bottom of this page.
16:36:51 But what it says is this formula six. Okay, so that was a struggle.
16:36:58 but
16:37:04 I have to go back and think about this.
16:37:07 And maybe I should write this up a little cleaner next time.
16:37:12 Just
16:37:14 because it is a bit confusing.
16:37:19 Anyway. Okay, so I consider that done.
16:37:25 And
16:37:29 I'm afraid to ask if there's any questions because it's such a botched performance but that's okay.
16:37:36 I think you get the idea. Let's go to the next file, we've already done some of this.
16:37:41 We did term six.
16:37:44 Well, not really. We just did the first part.
16:37:48 So here you have a nice well in part one, you have a nice month doesn't have two hours. One to one on two by Jackson, which is
16:38:02 nice and Marxism all the structure.
16:38:07 And the second part.
16:38:09 I'm going to skip. Even though, as I said last time it's kind of interesting.
16:38:15 But I wrote out the proof.
16:38:18 But because we're going to be finite dimensional it's something is subjective then it's, it's injected, I mean it's also by objective.
16:38:27 And so we're actually an ISO Morpheus and so the quality of moments here.
16:38:33 So this this paper was designed for arbitrary Alger was
16:38:41 not necessarily finance dimensional.
16:38:44 But for us, let's keep it simple.
16:38:46 So I talked about this group a little bit last time because it's kind of interesting as diagrams and some points but let's move on.
16:38:59 And we'll see where we are.
16:39:02 So now
16:39:08 I'm going back to this paper which we've already
16:39:13 done proposition 2.2 there's two proposition, 2.2, but they're different papers.
16:39:23 But anyway.
16:39:25 Um, okay. This one is going to be satisfied with, with just this right here.
16:39:32 So,
16:39:35 so proposition by one is about a trivial evolutionary run so that means that the structure matrix is.
16:39:48 Let's see,
16:39:54 a nonzero truly so it has a unit
16:39:58 and.
16:40:04 Okay.
16:40:10 So, um,
16:40:20 okay so let's, let's look at the structure matrix for this, this is now an evolution of finite dimensional.
16:40:23 And if you have an element, we ride coordinates with respect to the basis.
16:40:29 And this is the support this element in other words if some of these alpha 00, they don't contribute anything so we can define this to be the indices, where alibis not zero.
16:40:47 Um, okay then the multiplicative spectrum is given by this, this formula here.
16:40:55 So it's actually just the diagonal entries of this times the office.
16:41:02 So some of them could be zero.
16:41:06 Some of some of these alphas could be zero, but so if I take zero away.
16:41:12 Then I can write it this way, where I was in the support. Okay, that's not so important.
16:41:19 So that tells you what the multiplicative spectrum is.
16:41:24 So, this next remark is needed in the next proposition so.
16:41:29 And here's the proof of Proposition one, I want to give you an overview first one page here.
16:41:35 That's probably what we'll do in a minute.
16:41:39 And then now proposition 5.3 in this paper is finite dimensional non trivial evolution algebra.
16:41:51 And so,
16:41:54 so, in fact, a trivial one has a diagonal structure matrix.
16:42:03 But, non trivial one which is has a full matrix here.
16:42:10 And this tells you what the multiplicative the spectrum means.
16:42:18 And so that's just an overview. I need to go over, line by line and the way.
16:42:25 But before I do,
16:42:30 I'm probably not going to do both maybe just do one
16:42:34 need to see
16:42:45 need to just look at this file from last time, and just go to the very end.
16:42:53 And so here's this is a summary.
16:42:57 This paper.
16:43:00 This paper, it's a summary of this paper parts of it.
16:43:04 This is, of course, is the paper that we're, which is the main paper what we're working on.
16:43:10 But this summary, so theorem six, we just did.
16:43:16 If you have an ISO monotheism multiple color spectrum is preserved.
16:43:23 And we didn't do part two. So, the point of these propositions five one and five three, is that it tells you what the multiplicative spectrum is
16:43:37 tells you the calculated as ordinary eigenvalues of certain things.
16:43:43 So, if I just, if you just prove propositions 5153.
16:43:53 That'll. That'll be the proof of this
16:43:58 with timer and insured, I will do want to walk through
16:44:03 one of them.
16:44:06 Mainly than the first one.
16:44:09 So, going back to this. So we have a, we have a trivial evolution mountable so that means that the
16:44:24 the square, there's a natural basis the squares are just multiples of themselves.
16:44:31 Multiple square in the eye is a multiple of the.
16:44:35 And so the structure matrix is diagonal so it's not indicated here but anyway.
16:44:44 Okay, keep that in mind, this remark is needed for proposition by three so I'll skip it for now.
16:44:51 So,
16:44:56 okay, I'm gonna show my ignorance again now I don't remember why this is the case. So, so a as a unit member it's a trivial thing so that's the only one that has the unit.
16:45:11 And so we just need to look at the spectrum, the ordinary eigenvalues.
16:45:18 And so this is just linear algebra eigenvalues of Elsa bang.
16:45:25 So, that except this I forgot the exact reason why. Here we don't need to worry about zero which anything that's not not so important, this this is really what's important.
16:45:36 Look at the matrix, I just call it m of this operator. So they with respect to the basis.
16:45:45 So the way you get that is you multiply, you apply LA to ej.
16:45:54 And it's a time zj.
16:45:56 I should have written this out but, oh, I did. So here's a show to put a brand sincere time zj.
16:46:06 But he items ej zero so I must equal Jay.
16:46:11 And so it's equal to alpha just out the JVJ squared.
16:46:16 And ej squared is.
16:46:25 ej squared away. This is the definition of the structure matrix.
16:46:30 And so, the matrix. So, remember how you, you get the matrix of linear transformation you look at ej. The one.
16:46:42 So the coefficients of LA, he one would be this first column.
16:46:49 And so that that's what you get.
16:46:53 And if you if JS to you get
16:46:59 this column. But actually, this is crazy because this is a trivial nonzero. So this is a diagonal matrix I should have put that so the W two one and you know if i is not equal Ajay, these are zero.
16:47:16 So am is this.
16:47:19 This is the matrix. So, if we want to get the eigenvalues. Well, there they are there on the die and all it's a diagonal matrix so the eigenvalues are exactly on the diagonal.
16:47:32 So, no more precisely, if you take the subtract lambda times identity from this, and you take the determinant, and get that this is.
16:47:47 That's really the proof.
16:47:50 And so in other words, the multiplicative spectrum, which is the eigenvalues of LA. Because of this,
16:48:02 are just these numbers here alpha. wi. So let's go back and look at the statement.
16:48:12 So, the multiplicative spectrum is just awful Hi, can why, because I should have said here that this is already a diagonal structure matrix, that's a little sloppy for me not to say that.
16:48:31 Okay, I'm going to stop here.
16:48:35 So, next time, I'm going to take another look at the previous file, because I got a little confused there.
16:48:45 And then we can
16:48:51 do proposition 5.3 but I don't want to start now it's, it's, it's basically the same idea.
16:48:58 But we can, we can save that or just leave it as an as an exercise.
16:49:06 There's an exercise. Exercise but I'll probably mention it Next time,
16:49:13 so.
16:49:20 So I just want to close by showing you where this is coming from. So this is the paper we're in.
16:49:31 Remember we have these terms here from the previous paper by Roberts and brown.
16:49:39 And so here's the definition of the multiplicative spectrum.
16:49:47 With a unit.
16:49:50 And then without a marijuana.
16:49:54 So, here, here is them six.
16:49:59 Wow.
16:49:59 So
16:50:02 that's a kind of a very wordy proof here and there's some notation here which is not necessary because it's.
16:50:11 This is an infinite dimensional case so there is a point spectrum and subjective spectrum, but we don't need to worry about that, because we're finite dimensional and here's theorem seven.
16:50:23 This is the last thing I just did.
16:50:27 So you can you can have a look at this if you want to.
16:50:31 And.
16:50:33 Okay. So, that is going to be it for now.
16:50:37 Next time I'm not sure, while we'll try to go a little further, but it's starting to get into the graphs now.
16:50:45 And so, that's the next step.
16:50:52 Okay.
16:50:56 So that's it for today I'm going to stop sharing.
16:51:01 So now what do we do about the transcripts I'm just experimenting here.
16:51:08 Do I need to do something to save the transcript.
16:51:13 Let me see.
16:51:28 No.
16:51:28 Oops.
16:51:24 So bear with me for a minute. How did I do the scans.
16:51:38 I don't remember how I started the transcripts.
16:51:50 Oh, here live transcript. Okay.
16:51:54 Ah.
16:52:02 See, live chance to say bone.
16:52:04 Sign someone to type needs.
16:52:05 I remember reading that I need to save it.
16:52:08 It's not automatic but maybe it is because I don't see any place where I can do that.
16:52:17 So, I'll stop recording.
16:52:22 Okay.
16:52:34 I'm
16:52:34 okay. I'm done.
16:52:37 Either we have the transcript or not but it's not really necessary but it might be helpful, who knows.
16:52:43 Anyway, it's fun to try to figure out this technology.
16:52:49 Anyway, so I'm going to end now, and I promised you that I would send an email about them asking you for your update on the project.
16:52:59 I'll do that.
16:53:00 I'll try and do that.
16:53:02 After we're done.
16:53:07 Okay. Any, any anything.
16:53:16 Nathan any famous last words.
16:53:16 Professor.
16:53:17 Okay, so.
16:53:23 Yeah.