Birne Binegar: Okay, so.
Birne Binegar: For some logistical things.
Birne Binegar: Today, I wanted to course evaluations so i'm going to only going to lecture for about 20 minutes, then i'll stop i'll pass out the evaluation.
Birne Binegar: And if I could I need one volunteer who's willing to be to collect all the evaluations and take them up to the front office just in front of the elevators on the first floor you just.
Birne Binegar: Have you fill out the evaluation put them in the envelope and then somebody can drop it off, and as for a one now great.
Birne Binegar: So.
Birne Binegar: So what we'll do evaluations today homework set nine is posted on on web assign now I would try to get those done those problems done by Monday.
Birne Binegar: Because effectively next week is dead week and so all.
Birne Binegar: i'm going to finish up with diarization today i'll talk a bit about orthogonal ality.
Birne Binegar: In Friday on Monday and then Wednesday and Friday next week we'll just be preparing for the final exam so we're pretty much through the course except for one last topic, which will be orthogonal and i'll talk about that.
Birne Binegar: In two lectures on Friday and Monday, what I want to do today is briefly review.
Birne Binegar: How we diagnose major theories and then I want to i'll give you an example of that calculation and then I want to give you an application of.
Birne Binegar: Why it's useful to diagnose life matrices and what you will i'll show you today, using basically just linear algebra that we can solve systems of differential equations.
Birne Binegar: Once we can dive lives matrices.
Birne Binegar: Okay, so.
Birne Binegar: Douglas ation is about taking a square matrix and converting it into a diagonal form or the, the only non zero entries occur along that main diagonal.
Birne Binegar: So, in order to do that, but you need is an auxiliary matrix see that when you do this matrix multiplication on the lap a diagonal matrix is produced the topic this week has been studying this situation and what we found is.
Birne Binegar: If we have enough, I can vectors.
Birne Binegar: For the matrix a we can use those eigenvectors to create an inverted matrix and, in fact, what will happen is the.
Birne Binegar: Using the eigenvectors of a using them to create a matrix see we can get an identity like this, so long, and in fact the form of this diagonal matrix is already known once we know.
Birne Binegar: In in vertical or and linearly independent eigenvectors the diagonal entries of this matrix or just going to be the eigenvalues that correspond to the eigenvectors we use for this matrix see.
Birne Binegar: More precisely, a square matrix is dying advisable.
Birne Binegar: If, and only if we have an literally independent eigenvectors that linearly independence means you don't have parallel eigenvectors they have really distinct directions.
Birne Binegar: And then the way we get those matrix cce C and D is first of all, we have to solve diag the eigenvector eigenvalue problem for a so we get a bunch of eigenvectors in their corresponding organized each of these eigenvectors it's going to obey one of these eigenvalue equations.
Birne Binegar: And then, what we do is we simply use the eigenvectors we use those to form the matrix see and we use the eigenvalues to.
Birne Binegar: On the diagonal entries at the diagonal matrix, and these should be correlated, this is an Eigen factor, pay and so it's good and will be an equation like this.
Birne Binegar: Since it corresponds to the first column, we want to use that corresponding, I can tell you in that first slot.
Birne Binegar: The second column is also going to be an Eigen vector will want to use an eigenvalue say Lambda to in that second slot there, so the the columns here and the guys don't cheat here are correlated the equations like that.
Birne Binegar: Now, what we need to die idolizes.
Birne Binegar: an inverted matrix, and so this matrix see if I use the same vector over and over again, the columns would not be linearly independent and the matrix would not be convertible.
Birne Binegar: And so what we need our total of n linearly independent eigenvectors, then the matrix see we construct from them it's called are going to be linearly independent until which it will be inverted and we will be able to diagnose a using such a matrix.
Birne Binegar: Right now.
Birne Binegar: So the the basic condition in order to die lives and matrix a the basic thing you need are enough linearly independent eigenvectors there are several ways, you can detect whether or not you will have enough before you actually solve the eigenvector eigenvalue problems for a.
Birne Binegar: For example, yeah if it.
Birne Binegar: If you have a square matrix and you've calculated as to characteristic polynomial you do this in order to find the eigenvalues.
Birne Binegar: Once you've calculated this determinant you get a polynomial that will factor eyes and these numbers are ones are to up to rk these are going to be exactly your eigenvalues.
Birne Binegar: The number of factors of Lambda minus and eigenvalue that occurs in the characteristic partner meal that we've called the algebraic multiplicity of the of the eigenvalue.
Birne Binegar: And what we showed last time, that is if these eigenvalues are 128 K, are all distinct, meaning the course on a number of times they have.
Birne Binegar: These factors of Lambda minus and eigenvalue occur in the characteristic polynomial if these were all once.
Birne Binegar: Then, in fact, you're going to have an distinct eigenvalues each one is those is going to have a linearly independent eigenvector and so you will have enough.
Birne Binegar: Literally independent eigenvectors the diagonal is a job if each of these eigenvalues only occurs with multiplicity one then we're guaranteed that as diagnosable.
Birne Binegar: Alternatively, another way of checking yes, you have to dig a little bit deeper you're going to have to actually not only find the eigenvalues, but you have to find.
Birne Binegar: bases for the corresponding I against basis if those geometric multiplicity here this dimension, he subgroups our eyes.
Birne Binegar: This is the rei good space it's the space of all vectors in our in such that.
Birne Binegar: The vectors or I can vector a with eigenvalue are I know that's a subspace of our end we calculate this dimension by finding a basis for that subspace and this number Musa grip is exactly the number of linearly independent eigenvectors you have with that particular eigenvalues.
Birne Binegar: Now these multiplicity they're always going to add up the end, but the number of eigenvectors you can get for a given eigenvalue that could be less than the number of factors, you have here.
Birne Binegar: And in fact if any one of these numbers is less than the corresponding algebraic multiplicity, then you will not have enough eigenvectors to diagnose like a.
Birne Binegar: On the other hand, if these geometric multiplicity are all equal to the algebraic multiplicity then these numbers are going to add up the end just like these numbers do, and you will have enough linearly independent eigenvectors.
Birne Binegar: And a third condition that I mentioned just at the end of the lecture on Monday was that if.
Birne Binegar: A matrix is symmetric meaning when you convert it rose in the columns and you just reproduce the original matrix then those major cities are always diagnosable as well.
Birne Binegar: I point this example out because this condition is often very frequently natural in a physical context and so often in a physical problem where you have to die, like a matrix it's going to be guaranteed to be diagnosable because the matrix you're looking at is symmetric.
Birne Binegar: Alright, so that's reviewing what we covered on Monday and Friday last week let's do a quick example where we die guys a matrix.
Birne Binegar: The first thing i'm going to do this i'm going to answer the question of whether or not a is dying reliable before I actually fine to linearly independent eigenvectors in order to die.
Birne Binegar: So we're still looking for matrix see that will convert a to a diagonal matrix, the first thing i'll do is i'll solve the eigenvector i'll find the eigenvalues a so we take the determinant of a minus Lambda times I so called characteristic polynomial.
Birne Binegar: What happened here.
Birne Binegar: The matrix.
Birne Binegar: which was.
Birne Binegar: minus one.
Birne Binegar: To two and I calculate the determinant.
Birne Binegar: Of a minus Lambda times by that's going to be the determinant of five minus Lambda negative one to two minus Lambda.
Birne Binegar: And then, this determinant just the two by two matrix so it's going to determine is going to be the difference of the product along those.
Birne Binegar: The main diagonal in the opposite diagonal until here you have profit diagonal entries, and then the product at the cross that minus the difference of this diagonal product which will be two you'll get.
Birne Binegar: Five minus Lambda.
Birne Binegar: To minus Lambda plus two and then that simplifies.
Birne Binegar: To this quadratic polynomial and you can see sort of the factorization here you're looking for factors of this thing that that add up to this number and.
Birne Binegar: three and four out of the seven and they multiply together to produce 12 so.
Birne Binegar: that's the correct secularization of the characteristic polynomial and having achieved that factorization we can now identify the eigenvalues of a is either be three or four because those are the values of Lambda which that polynomial is equal to zero.
Birne Binegar: Right so we've got two different eigenvalues because we have a two by two matrix.
Birne Binegar: Each one of these eigenvalues is going to give us one linearly and at least one linearly independent eigenvector do I have to linearly independent eigenvectors the matrix will be diagonally usable so I didn't actually have to find the eigenvectors to conclude that is dying all Isobel.
Birne Binegar: Is because I have as many distinct eigenvalues, as I have dimensions of this matrix that's how I know which diagnosable.
Birne Binegar: let's go ahead and dive like that, then.
Birne Binegar: Once we found the eigenvalues.
Birne Binegar: So to diag allies.
Birne Binegar: n by n matrix a.
Birne Binegar: First of all, you find the eigenvalues.
Birne Binegar: And the case at hand for this matrix we found, we had to organize Lambda equal to three and Lambda equal four.
Birne Binegar: And then the next step is to find.
Birne Binegar: bases for the corresponding.
Birne Binegar: space.
Birne Binegar: So this he subscript our I see.
Birne Binegar: Our space.
Birne Binegar: And what that is.
Birne Binegar: it's actually all the vectors.
Birne Binegar: X inside.
Birne Binegar: Or, and so here i'm working with an invite and makes it so it's going to be the vectors in our in such a times X.
Birne Binegar: is equal to our times X is going to consist of all the eigenvectors of matrix a with that given eigenvalue but it's also equal to, and this is how we calculate it it's the no space of a minus r.
Birne Binegar: times the identity matrix and so actually to find these basis vectors for the Eigen space, we have to solve a linear system.
Birne Binegar: And that we do in the usual fashion, we take the coefficient matrix reduce to reduce electronic form, and from that reduce relational on form of the coefficient matrix we try to extract solution actors.
Birne Binegar: let's work it out and detail for this example here, so my first I can value is three so what i'm going to do, then, to find the corresponding eigenvectors i'm going to look at a minus three times the identity matrix and think of that as a coefficient matrix or linear system.
Birne Binegar: So that's that guy when I subtract three from the diagonal elements of a five minus three is to two minus three is negative one, so that explains those injuries, but what I want is the solution set of this matrix times X is equal to zero well what i'll do is all.
Birne Binegar: I didn't quite go to reduce your national form here but.
Birne Binegar: I can hear i've got it in row Echelon form.
Birne Binegar: A minus three I.
Birne Binegar: is equal to.
Birne Binegar: Two minus one.
Birne Binegar: Two minus one and we go all the way to reduce row for what happens is it first of all I can kill off this bottom row by subtracting the first row from it.
Birne Binegar: And then I make the pivot equal to one i'll have one minus one half 00 as the usual Echelon form of this matrix now i've got a coefficient matrix and reduced lunch form, I convert back to equations, and so the equations coming from this matrix are going to be X one.
Birne Binegar: minus one half X is equal to zero, and the second row is telling me to zero, with equal to zero, since this is in row Echelon form.
Birne Binegar: I know that the second call because it does not contain a pivot here's the only pivot net matrix since it does not contain a pivot it's going to X two is going to be a free parameter.
Birne Binegar: So I take the equations and I use them to express the fixed variable corresponding comes with it in terms of free variables homes without tickets, we get X one is equal to one half X two and then, once I know this relationship between the components.
Birne Binegar: Of the solution vector, then I write down X equals well for X one I can use one half X two X two I just use itself to pre parameter, and then I think of this is X two times one half one alright so there's a.
Birne Binegar: bunch of really tiny calculations that go from.
Birne Binegar: This writing down this matrix reducing it to real action line form and then getting the basis factors for the solution set of this matrix times X is equal to zero, and what we find.
Birne Binegar: Is those solution vectors are all proportional to this basis factor one half one that guy there the vector that you generate all the eigenvectors with eigenvalue three for the given matrix that's our eigenvector.
Birne Binegar: What you want often does, though, so we, we can deal with vectors without fractions in them is I can take this back your multiply by two to get a vector with just entered your entries, but this vector generates the same spaces there so it's still just as good a basis factor.
Birne Binegar: So we'll use the vector one to as the basis vector for the Eigen space, corresponding to the eigenvalue landed equal three and that takes care of the analysis of the Lambda equals three I can space, but we have another eigenvector so we'll have to do the same thing.
Birne Binegar: We now subtract the second eigenvalue from a long it's diagonal produces this matrix we reduce this matrix to reduce your actual on form will get this matrix and then.
Birne Binegar: We convert back to equations expanded.
Birne Binegar: Get equations for the components.
Birne Binegar: From that matrix and then express the three parameters in terms of specific express the.
Birne Binegar: variables corresponding columns with it, which we express them in terms of the variables course that don't corresponding two columns without pivots we get these relations we write down a solution vector and then we expand in terms of the free parameter to get our basis Center and so.
Birne Binegar: Similar calculation is repeated here and in this for this audience base, we find that it's generated by a single vector which components one one.
Birne Binegar: By the way, I want to point something out here.
Birne Binegar: You notice in both cases.
Birne Binegar: Whether i'm looking at the three Eigen space.
Birne Binegar: Or the for Eigen space.
Birne Binegar: In this row reductions you start off with the matrix that has lots of non zero entries, but what should always happen if you probably identify the eigenvalue when you reduce.
Birne Binegar: A minus it's eigenvalue types the identity matrix when you reduce that you should always get a zero at the bottom if you're not getting that go back and check your calculation because you've made some kind of numerical error.
Birne Binegar: You always have to have a zero row here, because the eigenvectors that we're looking for have to be non zero vectors if I end if this matrix ended up being reducible to the identity matrix to take a larger systems pose.
Birne Binegar: A minus Lambda times I suppose that reduces.
Birne Binegar: If it reduces to the identity matrix.
Birne Binegar: Then, when you convert back to equations it's going to tell us X one is equal to zero.
Birne Binegar: X two is equal to zero and X is equal to zero, which means you only have the zero vector so you don't want it to reduce to the identity matrix.
Birne Binegar: So how can you prevent this from reducing game well you start off maybe constructing thing, and maybe when you reach this call.
Birne Binegar: Suddenly, you expect to pivot here, but you don't have it there, so maybe you have a column like this, where you.
Birne Binegar: dropping down in the stairway staircase but then you get to a call or you skip over before you reach the next pivot or what's going to happen here, because these are square matrices you kind of hear you jumping down.
Birne Binegar: One step over one step down one step over one step down, but now you're jumping over one step over and two steps down.
Birne Binegar: So what's going to happen as you continue to fill out this diagonal matrix you're going to reach this end before you breach the bottom of the matrix and so, if you had a call with adaptation meaning, you had one three parameter and your real actual on or it would mean.
Birne Binegar: you'd be missing a pivot and it's got to come from one of the rows and when we go in relation to inform we put the zeros at the very bottom of a matrix So what should always happen.
Birne Binegar: If you're trying to find these Eigen spaces and you reduce a minus the Eigen daytime the identity matrix that row reduction should always produces URL at the bottom if it doesn't you've made a mistake and mistake would have been used improperly identified that the eigenvalue.
Birne Binegar: Okay, so so that's how you do it, and we do exactly the same thing is for the eigenvalue for us, we did for the eigenvalue three.
Birne Binegar: We express solutions that a minus four on four times I X equals zero, we see that they're all proportional to this vector so we call that vector the eigenvector up a eigenvalue for.
Birne Binegar: And now that i've got two eigenvalues that are corresponding I get vectors that's all I need to write down those make the matrix see that convert a into a diagonal matrix.
Birne Binegar: I form see like taking the eigenvectors and using them as calls, these are linearly independent vectors so this matrix that I hate that way it's going to be automatically be in vertical and moreover.
Birne Binegar: See, as you can use that now to convert the original matrix into a diagonal matrix that diagonal matrix is going to be exactly this matrix.
Birne Binegar: it's going to be diagonal and it's dying the entries are going to be the eigenvalues corresponding to the corresponding eigenvectors that first column.
Birne Binegar: gets you that, and what I use in this first diagonal entry is the eigenvector corresponding to that the column I use, so the I vector with three but that's why I put three there for the second column it's eigenvalues too, so I use.
Birne Binegar: A sorry, so I can values for so I use for for the second diagonal entry of D once I saw the eigenvector eigenvalue problem for a I can immediately write down the matrix see that convert a into a diagonal matrix and I can simultaneously write down that matrix that dive on matrix D.
Birne Binegar: Okay.
Birne Binegar: So what i'm going to talk about next is an application of this stuff.
Birne Binegar: will solve the differential a system of differential equations like diagonal lighting a matrix.
Birne Binegar: So here's the kind of system i'm interested in.
Birne Binegar: I have two variables two functions that i'm trying to find X, Y amp T and X 250.
Birne Binegar: i'm going to pass it like that they be a differential equation, or the time derivative X one depends on both X one X two X does the time derivative of X two it depends on both X one X two we call the couple differential equations because we actually have two unknown functions.
Birne Binegar: they're separated on the left hand side here, but their derivatives directly proportional to both.
Birne Binegar: functions.
Birne Binegar: they'll these kind of differential equations so really common in chemistry.
Birne Binegar: suppose you had.
Birne Binegar: Hydrogen interacting with with oxygen and you want to know the rate at which that reaction is proceeding.
Birne Binegar: Well, it sort of makes sense that the rate at which the hydrogen is burning up it's going to be proportional to how much hydrogen, you have but it's also needs to interact with the oxygen in order to form the water molecule, and so the rate at which.
Birne Binegar: The concentration of hydrogen is changing would be proportional to how much higher you have, but also the how much oxygen, you have the burden of it.
Birne Binegar: and central either the oxygen its rate of depletion would be proportional how much hydrogen is born and how much oxygen, there is the mix alyssa be sort of a natural differential equation that would govern.
Birne Binegar: The rate at which hydrogen and oxygen the weather concentrations are changing if they burn.
Birne Binegar: But a similar kind of system, you know you could think of in biology, or more precisely, maybe ecology, are you considering two species predator and prey and.
Birne Binegar: If if they were the two species left each other alone, then their birth rate, or the rate at which the population changes.
Birne Binegar: would be proportional to the size of the population at any given time so many rabbits are producing so rapidly.
Birne Binegar: But if there's a predator in the in the in the system as well the foxes can eat the rabbits and so that they could decrease the rabbit population and the other hand.
Birne Binegar: If X two represents the flocks population you'll have some great of food supply, so we could grow quicker and so system like this could also.
Birne Binegar: Be super predator and prey and, like so many things in ecology, to have a reflection in the stock market as well, if you have to equities and one.
Birne Binegar: sort of depends on the growth of another equity you get differential equations like this and financial situation.
Birne Binegar: You get differential equations at the same form when you have oscillators like a bunch of springs that are connected to one another they'll work out there equations emotion workout to be coupled systems electrical engineering, if you have.
Birne Binegar: circuits that consists of an inductor a resistor and a capacitor you trace around the voltages as you transverse that circuit you get.
Birne Binegar: You get differential equations for that as well, so the upshot is that these sort of simple.
Birne Binegar: systems of differential equations are really common but they don't teach you in math 20 to 33 that was how to solve the systems so i'll show you.
Birne Binegar: know the reason why they don't teach it in 20 to 33 is that they don't assume that you've had any linear algebra yet, but now let's use some linear algebra what i'm gonna do is i'm going to.
Birne Binegar: take these two functions and i'm going to encapsulate them as a single vector value function, so now for each value of tea, I have two component functions to determine.
Birne Binegar: And i'll take.
Birne Binegar: These coefficients a one one, so this is the coefficient of the first unknown function in the first equation coefficient of the second and don't function in the second equation.
Birne Binegar: I can use these the former coefficient matrix in the usual fashion and then my differential equation can be written and vector notation as the derivative.
Birne Binegar: Of the two components X one X two or equal to.
Birne Binegar: This matrix times, the thing called vector.
Birne Binegar: And that just restating.
Birne Binegar: The original differential equations component by component.
Birne Binegar: Alright, so we're going to solve the systems, but the first thing i'm going to do is i'm going to solve a very simple case.
Birne Binegar: When my coefficient matrix is a diagonal matrix in this situation, solving the corresponding differential equations is really easy when the matrix is diagonal we say that the system is decoupled because what's going to happen here.
Birne Binegar: here's.
Birne Binegar: The factor form of the differential equation equals a X or more explicitly the key X one, the first component edt to the second component is equal to this coefficient matrix a one one a one a one a two two times.
Birne Binegar: X one X two alright, so this is just.
Birne Binegar: Taking and, if you look at this component by component, you have to calculate this matrix product will be a one one X one plus a one two X two.
Birne Binegar: A two one X one plus a two X two, so this matrix reformulation of the problem it's the same set of differential equation, the first component.
Birne Binegar: dx one is equal to coefficient times X one plus million coefficient times X, to the equation, corresponding to the comparing the second component on both sides it's just the second of the original differential equations.
Birne Binegar: Right now, when this matrix is diagonal, though.
Birne Binegar: That means.
Birne Binegar: That term it equals zero and that term it equals zero what happens.
Birne Binegar: Is the a differential equations completely decouple meaning the derivative of X one only depends on X 100 X to only sorry, this is not the term that drops out this one drops out a to this one depends only on this The second thing and and so.
Birne Binegar: The differential equations themselves decouple and we can solve them independently.
Birne Binegar: These differential equations are really easy, you should know the solution of this from calculus because when you differentiate an exponential function you just bring down the.
Birne Binegar: Factor that's being multiplied by the variable inside the exponential function.
Birne Binegar: So any function of this form is going to satisfy this equation any exponential function of this form is going to satisfy the second equation, you have shot what i'm saying here is if the matrix is diagonal likeable there's not much work to be done in order to write down the solution.
Birne Binegar: But that's not the case we're interested in, we want the case when a is not diagonal Iser diagonal but let's assume that is at least diagonal eyes.
Birne Binegar: And so we doing with a matrix that may have.
Birne Binegar: Non zero off diagonal components.
Birne Binegar: What will assume that this matrix even though it's not diagonal it is dot it's capable of being guy idolize.
Birne Binegar: Like finding its eigenvalues and it's eigenvectors and using those to construct a matrix see that will die guys a so that using the eigenvectors of a we can create see and doing this matrix multiplication of a bi C and its inverse will produce that diagonal matrix.
Birne Binegar: Alright, so what i'm going to do.
Birne Binegar: To solve this problem.
Birne Binegar: Is i'm going to assume that.
Birne Binegar: I already know, the solution to the eigenvector eigenvalue problem for the matrix a so assume.
Birne Binegar: We found.
Birne Binegar: eigenvalues.
Birne Binegar: The one and Lambda to and their eigenvectors.
Birne Binegar: Say V1 and V2 and so with this data, we can diagonal is the matrix a, then we can create see.
Birne Binegar: The one.
Birne Binegar: The two.
Birne Binegar: With.
Birne Binegar: See the minus first time to take time to see is equal to a diagonal matrix weathers di GL entries are going to be exactly the eigenvalues of a.
Birne Binegar: So i'm gonna assume that i've done all of this calculation for the given coefficient matrix but that's not my issue my issue is to solve this differential equation.
Birne Binegar: So what I want to show you is how you can use this eigenvector I can value calculation or the solution of this diarization problem how I can use that to produce solutions of this equation and it's really pretty easy.
Birne Binegar: When I first do is instead of solving this differential equation.
Birne Binegar: I realized related to a different differential equation when I use this diagonal matrix so let.
Birne Binegar: y of T.
Birne Binegar: be the solution.
Birne Binegar: Not the equation, we want, but rather.
Birne Binegar: The dt y is equal to this diagonal matrix.
Birne Binegar: Times why.
Birne Binegar: The reason why.
Birne Binegar: That useful is because we already know, the solution here, if this is a diagonal matrix.
Birne Binegar: Zero Lambda two times, why one why to then this differential equation decoupled, and we know the solution vector.
Birne Binegar: it's going to be well, one of the equation for a while one is going to decouple from why to and a solution is going to be.
Birne Binegar: Corresponding exponential function times, an arbitrary constant.
Birne Binegar: In the first component and the second component can be see to that second diagonal entry is going to give me the right exponential function and this this function is going to satisfy this differential equation, I can write it down.
Birne Binegar: I can write down this solution as soon as I know what these diagonal entries are.
Birne Binegar: But now what i'm gonna do.
Birne Binegar: i'm gonna claim.
Birne Binegar: If I take the given solution that I just found and I multiply it by this.
Birne Binegar: diagonal matrix see.
Birne Binegar: So X T is going to be this diagonal icing matrix the one that diagnosis, the coefficient matrix times this simple solution to the decouple differential equation, this is a solution.
Birne Binegar: The X the T is equal to X.
Birne Binegar: So that's how I solve this original system where the matrix is non diagonal.
Birne Binegar: and easy to see how that works, first of all.
Birne Binegar: Because we have a diagonal system, we can readily solve this differential equation here's it solution now i'm going to take the solution we just found this particular vector and i'm going to multiply it by the matrix see the diagonal ISIS our original coefficient matrix.
Birne Binegar: though, let me show you that this factor, or this vector value function satisfied the differential equation, so the exit fee is defined to be this and then this derivative.
Birne Binegar: These.
Birne Binegar: These entries here are all constant they don't depend on time and so.
Birne Binegar: This differentiation is going to treat this constant matrix as if it were a constant you'll just move right through and you end up differentiating y of T but y of T.
Birne Binegar: It obeys this differential equation, so the derivative of Why is going to be detach why now replace that there now i'm going to do something.
Birne Binegar: I have.
Birne Binegar: To accept it.
Birne Binegar: i've got it set equal to.
Birne Binegar: See times that diagonal matrix times My simple solutions.
Birne Binegar: And what i'm going to do is i'm going to insert.
Birne Binegar: The identity matrix in between these two factors, but the way i'm going to write the identity matrix is a C minus first time see times y of T.
Birne Binegar: Right i'm fine, Sir, the identity matrix between these two factors nothing changes, let me write that again if you matrix as a product of it see in its inverse and then i'll just regroup things see the scene CD minus first hard to see.
Birne Binegar: Why T.
Birne Binegar: And I guess.
Birne Binegar: The C times y of T.
Birne Binegar: that's what that's how we form this guide.
Birne Binegar: suppose I declared this to be the conjugating matrix times nice solution to the diet, a couple problem and.
Birne Binegar: So that's just X and then this here turns out to be exactly a because you have seen the minus version, a time see is equal to D that's how we diagonal is a but if I multiply from the Left, I see on both side and I multiply both sides from the right by the inverse.
Birne Binegar: what's going to happen here we'll get a factor of one here see time see the miners first you can give a factor of one and then here will have CD CD minus first or a is equal to see see see the miners first they'll basically i'm taking.
Birne Binegar: The diarization condition which expresses.
Birne Binegar: The diagonal matrix as a product of C minus first times eight times see and i'm solving for a That means a is equal to see the see the minus first.
Birne Binegar: that's exactly this factor, and so what i've shown you is that, when I solve the decouple problem that's easy and I multiply it by the matrix that converts.
Birne Binegar: A into a diagonal form, I simultaneously end up.
Birne Binegar: Solving the original differential equations because now.
Birne Binegar: This vector I create by this is the solution of the decoupled system multiplying by the matrix that converts the coefficient matrix into the diagonal matrix I end up satin coming up with a function of vector value function that satisfies the original in your system.
Birne Binegar: So let me phrase this whoops i'm running out of time i'm let me phrase this or generally if you had a system, a couple of ordinary differential equation, the way you would solve it is first of all.
Birne Binegar: You would diagonal is the coefficient matrix that will give you a bunch of linearly independent I give vectors and their corresponding eigenvalues and you can create the matrix seed that converts a into the diagonal matrix usual way.
Birne Binegar: Then you solve the decoupled system but that's really easy because for each dive a little entry, you can have an exponential solution, and you simply write the exponential solution in the same order as they appear these numbers appear diagonally.
Birne Binegar: And then to get the solution of the original problem where you have a non diagonal matrix you take the solution here, which is easy to write down you multiply it by your diag lighting matrix and then you get a solution to the system.
Birne Binegar: The activity is equal to this matrix a times.
Birne Binegar: And you're done.
Birne Binegar: Okay, so I just wanted to.
Birne Binegar: give you a concrete example of how outside this course you'll begin to use some of this apparatus i'll stop here today.
Birne Binegar: Let me pass out.
Birne Binegar: course evaluations.
Birne Binegar: And then, whoever is last to finish their evaluation.
Birne Binegar: If you could.
Birne Binegar: Still.
Birne Binegar: Once you're done stuff stickier evaluations inside this yellow envelope.
Birne Binegar: And then the last one to turn it in if you can simply walk the envelope up to the end of the hallway up there in front of the elevators is the math office and just drop off the envelope there that was great.