Birne Binegar: continue talking about dying likability.
Birne Binegar: Last time we learned.
Birne Binegar: That essentially if you want to dying, like the matrix.
Birne Binegar: You need to solve the eigenvector eigenvalue problems for the matrix that you wish do diagonal is over there was.
Birne Binegar: A caveat to that.
Birne Binegar: To that construction and so what i'm going to do today is both focus on that caveat, so let me remind you quickly.
Birne Binegar: The idea behind diarization and then we'll get to the caveat and then we'll show you how you work around it.
Birne Binegar: So the idea that we've been pursuing is given a square matrix can we find a simpler matrix that somehow represents that matrix the ideas, well, we want to construct a matrix see that effectively converts a non diagonal matrix into one this diagonal because.
Birne Binegar: What this will supposedly do and i'll show you in examples later on Wednesday, that this simplifies not, this is a great simplification of the matrix a in and how we use it.
Birne Binegar: But right now we're just going to concentrate on.
Birne Binegar: How we find a diagonal matrix that, in this sense, equivalent to the original matrix the fact that a can be converted into D and.
Birne Binegar: I will show you not only how to.
Birne Binegar: What that diagonal matrix going to be this can be actively determined by the eigenvalues of a and the matrix see that converts a into a diagonal matrix that's going to be constructed from the corresponding eigenvectors of a.
Birne Binegar: Listen, how it works.
Birne Binegar: If you have a matrix see what these two llamas do is they.
Birne Binegar: tell you how matrix multiplication work when you think of a matrix.
Birne Binegar: As being specified by listing it's called.
Birne Binegar: So take a make an n by n matrix see Well, I can fill in all the entries or I can fill in the entries column by column we're taking this latter point of view here.
Birne Binegar: And what this first identity tells me is that, if I have a matrix expressed in terms of its columns and I multiply it from the left by another matrix a well.
Birne Binegar: The new matrix the new product can actually be computed column by call and all you have to know is how the matrix a X on the individual columns of see and.
Birne Binegar: act individually on the columns you get the action on the whole matrix so that's one identity.
Birne Binegar: And a second matrix a second identity is one we multiply given matrix see expressed in terms of the columns what happens when we multiply from the right hand side by diagonal matrix.
Birne Binegar: And here's the identity is my multiply see from the right by a diagonal matrix what that multiplication effectively does is it takes the.
Birne Binegar: cons of the matrix and it multiplies them by the corresponding diagonal elements, so the second example the diagonal matrix.
Birne Binegar: is one where you have numbers when the one and the two to Atlanta in along the diagonal and everywhere else you have zero.
Birne Binegar: And the rule is, if you multiply see from the right by D what happens is is that the columns of see just get multiplied by the corresponding diagonal entries so that first call gets multiplied by Lambda one second column gets multiplied by Lambda to etc.
Birne Binegar: These identities followed directly from the definition of what it means to multiply two matrices together the fact that each entry in this product is a dot product of a row of a in a certain column to see.
Birne Binegar: Right now, those two identities.
Birne Binegar: lead to.
Birne Binegar: A condition for dialogue liability of a matrix.
Birne Binegar: Let me state thing if you want to have a matrix see that diagnosis, a matrix a well, then you start off with this identity.
Birne Binegar: And then, if I multiply both sides of this by the matrix see well what's going to happen see time see the miners first is just going to produce the identity matrix so on the right hand side i'm going to get a see.
Birne Binegar: On the left hand side it's still going to be just see times D and you're So you see that this identity, the one we want the identity that convert a to a dial diagonal matrix is a call into this identity we're now see us living on both sides of the equation.
Birne Binegar: But reason for writing this.
Birne Binegar: Is I have expression for each side of that equation, a time see and a diagonal matrix time see.
Birne Binegar: And so.
Birne Binegar: When I exploit those identities, what happens is if if this is true, and this is true, and then that means that.
Birne Binegar: At times, the first column of C has to be equal to the first diagonal entry of D times the same column, in other words, see one has to be an Eigen vector of a eigenvalue Lambda one and that's going to be true for each call So if you do have a matrix that converts to a to a diagonal matrix.
Birne Binegar: What we see here is that each column of see has to be an eigenvector of a and, in fact, each diagonal entry of the matrix D has to be an eigenvalue a.
Birne Binegar: Until that gives us a way of.
Birne Binegar: Perhaps constructing both the matrix CC that diving license a and the matrix D, that is the result of dying lighting a.
Birne Binegar: What we do we take a bunch of eigenvectors a use them as columns of the matrix we use their corresponding eigenvalues as a diagonal entries at the matrix D and then this condition will be automatic we.
Birne Binegar: Basically constructed two matrices such that the columns of one are always eigenvectors corresponding to the diagonal entries, of the other.
Birne Binegar: But the the thing I was alluding to it a startup lecture is that wow or diagonal ization condition in prices.
Birne Binegar: This identity we're see sits on opposite sides of the equation rather on the same rather than seeing its inverse on the same side, unfortunately this identity does not necessarily imply the diarization condition that we want because.
Birne Binegar: there's nothing i've assumed so far that tells me that this inverse matrix actually exists and matrices do not always have been versus.
Birne Binegar: So today what we're going to focus on is specifying conditions that will tell us whether or not we have.
Birne Binegar: an inverted matrix see such that we can diagonal is a given matrix So the idea is still going to be to make see from the eigenvectors of a, but the question is, can we just take arbitrary eigenvectors throw them in a matrix and get a diamond matrix or do we need something else.
Birne Binegar: And in fact we do need something else.
Birne Binegar: If you have a matrix it's convertible.
Birne Binegar: Well let's review that a little bit.
Birne Binegar: Last time I presented it.
Birne Binegar: by citing the fundamental not essential conditions we need proceed to be convertible by reciting the fundamental theorem of convertible matrices today i'm going to take a different tack, I just want to remind you of some calculations.
Birne Binegar: First of all, I.
Birne Binegar: See is convertible.
Birne Binegar: Well, that means that see can be row reduce.
Birne Binegar: To the identity matrix, why is that or remind you, the calculation, the way you find the inverse of the matrix see.
Birne Binegar: Is you a join to it, the identity matrix of the same size and you reduce.
Birne Binegar: Then one of two things happens.
Birne Binegar: Either the left hand side row reduces to the identity matrix, in which case the right hand side you're going to get transformed.
Birne Binegar: into the inverse of the matrix you started off with i'm using see now or you get something else.
Birne Binegar: Something else on the left hand side here.
Birne Binegar: And then that case there's no inverse.
Birne Binegar: that's calculation, hopefully, everyone recalls.
Birne Binegar: The second thing i'm going to point out is.
Birne Binegar: How to find.
Birne Binegar: A basis for the called space.
Birne Binegar: Of a matrix well here's our algorithm what we do is we reduce.
Birne Binegar: The given matrix to.
Birne Binegar: Reduce rope.
Birne Binegar: automatically sufficient to go to row Echelon form.
Birne Binegar: Let me state it as taking it all the way to reduce all Ashkelon form and then what we do is we.
Birne Binegar: Look at where the ticket slip let's call that see double prime.
Birne Binegar: So the columns.
Birne Binegar: Of the original matrix see that correspond to.
Birne Binegar: poems about relational and form.
Birne Binegar: That have picnics.
Birne Binegar: end up being.
Birne Binegar: A basis.
Birne Binegar: For.
Birne Binegar: The original matrix.
Birne Binegar: Suppose see us in vertical that means he can be reduced to the identity matrix but the identity matrix.
Birne Binegar: has always has a pivot athlete it's one all along this main diagonal and so every column is going to have a pivot.
Birne Binegar: That means every call must the original matrix see it's going to be a basis factor for the column space now one thing about basis vectors so they have to be linearly independent So what is this telling us it telling us that.
Birne Binegar: In order to create an advertorial matrix yeah we need we need all the column vectors to be literally independent your argument goes, yes, is that.
Birne Binegar: If the real actual on form has a pivot and every column then every column of the original matrix must be a basis director.
Birne Binegar: basis, these are always linearly independent therefore every column of see must be literally indeed tended if they can be reduced to the identity matrix so that's the argument I make here.
Birne Binegar: And what it allows us then conclude is is that we have a way.
Birne Binegar: Of.
Birne Binegar: Constructing a matrix see so that this identity hold all we have to do in order to get this identity to hold is to make sure that each column a C is an Eigen factor of a and we want the corresponding diagonal entry of D to be.
Birne Binegar: The same have the same eigenvalue so One example I can, if.
Birne Binegar: It.
joshl: Were.
Birne Binegar: let's say V one.
Birne Binegar: Is and I can affect your.
Birne Binegar: Feedback of see have a.
Birne Binegar: With.
Birne Binegar: eigenvalue Lambda Lambda one, then what I could do so, I could write the one.
Birne Binegar: The one.
Birne Binegar: You want I can use that same eigenvector over and over again, I could call that C and I could.
Birne Binegar: Like Lambda one.
Birne Binegar: And the one as a diagonal entry so matrix D and this identity will hold because each column, the first column here is going to be the eigenvector corresponding to the corresponding diagonal entry, but the problem is, is that.
Birne Binegar: This matrix is not going to be in vertical because i'm using the same vector over and over again, these these columns are not going to be linearly independent and so see the minus first will not exist.
Birne Binegar: So it's not going to be enough to just throw a bunch of eigenvectors into a matrix and hope that that's going to be able to die analyze your matrix a.
Birne Binegar: If you just use the same eigenvector over and over again, you end up with a matrix that satisfies this identity, but you can't get the identity one.
Birne Binegar: See the miners first AC equals B that's not going to follow because yeah for this matrix the inverse does not exist alright so that's that's the issue we're dealing with and the purpose of today's lecture is to.
Birne Binegar: avoid the sort of complications, we want to be able to always be able to invert them.
Birne Binegar: The matrix we construct out the I vector so that we can actually proceed with the diagonal ization.
Birne Binegar: stuff here's the basic question that will answer today, how do we know if a square matrix has enough linearly independent Ireland values, so that when we create the matrix see it's going to be inverted.
Birne Binegar: Right, so the first answer.
Birne Binegar: i'll give you to this question is going to rely on our.
Birne Binegar: Our solution up the eigenvector eigenvalue problem where we not only down eigenvalues an idea vectors, but we also calculate their algebraic in geometric multiplicity.
Birne Binegar: So let me remind you what those guys are.
Birne Binegar: If you have a n by n matrix you can calculate it to characteristic polynomial it's going to be a minus Lambda times the identity matrix and the terminus that matrix setting this polynomial Atlantic equals zero is how you find the argument.
Birne Binegar: But according to the fundamental serum of algebra this is going to be a polynomial degree and so it's going to have a complete factorization in terms of linear polynomials we call these what what i've done here is i've completely factor is this poll normal.
Birne Binegar: In terms of linear polynomial and I collect together all the similar factors, so this term here is going to be all the factors of the characteristic polynomial that are up the form Lambda minus the first eigenvalue.
Birne Binegar: These powers here are called the algebraic multiplicity.
Birne Binegar: there's a different kind of multiplicity as well, and this is actually the one that's more relevant to our current problem, our problem of finding enough linearly independent eigenvectors to construct and vertical matrix.
Birne Binegar: The geometric multiplicity of an eigenvalue what it is it's exactly the dimension of the corresponding I can space.
Birne Binegar: The space he subscript our I is the space of all vectors such that they have eigenvalue are I.
Birne Binegar: respect to the matrix egg.
Birne Binegar: On the other hand.
Birne Binegar: We can calculate this number it's the number of.
Birne Binegar: it's effectively dimension of the solution space.
Birne Binegar: Of this, not homogeneous linear system with this coefficient matrix, so this is also the number of free parameters, you get when you reduce this matrix it's a number of pivots you get when you reduce this matrix to real Echelon for.
Birne Binegar: All right, but here's, the key point, though, what i'm doing when I count up the individual.
Birne Binegar: dimensions of these spaces i'm counting the number of linearly independent vectors it needs to generate this entire space, so I am I do end up counting linearly independent vectors except I do it subspace by subspace I pick and I can value eigenvalue.
Birne Binegar: Once I calculate the corresponding geometric multiplicity I know exactly how many linearly independent vectors I have with a given eigenvalue.
Birne Binegar: Then I can proceed to another eigenvalue and i'll pick up some more I can vectors So when I count up all these geometric multiplicity what that should be is, it should be the number of literally independent eigenvectors I have.
Birne Binegar: Not just in a single subspace when I sent over these steps spaces, I get all the not possible linearly independent eigenvectors for the matrix a.
Birne Binegar: And in fact.
Birne Binegar: One property of these algebraic and geometric multiplicity is first of all.
Birne Binegar: You look at this definition of the algebraic multiplicity for n by n matrix.
Birne Binegar: This determinant it's going to be a polynomial degree in so the characteristic polynomial polynomials degree in.
Birne Binegar: The claim is it factor is like this well if this is also going to be a porno me or degree in well, that means that these powers in one plus empty tube etc plus mk they have to add up to exactly and so that we have polynomials degree and on both sides.
Birne Binegar: On the other hand.
Birne Binegar: i'm these.
Birne Binegar: Corresponding algebraic or sorry the geometric multiplicity.
Birne Binegar: Sorry.
Birne Binegar: there's a condition on them as well, and that is that there, they have to be at least one.
Birne Binegar: Otherwise you.
Birne Binegar: If there was not equal to, if you didn't have a factor of Lambda minus R I at least one factor, then you wouldn't have an eigenvalue.
Birne Binegar: Right, you need at least one of these factors in order to say that are one is, and I can value.
Birne Binegar: Until but the geometric multiplicity or sort of bound, they have to be at least one.
Birne Binegar: But they also have.
Birne Binegar: Less than or equal to the corresponding algebraic multiplicity.
Birne Binegar: So what I want you to observe if these numbers, when I add my some over all all the eigenvalues they have to add up the end because I have to get a polynomial degree in this fashion.
Birne Binegar: On the other hand, because the geometric multiplicity are always bounce from above, it algebraic multiplicity if one of these numbers is strictly less.
Birne Binegar: than the corresponding algebraic multiplicity it then i'm no longer going to add up the no added some number less than in that's going to mean we don't have enough linearly independent eigenvalues.
Birne Binegar: linearly independent eigenvectors.
Birne Binegar: What i'm suggesting here is we.
Birne Binegar: We count up these eigenvectors I can space by Eigen space so maybe the first two I conductors belong to the first eigenvalue and then we might have another Eigen space and let's use, third and fourth and fifth one coming from this, I can space and I.
Birne Binegar: run to the eigenvalues and then I can write down linearly independent eigenvectors and form a call and the total number it's kind of be the total dimension of these Eigen spaces or.
Birne Binegar: Exactly the some of the geometric.
Birne Binegar: These geometric multiplicity and what this says is any one of these.
Birne Binegar: Any one of these these numbers that the geometric multiplicity is less than the corresponding algebraic multiplicity then it's impossible for the number of.
Birne Binegar: linearly independent eigenvectors to add up in, and so I won't have enough linearly independent called you create an invisible matrix.
Birne Binegar: Right, so that leads.
Birne Binegar: To the following.
Birne Binegar: Conditions for non diag liability if you have a matrix with on who's eigenvalues or say Lambda one Lambda K.
Birne Binegar: Let him, be the number of factors of Lambda minus a given eigenvalue and let me mute be the dimension of the corresponding I get space.
Birne Binegar: If any one of these geometric multiplicity is less than the corresponding algebraic multiplicity then a if not going to be diagnosable because you will not have enough linearly independent eigenvectors to create.
Birne Binegar: A matrix see that it will be successful in diagonal icing you're given matrix a.
Birne Binegar: Alright, so this dish repeating the argument.
Birne Binegar: But that sort of a negative result it tells us that and, but in fact we have to find all the wagon vectors and I liken it to in order to draw that conclusion, you have to actually explicitly calculate each of the geometric multiplicity in order to see what they add up to you.
Birne Binegar: But there's a case where you don't have to do that.
Birne Binegar: Suppose be one you have a set of eigenvectors.
Birne Binegar: But, for which there eigenvalues are all distinct.
Birne Binegar: So i'm assuming that.
Birne Binegar: For both these factors for all the factors, do you want to bk i'm going to have a the eye is equal to Lambda I see I that I want Lambda I not to equal Lambda J, yes I that's not equal J show.
Birne Binegar: So i'm.
Birne Binegar: i'm working with a set of eigenvectors that are tied to particular eigenvalues hypothesis here is that the vectors.
Birne Binegar: A vector with a different index here is going to correspond to a different eigenvalue.
Birne Binegar: And what i'll show you.
Birne Binegar: And said in this situation.
Birne Binegar: If to eigenvectors correspond to different eigenvalues, then there are also linearly independent.
Birne Binegar: Now that's going to be true, no matter how many eigenvectors I use but i'm just going to prove this, when I have to eigenvectors.
Birne Binegar: So here i'm going to assume the one in D to or I can factors of a matrix a and.
Birne Binegar: There I corresponding eigenvalues are different, so I guess, let me stay my assumption here explicitly.
Birne Binegar: Do you want to.
Birne Binegar: satisfy a B one.
Birne Binegar: b one B one a B is equal to Lambda to details that's the fact that they both I can factors and.
Birne Binegar: Lambda one does not equal Lambda to.
Birne Binegar: And what i'll try to show is that necessarily these two vectors are going to have to be literally independent.
Birne Binegar: Alright, so these are my assumptions on V1 and V2 let's suppose there is a dependence relation suppose I can find nonzero numbers X one X two so set this dependence relation can actually be solved.
Birne Binegar: well.
Birne Binegar: We can fact use a dependence relation to express what are the factors in terms of the other see if X one is not equal to zero i'll be able to divide by it and then use this equation, to express the first factor in terms of the second.
Birne Binegar: But my hypothesis that the these guys are also eigenvectors john particular eight times the one it's going to be it's eigenvalue types of actors.
Birne Binegar: On the other hand, because these guys are assumed to be linearly linear dependent, I can express the one in terms of the two well that's what i've done here, but now this matrix a can go right through the scale of multiplication and it can attack.
Birne Binegar: A can attack the vector D to directly, but the two.
Birne Binegar: is also assumed to be an icon vector a but with a different eigenvalue.
Birne Binegar: So what i'll get when he hits the two is i'll get land, the two times each you.
Birne Binegar: But now the two.
Birne Binegar: Well, I just reverse the order, though scale or multiples and then i'll recognize minus X divided by.
Birne Binegar: X one is another expression for the one.
Birne Binegar: And now, when I look at the extreme.
Birne Binegar: Side Lambda one times, one is equal to Lambda two times.
Birne Binegar: But my hypothesis is that these numbers Lambda Lambda two are different, so this is a contradiction.
Birne Binegar: I can't have Lambda one time fee will also be equal and the two times do you one unless be one is the zero vector otherwise I have something times a non zero vector is equal to something else time that non zero vector and that can happen.
Birne Binegar: But remember I can vectors are always non zero vector so effectively.
Birne Binegar: This is a contradiction.
Birne Binegar: it's a contradiction that my to Ivan values are not the same.
Birne Binegar: In short, if we have to I can vectors corresponding is different eigenvalues then they can offset.
Birne Binegar: Still, the must be literally independent.
Birne Binegar: Right now, I shown that when I have two different I get actors, having different eigenvalues but the same thing can be shown by using a proof by induction that no matter how many eigenvectors I have if they all have different eigenvalue Sir necessarily literally independent.
Birne Binegar: Okay, so that gives us.
Birne Binegar: A criteria suppose we have a matrix a with end distinct eigenvalues, then the matrix a is always going to be diagonal likeable.
Birne Binegar: I guess there's a number of ways of thinking that.
Birne Binegar: If I.
Birne Binegar: A.
Birne Binegar: distinct eigenvalues.
Birne Binegar: That means.
Birne Binegar: Lambda the characteristic polynomial it's going to be Lambda minus or one well, I mean to get us a factor like this for each eigenvalue it's all have Lambda minus R N, and that means.
Birne Binegar: We have n distinct eigenvalues That means we have total and distinct factors and the characteristic Paul o'neill and they all curve with algebraic multiplicity one.
Birne Binegar: If I had a factor to hear that would mean that to have my eigenvalues who are effectively the same two of these factors, would be the same.
Birne Binegar: So that means.
Birne Binegar: summation of.
Birne Binegar: All these yeah that effectively meet all the M eyes are equal to one.
Birne Binegar: equals one the end all the algebraic multiplicity is equal to one, but we also know.
Birne Binegar: That when we sum all the algebraic multiplicity.
Birne Binegar: We always get the size of the matrix.
Birne Binegar: And we also know that one is less than equal to the corresponding geometric multiplicity is less than or equal to the corresponding algebraic multiplicity So these are properties of algebraic and geometric multiplicity.
Birne Binegar: Now these guys are all equal to one.
Birne Binegar: But the corresponding geometric multiplicity has to be greater than equal to one if these guys are all equal to one, that means that all the geometric multiplicity these are equal to one, so we need ui equal to one.
Birne Binegar: And these guys out.
Birne Binegar: The number.
Birne Binegar: Of linearly independent eigenvectors.
Birne Binegar: value.
Birne Binegar: I well.
Birne Binegar: Well yeah what am I calling those yeah i'm calling them or I and so these are all equal to one, and so what they're going to add up to his end.
Birne Binegar: So that's going to mean that the total number of linearly independent eigenvectors is and and that's exactly how many literally independent I conduct, you need in order to create an inverted matrix.
Birne Binegar: And so the.
Birne Binegar: A is going to be diagnosable because you will have enough linearly independent eigenvectors to diagonal is it.
Birne Binegar: Alright, so you just use for each of these eigenvalues you'll have a an eigenvectors you use those factors these guys are all going to be linearly independent because they correspond to distinct eigenvalues and so that matrix will be in vertical.
Birne Binegar: Okay, so let me do some examples.
Birne Binegar: Now the rest of the lecture is just going to be some computation.
Birne Binegar: Well, the first thing I want to do.
Birne Binegar: Is look at a two by two matrix and without actually explicitly calculating the eigenvectors determine whether or not this matrix is can be diagonal.
Birne Binegar: So I will have to find the eigenvalues so let's calculate the characteristic potent the determinant of the matrix with a parameter Lambda subtracted from its diagonal elements.
Birne Binegar: determine if this matrix we can get by the cross the difference of the diagram of products one minus Lambda squared here, you see, you have one eigenvalue with algebraic multiplicity to but they only have one distinct factor here, so I only have one distinct eigenvalues.
Birne Binegar: feel this means.
Birne Binegar: That diagonal eyes ability is not guaranteed if I had two distinct eigenvalues, then I know the matrix should be able to be diagnosed by the proceeding theorem.
Birne Binegar: If you for two by two matrix I would need two distinct eigenvalues but in this example I only got one distinct eigenvalue so it doesn't mean that the matrix is not diagonal Isobel but it doesn't mean we have to check further.
Birne Binegar: So.
Birne Binegar: What we'll do is we'll see how many eigenvectors we have.
Birne Binegar: Since we only have one I can space, you only have just we only have one eigenvalue we have only one I get space, but it could be that within that I can space there's two linearly independent eigenvectors.
Birne Binegar: In that case we would have enough linearly independent eigenvectors to construct and in vertical two by two matrix that could die guys the original matrix.
Birne Binegar: But we don't know that yet will need to calculate this space so.
Birne Binegar: We calculate the Lambda equals one Eigen space it's going to be the north face this coefficient matrix when I plug in the eigenvalue I get that and then.
Birne Binegar: yeah as soon as I calculate it I get a matrix it's already in reduce your Ashkelon form and what you see is that what's the dimension of this no space.
Birne Binegar: we've got one for one column about a pivot one column, with a pivot it's going to be a one dimensional solution set.
Birne Binegar: That means this Eigen space is only one dimensional and that's the only I good space, I have, so I need to linearly independent eigenvectors but they only have a one dimensional I can space to work with so i'm not going to be able to create an embeddable matrix see.
Birne Binegar: From a space that's only one dimensional.
Birne Binegar: And so, in this example the matrix we started off with is not diagnosable.
Birne Binegar: Alright, let me let's do another example.
Birne Binegar: Where we use the theorem.
Birne Binegar: They actually show.
Birne Binegar: That a matrix is diagonal Isobel simply by looking at this eigenvalues.
Birne Binegar: Here we have a three by three matrix.
Birne Binegar: We calculate this characteristic Paul no meal.
Birne Binegar: turns out, you can do a Co factor expansion, on the second row and you'll find that this is the determinant a minus Lambda times the identity matrix.
Birne Binegar: This polynomial can be factor is its cubic if, after ISIS into three linear polynomials this one corresponds to an Eigen value of zero this one corresponds to an Eigen value of negative one, and this factor corresponds to an eigenvalue have to.
Birne Binegar: So we see that this matrix has three distinct eigenvalues zero minus one and two it's a three by three matrix, and so I can apply the theorem.
Birne Binegar: If I have an n by n matrix with and distinct eigenvalues it's guaranteed to be diagnosable.
Birne Binegar: And so we know the answer this matrix can be diagonal eyes and I see that as soon as I factor is the characteristic polynomial.
Birne Binegar: But I have not yet diagnosed a so i'll do that next.
Birne Binegar: What we need to do in order to diagnose a is, we need to find since as three by three, we need to find three linearly independent I can factors.
Birne Binegar: And then, this construction will work.
Birne Binegar: So we look at the first audience page where the eigenvalue with zero That means we look for the no space of a minus the eigenvalue time see again injury matrix this matrix.
Birne Binegar: We reduce to reduce rational on or because actually what we want to do is, we want to fold solve the linear system with this matrix is the coefficient matrix so it reduces to this, and now you see that effectively the equations.
Birne Binegar: Of the solution art X one plus x is equal to zero second row tells me that X three have two equals zero and from those equations you can quickly deduce that.
Birne Binegar: In order to live in this solution set the first component has to be the negative of the second component and the third component has to equal zero.
Birne Binegar: So.
Birne Binegar: This Eigen space is one dimensional it only has one basis factor only takes one factor to generate the entire set.
Birne Binegar: Alright, so i'll use the vector one minus one zero as a basis factor for the Lambda equals zero space.
Birne Binegar: I did switch the sign here.
Birne Binegar: that's because if you have a basis factor, you can always do a scale or multiple of it, and it will still be a basis factors long as I don't multiply by zero this vector it's just as good a.
Birne Binegar: basis factor is this space is this guy.
Birne Binegar: Right and then we have tomorrow again spaces, we had an eigenvalue of negative one, when I calculate the corresponding against based.
Birne Binegar: I look at this coefficient matrix that's that it reduces to this matrix in reduce your actual on form, if this guy.
Birne Binegar: And then from this matrix and reduce your rational on board, I can calculate the solution directors.
Birne Binegar: In this case, the first component is going to happen equals zero and the second component is going to have to be the negative of the third component, so you get that as a basis factor and we only get one factor so again, you have a one dimensional item space.
Birne Binegar: Then, lastly, we have last eigenvalue Lambda equals to we find it I get face by calculating the north face of this matrix same calculation.
Birne Binegar: First, you substitute in for the eigenvalue, then you reduce to reduce your national on form and then you've tried to write down a solution vector.
Birne Binegar: And in this case, what will need is X one equals three half X three x to we'll have to equal one half X three and last row just going to say zero is equal to zero, and you end up with a single basis director for the entire solution set or currently a single eigenvector.
Birne Binegar: That generates the entire land equals to Eigen space.
Birne Binegar: What we'll do the simplified the expression of this basis vector is on multiply it by two to get rid of those denominators and now i've got a vector with integer entries that generates the entire Lambda equals two Eigen space.
Birne Binegar: Right, so what we found our three linearly independent eigenvectors.
Birne Binegar: First of all, we had three distinct eigenvalues that already told me that I would have enough linearly independent I conductors to create the.
Birne Binegar: matrix see that would trance.
Birne Binegar: transform a into a diagonal matrix to write down that matrix see I choose and I can I choose an eigenvalue and the corresponding eigenvectors I use.
Birne Binegar: The eigenvector as the first column of the matrix see and I use the the eigenvalue as the first diagonal entry of the matrix D.
Birne Binegar: Then I go to the next Eigen value I use it to format, the next dive you know entry and the corresponding call over here i'll use the eigenvector corresponding to that eigenvalue.
Birne Binegar: Then, finally i'll use the last eigenvectors the last call and last eigenvalues to last entry of this diagonal entry, so I can vectors that I used for columns are going to be correlated with their corresponding eigenvectors along this diag corresponding eigenvalues will on this diagonal.
Birne Binegar: Then, what will happen is that this identity is going to be automatic.
Birne Binegar: The identity AC equals DC that's automatic whenever you.
Birne Binegar: form a matrix see using I conductors and you form an analogous matrix D by inserting their eigenvalues and the corresponding slot but that if these columns are not linearly independent I won't get an inverted matrix.
Birne Binegar: In this example, though I know I have three linearly independent eigenvectors to work with, and I can use those three linearly independent I get back just literally independent columns so that matrix is in vertical.
Birne Binegar: Right so.
Birne Binegar: here's like.
Birne Binegar: The matrix see that diagonal is the original matrix a and this matrix D is what's going to result when I do this matrix multiplication.
Birne Binegar: You don't have to calculate see the miners first to verify that this is the right diagnosis matrix that's already implicit from the fact that this first column has.
Birne Binegar: eigenvalue zero the second column has eigenvalue minus one in the third call, as I can tell you too.
Birne Binegar: All right, there's one last thing I want to touch on today.
Birne Binegar: And that's another special place when we can tell when we have a diagnosable matrix and this one occurs quite frequently and physical applications.
Birne Binegar: A matrix is symmetric whenever it's equal to a trance those.
Birne Binegar: Even example, you have a equals one to two one then it's transpose you get this matrix by taking, for example, the rows of a and rewriting a miss calls, so this first row would become the matrix.
Birne Binegar: One because one to the second row would become a call to one and notice that, for this matrix a is exactly equal to X transpose that's that's what we mean by a set symmetric matrix more typically a matrix it's going to be symmetric.
Birne Binegar: it's going to be that way, if you look along the main diagonal and you reflect the crosses the list would be big entry entry me say I throw in jail calm.
Birne Binegar: When you reflect across this angle, a process main diagonal again an entry in the J throw an icon what you're requiring are these numbers to be the same like here if I reflect on this main dangle this to go to here, and it must be the same.
Birne Binegar: Alright, so make sure sees that are symmetric actually occur quite frequently in applications, and here we have a nice theorem.
Birne Binegar: anytime you have a symmetric matrix the matrix is always dying legible and, in addition, it has another nice property.
Birne Binegar: The characteristic polynomial.
Birne Binegar: Which is going to look like Lambda minus.
Birne Binegar: Say r1 to an algebraic multiplicity of the minus R K algebraic multiplicity can take.
Birne Binegar: The factorization that the characteristic polynomial were reason this life, you have such a factorization is because of the fundamental theorem of algebra that says that every polynomial degree in can be factor is in terms of linear polynomials but.
Birne Binegar: These assets here, these roots are, in general, complex numbers, rather than real numbers, for example, if you had a equal to.
Birne Binegar: 00 minus one one notice this matrix is not symmetric it's anti symmetric that is transpose is going to be the negative one, which started off with characteristic polynomial for that matrix it's just going to be.
Birne Binegar: minus Lambda squared plus one or it's going to be Lambda squared plus one, and you know that this.
Birne Binegar: This doesn't have.
Birne Binegar: any real root, because if Lambda squared plus one is equal to zero Lambda squared would have to equal negative one, but that's not true for any real number so.
Birne Binegar: When you even if you have a matrix with real entries it's possible that it's eigenvalues are going to be complex numbers what's nice about symmetric matrices is that.
Birne Binegar: it's always going to be dying advisable you'll always have enough I conductors to diagonal is it and, moreover, you don't ever you don't have to leap up into the complex numbers to solve the eigenvector eigenvalue problem your eigenvalues are always going to be real.
Birne Binegar: I guess i'll give you one application this.
Birne Binegar: And then i'll let you go.
Birne Binegar: rotational dynamics.
Birne Binegar: So this is the.
Birne Binegar: freshman physics and what you're talking about now rotational motion rather than linear emotions.
Birne Binegar: and
Birne Binegar: what's relevant here it's a certain matrix.
Birne Binegar: usually called it called a moment of inertia matrix and what it is, as you integrate over the entire volume of the body you integrate the density at a point X multiplied by a factor X I xj minus all right at this point, do I do.
Birne Binegar: Next, one squared well del y squared times the magnitude of this actress.
Birne Binegar: integrate that over the entire body so.
Birne Binegar: What this kind of measures is first of all, the density tells you how the weight is distributed over the body and what this these terms here this short indicating is.
Birne Binegar: How that math is dispersed throughout the body it's weighted in a certain way, if I look at the diagonal components, what happens here is that.
Birne Binegar: A I, for example, it's just going to be that density times.
Birne Binegar: Well it's going to be the distance squared from the origin will be X ah.
Birne Binegar: Let me do it this way, one one then it'd be.
Birne Binegar: X two squared plus x three integrated against that, and you can sort of see the point of this matrix is it.
Birne Binegar: It has a certain contribution, due to the how the masses distributed what this is is.
Birne Binegar: What this factor involving coordinates is doing is telling you the distance that the given point is from the X axis here the square at the second component and the third component where that's the different distance.
Birne Binegar: From a given point and the X axis, so this is sort of measuring how the mass is distributed away from a given access but more generally use this construction it tells you how the influence in one mass distribution in one direction kind of is affected by other directions as well.
Birne Binegar: What I just want to point out is that this matrix that tells you how the masses distributed and it's used in rotational dynamics is a symmetric matrix because you reverse the roles of I nj here, you get exactly the same expression.
Birne Binegar: Those inertia.
Birne Binegar: Those inertia matrix these are always going to be symmetric until they're always going to be dying or sizable and they're eigenvalues we're going to be real numbers.
Birne Binegar: that's going to be important in rotation all again damage, where you actually have to interpret this matrix it turns out that the eigenvectors.
Birne Binegar: At this matrix they correspond to.
Birne Binegar: The so called natural axis of rotation.
Birne Binegar: Those are going to be the unique directions and which, like a basketball, but you can take a more irregularly shaped object as possible to spin it in such a way that you can balance it on one finger.
Birne Binegar: Those those particular directions, the those natural axis of rotation can be calculated by calculating this matrix and then diagonal is to get the.
Birne Binegar: Corresponding eigenvectors are going to be the natural axes rotation and the eigenvalues they have a nice meeting as well, these are going to be a moment.
Birne Binegar: of inertia.
Birne Binegar: Along the corresponding.
Birne Binegar: axis of rotation.
Birne Binegar: In other words, what those moments of inertia there the analog of mass if you think of mass as the resistance to acceleration say F equals ma you need a stronger force it.
Birne Binegar: To do the same acceleration for a given mass these moments of inertia are analogous to that for rotational motion they they correspond to the resistance to accelerating a rotation.
and
Birne Binegar: And that sense they're just as important as mass is to partner director linear motion of a particle.
Birne Binegar: turns out that both this.
Birne Binegar: Both the natural axis of rotation and the corresponding moment of inertia dirt pocketable so I die analyzing a certain matrix that happens to be symmetric.
Birne Binegar: These two properties are therefore kind of really nice, first of all you're guaranteed that you can die analyze it so you can find eigenvectors in eigenvalues.
Birne Binegar: and, moreover, you never have to jump up in the complex numbers because for symmetric matrix it always turns out that the Eigen values are real numbers until they're eigenvectors will also be real numbers real doctors.
Birne Binegar: About okay so i'll stop here today what I intend to do on Wednesday is do another application of.
Birne Binegar: diarization of matrices and this one is going to be kind of Nice because it's going to take us.
Birne Binegar: A little bit of far we're trying to solve the differential equation using linear algebra and it will turn out that a key step and that solution is going to be to diagonal I see a certain matrix so i'll do that on Wednesday.
Birne Binegar: And with that i'll guess i'll let you guys go for today.