Birne Binegar: Okay, so um.
Birne Binegar: A couple things first of all.
Birne Binegar: homework set eight was to be done on Monday, now there any difficulties in that I can address them now.
Birne Binegar: i'm going to make available this afternoon problem set nine it's going to deal with eigenvectors and eigenvalues and so today's lecture after afterwards, you should be able to enter all the problems on problem set nine.
Birne Binegar: i've also posted the solutions.
Birne Binegar: Including the next assignment homework assignment 10 on canvas we're getting close to the end of the Semester so i'm going to have all the material available to you at all time.
Birne Binegar: And in fact not only the Homer problems, the solutions, I post, are not going to be exactly the problems that we have assigned a few to solve because they change the numbers.
Birne Binegar: But you can at least look at my solutions if you get stuck in figuring out how to approach a problem that or what the procedure is.
Birne Binegar: Today, we will continue with the eigenvector and eigenvalue problem if you look on problem set nine.
Birne Binegar: you're typically given a matrix and you're asked to find it eigenvalues and it's eigenvectors but there's two more additional bits of data that you will cover today so.
Birne Binegar: Oh i'm not sure i'm not just we'll just start with the lecture there'll be two additional numbers one associate with an Oregon value, one would be a so called algebraic multiplicity that's going to depend on how that characteristic polynomial factor rises and then there's another.
Birne Binegar: associated multiplicity it it it's basically the dimension of the corresponding Eigen space so once you have an eigenvector.
yeah.
Birne Binegar: it's going to have a certain algebraic multiplicity because it comes as the roots of us up the characteristic polynomial.
Birne Binegar: that's the algebraic multiplicity the geometric multiplicity, you know as the name implies, it has to do with more geometry, and in fact it's dimension of the corresponding I can space all right, so let me.
Birne Binegar: i'll show you know, in the course today's lecture how these guys come up but let's review what we know so far.
Birne Binegar: You have a square matrix call it a n by N, then.
Birne Binegar: The eigenvector eigenvalue problem for that matrix is finding vectors D and numbers Lambda so, but this condition, where in matrix multiplication by a can be replaced by scale or multiplication by the number of Lambda that's what we're seeking to do, and I should point out here that.
Birne Binegar: If you look at this equation.
Birne Binegar: A given I can vector V that satisfies the equation is only going to do so for one value of Lambda.
Birne Binegar: Because if I change Lambda the entire right hand side of the equation is changing, but on the left hand side is not changing.
Birne Binegar: So there's only going to be one eigenvalue for a given eigenvector In contrast, for a given for a given eigenvalue may have more.
Birne Binegar: you'll have infinitely many eigenvectors because I can take this whole equation, and I can replace the by a scale or multiple itself and it's still going to satisfy the same condition with the same eigenvalue.
Birne Binegar: So it might seem that maybe the first thing we should do is find the eigenvectors and then they'll term determine all the eigenvalues, but in fact we do just the opposite what we do is we figure out the possible numbers Lambda first and then we figure out the eigenvectors that work.
Birne Binegar: Alright, so.
Birne Binegar: Now the nomenclature here is that vectors D that satisfy such an equation are called item vectors and the numbers Lambda appropriate for a given eigenvector it's called the eigenvalue corresponding to that eigenvectors.
Birne Binegar: With one caveat to remember one way to satisfy an equation like one is to set D equal to the zero vector that particular solution is irrelevant.
Birne Binegar: Because, as soon as you substitute zero for the on both sides, you getting a triple equation that just says zero is equal to zero or more precisely the zero vector is equal to zero adapter that's not really a condition.
Birne Binegar: Alright, so here's how we approach the problem, we take the eigenvector condition and we transform it into something a little bit more recognizable in the context of this course.
Birne Binegar: This condition implies that this equation has to be satisfied i'm just taking the lamb to be to the opposite side and then putting in a factor of the identity matrix but then.
Birne Binegar: Think of X is just being the.
Birne Binegar: This condition is equivalent to say that the have to satisfy this linear system, and so the way we actually will find eigenvectors is by solving linear systems, which means reducing a matrix to reduce your national informing etc.
Birne Binegar: But.
Birne Binegar: We can do that, once we know what the eigenvalues are So the first thing we do to solve and eigenvector eigenvalue problem is to figure out the appropriate.
Birne Binegar: Values up and parameter Lambda and.
Birne Binegar: Since we're looking for non zero solutions of this we require the coefficient matrix of this homogeneous linear system, we need its coefficient matrix D have determinant zero.
Birne Binegar: Because back.
Birne Binegar: When we were studying in invert ability of matrices we discovered that.
Birne Binegar: You can have solution to other than X equals zero only when the determinant of the coefficient matrix is equal to zero.
Birne Binegar: Alright, so that led us to the following procedure.
Birne Binegar: given him a creek a we construct.
Birne Binegar: A certain polynomial not from a itself, but what we do, which we subtract from a along its main diagonal a parameter Lambda.
Birne Binegar: Then we take the determinant of that we're going to get a polynomial in all the entries of a but, in particular in this.
Birne Binegar: All along the diagonal you're going to have things that are proportional for terms of Lambda this will end this polynomial is going to end up being a polynomial degree in in that parameters.
Birne Binegar: What we want is that.
Birne Binegar: The determinant of this matrix two equals zero that's going to give us a polynomial condition on Lambda we call the polynomial the characteristic polynomial.
Birne Binegar: And the solutions, a P, a Lambda equals zero those are going to be exactly our eigenvalues, and so our focus is going to be on the equation, we get once we calculate this polynomial and that will be the characteristic equation of the matrix a.
Birne Binegar: By the way, there's something interesting thing that happens here that I don't you don't really see until advanced linear algebra but.
Birne Binegar: When you're looking at solving a polynomial equation you're looking for numbers so once you substitute a solution, a solution in the evaluation of the right hand side produces zero right that's how you know you have the solution, but those polynomial.
Birne Binegar: He sorry.
Birne Binegar: This characteristic polynomial he a Lambda.
Birne Binegar: equals determine a minus Lambda times the identity matrix this is going to be a part A, is n by n.
Birne Binegar: Then this can be a polynomial of degree and she can look like when am to the end some coefficient Lambda N minus one etc.
Birne Binegar: Plus, maybe a one Lambda the first hour, plus a constant need zero so as soon as you have an explicit matrix a you compute this determinant it's going to work out to be.
Birne Binegar: A polynomial degree and and the way you verify that Lambda is actually not good value is you substituted into this polynomial and evaluates zero.
Birne Binegar: Presumably, because some of the signs are different and you get cancellations, but what's actually surprising is instead of just substituting numbers into this equation, you can actually submit to a itself because.
Birne Binegar: Since a is an n by n matrix you can take powers and you get I mean a time hey it's just an n by n matrix times n n by n matrix this matches so that multiplication that's defined.
Birne Binegar: And what you get out of it it's another n by n matrix so rather than substitute numbers into the characters the polynomial it's also possible to substitute a matrix into this polynomial and what you get is a to the power and plus some coefficient i'm a to the power in minus one surprises.
Birne Binegar: This is so called kaylee Hamilton theory.
Birne Binegar: That, if you do so, if you take the characteristic polynomial for the given matrix that's going to determine all these coefficients and instead of substituting a number Landa you.
Birne Binegar: Put the matrix in there.
Birne Binegar: You get zero automatically so, in fact, the matrix satisfies the same equation, as the as the eigenvalues do these are numbers, you know we will use it, but if we simply replace them by the matrix a that defined as polynomial it automatically equals zero.
Birne Binegar: So that's um.
Birne Binegar: that's kind of a big surprise and.
Birne Binegar: But it's not so relevant to this course.
Birne Binegar: What we're gonna do today is just.
Birne Binegar: work out the details of how we find not only the eigenvalues but also the corresponding eigenvectors so what we're doing so far is we're setting up this polynomial.
Birne Binegar: And it's going to have the property that the route to that polynomial the solutions of P subscript a of Lambda equal to zero, those are going to be exactly the eigenvalue so the solution to this equation when we interpret Lambda is a number are going to be the eigenvalues today.
Birne Binegar: Alright, so just a little bit about solving polynomial equations every polynomial degree in can be fact rise.
Birne Binegar: Over the complex numbers, though these this coefficient and these numbers are one through our end, you have to take them as being complex numbers in general in order to achieve this goal factorization.
Birne Binegar: But.
Birne Binegar: When you can achieve this full factorization, then the numbers are one who are, and those are effectively the roots exactly the values of extra which the polynomial vanishes.
Birne Binegar: And so yeah so fundamental fear of algebra just says.
Birne Binegar: In order to be a solution, a P of X equals zero.
Birne Binegar: Given this factorization X is X equals hard only going to be a solution of this equation when X minus R actually divides that polynomial.
Birne Binegar: So the strategy for solving polynomial equations is to work out this factorization.
Birne Binegar: So let's let's do it within a particular example i'm choosing a real simple matrix I don't have to do a lot of row reduction or on yet so here, I take a upper triangular matrix.
Birne Binegar: reason why I did, that is because I want to do a simple determinant calculation when a matrix is upper triangular the determinants just a product of the diagonal elements.
Birne Binegar: So for this particular matrix when I calculate the characteristic polynomial it's going to be trivial, I just have to the matrix remains upper triangular so I just have to multiply along the main diagonal calculate that determining and so what we get.
Birne Binegar: In the end, it's a polynomial that in fact nicely factor is already forth, and we see that the only way this polynomial is going to equal zero.
Birne Binegar: Well, one of these factors has to equal zero so either Lambda has to equal one or Lambda has to equal to Orlando has to equal one again we really only have two distinct.
Birne Binegar: Solutions a P amp a of Lambda equals zero those are going to be the eigenvalues of a will have one eigenvalue one in a second eigenvalue to.
Birne Binegar: All right now.
Birne Binegar: Now i'm going to introduce the first multiplicity it's already apparent.
Birne Binegar: When first let's talk about polynomials in general, when you have a polynomial up degree in and you've got it factor is it best you can, in terms of linear polynomials.
Birne Binegar: you'll end up with something that looks like this, what are these extra elements when i'm assuming here is that maybe some of the.
Birne Binegar: Previous example, maybe some of the factors are exactly the same so i'm going to collect them all together.
Birne Binegar: And so, when I write an expression down like this are one in our two are going to be different numbers, corresponding to different solutions, a P of X equal to zero, and the totality of them are one two rk those are going to be all the solutions of P of X equals zero.
Birne Binegar: We.
Birne Binegar: an algebraic would call these numbers roots of the polynomial.
Birne Binegar: And these exponents here would be the multiplicity of those roots.
Birne Binegar: Okay now linear algebra we're going to have a different kind of multiplicity, so we need a little more specialized language was still going to get a calling the roots of the characteristic equation.
Birne Binegar: route well well identify them as as eigenvalues so I can values are the route to this polynomial but rather than introduce a new route i'll just continue to call them eigenvalues.
Birne Binegar: But these the multiplicity of a route.
Birne Binegar: The number of times, it appears in the factorization of the characteristic polynomial rather than just calling that the multiplicity of that group i'm going to call it the algebraic multiplicity of the corresponding Eigen Daddy So if I say.
Birne Binegar: If.
Birne Binegar: Lambda is.
Birne Binegar: An eigenvalue.
Birne Binegar: A with algebraic multiplicity.
Birne Binegar: let's see em what that means let's make changes to our so if our is an eigenvalue of a matrix a with eigenvalue m What that means is that there was a factor.
Birne Binegar: Of Lambda minus R to the m.
Birne Binegar: and
Birne Binegar: ta of Lambda.
Birne Binegar: Though that that algebraic multiplicity just tells you what the multiplicity of the corresponding route is.
Birne Binegar: Alright, but as i'm hitting there's another kind of multiplicity as well.
Birne Binegar: Alright, so in the last example.
Birne Binegar: I had a specific matrix this matrix and I calculated its characteristic polynomial and what I found is that there that polynomial had two factors are one minus Lambda in a single factor of two minutes.
Birne Binegar: Okay that's one i'm catching up with this and now.
Birne Binegar: Because we have two distinct route Lambda equals one and Lambda equal to those who are going to be precisely or eigenvalues but now I can ascribe to them not only the fact that there a solution of P as Lambda equal to zero, but often I can also identify their algebraic multiplicity.
Birne Binegar: The route Lambda equals one has algebraic multiplicity to the root Lambda equals two has algebraic multiplicity one because it only occurs one time in that factorization.
Birne Binegar: alright.
Birne Binegar: So.
Birne Binegar: We found all possible eigenvalues of the matrix a there's only two of them now we're going to look for the corresponding eigenvectors, and for that.
Birne Binegar: we're going to end up plugging in eigenvalues as numbers, not as parameters and demand that are eigenvector satisfy the Eigen.
Birne Binegar: eigenvalue condition.
Birne Binegar: This space here the space of all vectors with eigenvalue our it's called the are I good space with a and we can calculate it because this condition.
Birne Binegar: Is the equivalent to saying that this matrix a minus R times I times the vector X is equal to zero so looking for the North face of this matrix.
Birne Binegar: And now we can calculate in the usual fashion, we have a coefficient matrix will reduce it to the reducer lesson learned for and then.
Birne Binegar: When we convert back to equations and we get we can write down a typical solution vector and expand it in terms of the three parameters will have certain content factors in that expansion those content vectors are going to be the eigenvectors we seek.
Birne Binegar: Alright, so.
Birne Binegar: Before we do in this example.
Birne Binegar: This the set of all possible.
Birne Binegar: vectors in our end with eigenvalue are for a given make things there's general going to be infinitely many of those.
Birne Binegar: Because, if you have.
Birne Binegar: If you have one vector if D satisfies.
Birne Binegar: A the equals, let me put a specific I can die, you are V, then.
Birne Binegar: So does.
Birne Binegar: Any scale or multiple.
Birne Binegar: Except for example by multiply by two, I can still maintain the same equation, and so.
Birne Binegar: there's always if you have.
Birne Binegar: If you have an eigenvalue there's always going to be infinitely many eigenvectors with that I can tell you, we can't write them all down, but what we can do is instead.
Birne Binegar: Look at the space that generated by the eigenvalues that's going to be a finite dimensional space it's going to have a basis, and so we satisfy ourselves that we've got all the eigenvectors.
Birne Binegar: If we can write down the basis for the corresponding audience base, we think of this as a subspace of art in and we try to find a set of vectors that not completely generates the space, by taking linear combination and which, once we write a vector and be.
Birne Binegar: A vector in the subspace as a linear combination of these guys that the coefficients and expansion of that vector of respected the basis vectors is unique.
Birne Binegar: Alright, so.
Birne Binegar: What we're going to do is once we have the eigenvalues we're going to examine the spaces and we're going to try to find a basis for these sub spaces those basis vectors are what we call the eigenvectors corresponding to that, I do that.
Birne Binegar: Alright, so here's the nomenclature, that will be using.
Birne Binegar: The polynomial we create by shifting a along the diagonal by a parameter Lambda and taking us determine that's the characteristic polynomial as a polynomial degree in in.
Birne Binegar: it's going to have a factorization we call these offset from lagonda the eigenvalues these again are one or two rk are going to be the roots.
Birne Binegar: Of the polynomial and or, in other words they're going to be the eigenvalues of the matrix a the numbers m one m to up to mk telling us how many times, a factor Lambda minus a given eigenvalue occurs in the characteristic polynomial those are going to be.
Birne Binegar: referred to as the algebraic multiplicity eigenvalue so for for the argue value are to the corresponding algebraic multiplicity is this Tower empty.
Birne Binegar: And then distinct from that.
Birne Binegar: or this.
Birne Binegar: The subspace of our end that consist of all possible eigenvectors with a given eigenvalue that's going to be a finite.
Birne Binegar: finite dimensional space of art in we call it the art I in space in a and our goal would be to find a basis for each of these subs basis the basis vectors we're going to refer to as the eigenvectors of a with a particular eigenvalue.
Birne Binegar: Now.
Birne Binegar: If you have a subspace and you have a basis for this up space, the dimension of that subspace is the number of basis factors, you have it's the number of independent directions that span that space so we're going to.
Birne Binegar: grab that bit of data, but we'll call it the geometric multiplicity of an ID value then once I figure out what these spaces are.
Birne Binegar: account the number of basis vectors i'm going to get an integer that i'm going to call the geometric multiplicity of the corresponding eigenvalue so i'm going to use.
Birne Binegar: You subscript I indicate the dimension of the ID spaces and m subscript I to indicate the number of factors of our Lambda minus R I that occurs in the characteristic polynomial.
Birne Binegar: These are the definition she'll need to do.
Birne Binegar: problem set nine on web assign I think you have about six problems from which, in each of them, you have to.
Birne Binegar: calculate the characteristic polynomial find the eigenvalues and then identify both the algebraic and geometric multiplicity of the Oregon value and you're also going to have to find the basis sectors.
Birne Binegar: So let me.
Birne Binegar: Oh well, before I work out some explicit examples, let me make one remark.
Birne Binegar: About algebraic in geometric multiplicity.
Birne Binegar: keep trying keep this up on the blackboard so key a Lambda this polynomial.
Birne Binegar: A minus land I this is used.
Birne Binegar: To find the eigenvalues.
Birne Binegar: and
Birne Binegar: separate from this, we have.
Birne Binegar: Am I, yes, the.
Birne Binegar: number of times.
Birne Binegar: Lambda minus R I occurs.
Birne Binegar: In P, a of Lambda this is going to be the algebraic multiplicity.
Birne Binegar: of our eye and then.
Birne Binegar: distinct from this is the number of you, I this is going to be the number of basis factors.
Birne Binegar: For.
Birne Binegar: The audience based corresponding to that eigenvalue.
Birne Binegar: quibbling dimension of that space.
Birne Binegar: And then, what i'm going to point out on this slide is some connections between these two numbers.
Birne Binegar: The first one.
Birne Binegar: This first identity.
Birne Binegar: restating the definition here's the one identity that when you add up all the algebraic multiplicity they happy.
Birne Binegar: They sum up to a number that's identical, the number of rows and columns of the original matrix that's because.
Birne Binegar: This determinant as the determinant of an n by n matrix is a polynomial of degree in.
Birne Binegar: Now.
Birne Binegar: PA.
Birne Binegar: Lambda equals a polynomial.
Birne Binegar: Up degree in if we started off with an n by n matrix but from algebra, we know that every polynomial has a factorization and we've been writing it now is Lambda minus.
Birne Binegar: or one some power in one.
Birne Binegar: Lambda minus maybe the last eigenvalue arcade or some power mk this polynomial is going to have a complete factorization but it's still going to have to be a polynomial degree in.
Birne Binegar: what's that going to mean what's the highest power of landed that can come out of such a product like this.
Birne Binegar: Well, the leading term of this polynomial is going to be Lambda Lambda to the power in one same thing to a second factor and third factor in the last factor, and then, when I multiply.
Birne Binegar: This is going to look like Lambda to the M one plus some other stuff in this to get a multiply Lambda to the two plus some other stuff once I expand these.
Birne Binegar: Products of Lambda miners are with itself and then, when you multiply these guys together what i'm not writing here are the lower order terms in this expansion, and so this is going to start off as Lambda to the M one plus two, plus am K plus low order terms.
Birne Binegar: This polynomial is of degree and one plus am to up to mk but it's got to be a polynomial degree in that's why those multiplicity is have to add up in.
Birne Binegar: Now is.
Birne Binegar: It turns out that the geometric multiplicity do not have to add up to any longer rather there's a relationship between an algebraic multiplicity.
Birne Binegar: And the corresponding geometric multiplicity there's a relationship between the number of times a given.
Birne Binegar: Factor occurs in the characteristic polynomial and then the dimension of the corresponding I get space.
Birne Binegar: But that relation is a kind of a week one they're not exactly equal and they're not always less than there's just less than equal the geometric multiplicity to always less than or equal to the corresponding algebraic multiplicity so what's going to happen.
Birne Binegar: And this is a good check to make sure you're calculating things correctly, the algebraic multiplicity should always add up to N and but the geometric multiplication of geometric multiplicity are going to add up to a number that's less than or equal to end.
Birne Binegar: Alright, so let's continue with the example.
Birne Binegar: Here was that triangular matrix we started off with we've already identified that the route to the characteristic equation on one and two, and now what we're going to do is for each eigenvalue.
Birne Binegar: One or two we're going to calculate the corresponding Eigen space now that's done by.
Birne Binegar: Effectively solving a homogeneous linear system where the coefficient matrix is matrix you started off with with the art corresponding eigenvalue subtracted along the main diagonal.
Birne Binegar: So here's the first one for the eigenvalue one we should track the number one along the main diagonal that's now explicitly computed this matrix.
Birne Binegar: We want to find in small space, so we do reduction until we reach the matrix in reduce row Echelon form, here it is.
Birne Binegar: The pivots I have to in the first and second row they're both equal to one and above and below the pivot I have only zeros.
Birne Binegar: So the matrix on the third line is the reduce row Echelon form of this matrix one minus the eigenvalue time see identity matrix now.
Birne Binegar: i've got the coefficient matrix and reduce cholesterol and form, I can use this to write down the equations of the solution, and in fact.
Birne Binegar: I can offer us this matrix to interpret the first call corresponding to the variable X one that's that's going to be a free variable and then.
Birne Binegar: The second column in the third column, corresponding to X two X three those are going to be fixed variables, the first.
Birne Binegar: row at this matrix tells me that X two has to equal 00 times X one one times X two plus zero times X three has to equal zero or simply X is equal to zero, the second row tells me X is equal to zero, and the third row tells me that zero is equal to zero or.
Birne Binegar: sort of the triviality so really the conditions on a vector such that live inside this the subspace is that needed second component equals zero the third component equal to zero.
Birne Binegar: But its first component is unfixed because it's a free parameter in the solution, as it corresponds to a call we've had a pivot.
Birne Binegar: And so, our solution vector I have to set X two X three is equal to zero X one was free I expand in terms of the free parameter this guy here that's what you should regard if the I can vector of the original matrix a corresponding to the eigenvalue one.
Birne Binegar: Alright, so i'll write it this way.
Birne Binegar: i'll grab that vector, and here the subscript it's not indicating its position inside this vector rather I want this subscript here to indicate that the eigenvalue corresponding to this vector respect to the original matrix it's eigenvalue is one.
Birne Binegar: All right, and then, since I only have one basis factor my one I didn't space it's going to be one dimensional, and so the geometric multiplicity of the eigenvalue one is one because the corresponding Eigen space is one dimensional.
Birne Binegar: All right, you had a second eigenvalue it was the number two so we'll do the same thing we subtract the number two along the diagonal elements of a creating this matrix and now we calculated, a small space, by reducing the matrix to reduce relational and for.
Birne Binegar: i'm done that, on the third line to reduce your national on form, and so I can begin writing down the corresponding.
Birne Binegar: vectors that live in this no space first row is telling me to X one plus x two has to equal zero second row is telling me that X three has to equal zero on the third row not telling me anything.
Birne Binegar: Until what we get is that.
Birne Binegar: From this coefficient matrix we get these conditions which.
Birne Binegar: When you try to write down a vector that satisfy these three conditions, you can think of X one is being a free parameter X two X three is being the fixed variable and your general solution vector it has this form or simply X two times.
Birne Binegar: This constant vector 110.
Birne Binegar: So this is going to be basis factor for this space it's going to be your your the eigenvector of the matrix a with eigenvalue to.
Birne Binegar: Let me step back a little bit here.
Birne Binegar: You see this zero here.
Birne Binegar: When you're calculating eigenvectors.
Birne Binegar: you're going to construct this coat that like coefficient matrix inserting the eigenvalue of interest into subtracting along the main diagonal and you can reduce.
Birne Binegar: You should always be getting a zero with the bottom, because the way we arrived at this number two was by demanding that determinant.
Birne Binegar: Of this matrix was equal to zero, so when I submit to the eigenvalue into this i've got to get a matrix it's not real reducible to the identity matrix.
Birne Binegar: And so, this is a check your calculation if you're not getting zero at the bottom after you reduced through actual form you've made a numerical error somewhere.
Birne Binegar: right but seeing that zero that's good sign that's good and i'm fat.
Birne Binegar: Once we reach reduce rational and form it's pretty simple to actually write down the solution, and so the basis factor for that I can space corresponding the Lambda equal to two is that vector.
Birne Binegar: Alright, so let me summarize what happened for this particular matrix.
Birne Binegar: Without calculated as characteristic polynomial and we identify the to route Lambda equal to one and London equal to two.
Birne Binegar: Nobody how it factor if we can actually extract some more information, because we have two factors one minus Lambda and the characteristic polynomial.
Birne Binegar: The algebraic multiplicity of the room Lambda equals one it's going to be to that two points there and then we only have one factor of two minus Lambda to the algebraic multiplicity eigenvalue two is one.
Birne Binegar: separate from the algebraic multiplicity we have the geometric multiplicity these these numbers have to do with the size of the Eigen space it's effectively dimension.
Birne Binegar: Because you only have one on one linearly independent eigenvector with eigenvalue one the geometric multiplicity is one and here again we only have one basis factor and so.
Birne Binegar: The geometric multiplicity is again going to equal one, and what I want to point out is those two.
Birne Binegar: identities that I.
Birne Binegar: stated earlier hold for this example.
Birne Binegar: Some of all the algebraic multiplicity says equal the end.
Birne Binegar: We started off a three by three matrix two plus one is equal to three.
Birne Binegar: And then.
Birne Binegar: The other thing I stated about not these multiplicity is that the geometric multiplicity is always less than or equal to the corresponding algebraic multiplicity and D one is less than or equal to two in one is less than or equal to one, so this is all working out according to plan.
Birne Binegar: So let's do another example well i'll pause here your questions.
Birne Binegar: let's do another quick example show you how this these calculations proceed.
Birne Binegar: here's another.
Birne Binegar: Three by three matrix but now i'm going to start off with a more complicated one upper triangular and Nevertheless, I can still go ahead and calculate the determinant of the soap opera so the matrix a minus Lambda times I I didn't do the calculation explicitly here for you.
Birne Binegar: When you're reducing when you're trying to find determinants of major cities that have variables in as entries.
Birne Binegar: you're probably not going to want to do row reduction as a means of calculating determining what I actually did here was a Co factor expansion along that first row so tough determinant of this matrix would be the entry in the one one slot.
Birne Binegar: Times assign factor minus one to the one plus one plus the determinant of the corresponding minor, which is this matrix in the bottom.
Birne Binegar: right hand corner of the matrix and you can see the determinant that two by two matrix is now the difference of the diagonal product it's just going to be two minus Lambda squared.
Birne Binegar: And then have two more terms in the co factor expansion, but they're being multiplied by zero so that's why the calculation is actually done after we calculate the.
Birne Binegar: term corresponding to the one one slot all right, and now we see, we have to eigenvalues either Lambda equals to algebraic multiplicity to or Lambda is equal to negative four with algebraic multiplicity one only one factor of minus four minus Lambda.
Birne Binegar: Alright, so there are eigenvalues.
Birne Binegar: And our algebraic multiplicity or respectively, one for Lambda equal minus four and two for Lambda equal to.
Birne Binegar: All right.
Birne Binegar: let's.
Birne Binegar: we've got the eigenvalues now we'll look at the corresponding Eigen spaces these you're gonna have to do, one by one.
Birne Binegar: We take the first eigenvalue negative four and we'll try to calculate with a basis for this no space, so that means, first of all, writing down the matrix explicitly.
Birne Binegar: Reducing to reduce cholesterol on form and all but what happened in this case and so we'll get the fractions appearing that last slot that these fractions arise sort of inevitably when we try to make our pivots equal to one.
Birne Binegar: So they are even so that the, even though I have fractions appearing in that matrix it doesn't prevent me from writing down the solutions of this matrix times X is equal to zero, and what you'll find is that you have a one dimensional solution space it's going to be exactly.
Birne Binegar: How do I know which one dimensional because I only have one called we've got a pivot that's number of free parameters in your solution set and what we see is that the vector that generates the solution set is the vector minus six seven minus 5771.
Birne Binegar: All right, one year.
Birne Binegar: i've been saying.
Birne Binegar: that once you have an eigenvalue you have influence in many eigenvectors with that eigenvalue because you can always multiply and eigenvector by a scheduler and get another eigenvector.
Birne Binegar: Well, we can use that property to simplify this if I multiply this vector by seven I can convert all these entries into integers and I think that's what web assign likes you to do so.
Birne Binegar: You can either write your answers in terms of fractions or equivalent Lee you can say that the eigenvector corresponding to the eigenvalue negative four is negative six negative, five and seven just by performing a scale multiplication.
Birne Binegar: Okay, but at any rate, what we've done is we've for the eigenvalue minus four we found all the eigenvectors corresponding to that eigenvalue they're all multiples of this specters.
Birne Binegar: All right, we had a second eigenvalue.
Birne Binegar: will do the same thing we subtract the eigenvalues too, so we subtract the number two from the diagonal entries at a then we reduce to reduce show Echelon form in this case, we see that the equations.
Birne Binegar: corresponding to this matrix and reduce us long form or X one is equal to zero X three is equal to zero this last row.
Birne Binegar: It just says zero is equal to zero don't the equations are actually going to be excellent is equal to zero X two is equal to 00 sorry X is equal to zero and X to corresponding to the second column is going to be a free parameter in fact your.
Birne Binegar: solution vector is going to be a multiple of the standard basis vector a corresponding to the.
Birne Binegar: Second call or.
Birne Binegar: corresponding to the variable corresponds to the second column alright, so our Lambda equals to Eigen space is generated by the single vector.
Birne Binegar: Alright, so what happened in this problem here's the matrix here's the characteristic polynomial We calculated and that calculation displayed minus four and two as the eigenvalues of a.
Birne Binegar: More over it shows us that the eigenvalue to have algebraic multiplicity to and the eigenvalue negative for has not algebraic multiplicity one.
Birne Binegar: Once I had the eigenvalues or use that to calculate the corresponding I am faces for the eigenvalue minus four, I had a one dimensional solution set it was generated by this factor.
Birne Binegar: And for the eigenvalue to I also had a one dimensional.
Birne Binegar: solution set it was generated by this factor and now i'm putting everything together in a table the eigenvalue minus four it's.
Birne Binegar: here's the the Eigen space corresponding this eigenvalue is generated by this one factor, because I have one factor of minus four minus Lambda the algebraic multiplicity of that eigenvalue is one.
Birne Binegar: Because I only have one basis specter and the corresponding I can space, the geometric multiplicity of that eigenvalues minus for.
Birne Binegar: The second row we're looking at the Eigen Daddy to.
Birne Binegar: To minus Lambda only occurs twice in the characteristic polynomial so which I can algebraic multiplicity is too.
Birne Binegar: But the corresponding Eigen space turned out to be one dimensional every vector in the Lambda equals two I can space is a skill or multiple of this factor so.
Birne Binegar: The dimension of that I can space, since it generated by a single vector is going to be equal to one and, again, we see that the algebraic multiplicity.
Birne Binegar: Those add up to the dimension of the original matrix one plus one is equal to three because we started off with a three by three matrix the geometric multiplicity however don't necessarily add up the three they only add up to two in this case, but.
Birne Binegar: that's a sort of analysis you're going to do, I think, maybe six times in the problems and holbrooke's at nine each for a given matrix.
Birne Binegar: Where you proceed on the problem set and should calculate the characteristic polynomial it will try to factor eyes it and, once you achieve the full linear factorization, then you can read off the eigenvalues in the algebraic multiplicity.
Birne Binegar: Once you have those you have to individually work your way through you're allowed eigenvalues calculate the dimensions of those no spaces.
Birne Binegar: Those will be the dimensions of the corresponding against basis those dimensions are exactly the geometric multiplicity of the eigenvectors the basic factors themselves are going to be the eigenvectors correspond to the eigenvalues.
Birne Binegar: Okay, so.
Birne Binegar: Oh i'm finishing a little bit early today I have certainly have time for questions.
Birne Binegar: um well, so what we're going to consider next is when can we.
Birne Binegar: When we can we diagonal is a matrix that that's going to be related to the problem of when can I find a good set of coordinates well which my coordinate directions are actually.
Birne Binegar: are actually eigenvectors for the given matrix.
Birne Binegar: And now, what you what you want, is that when you coordinate vector so I conjectures the matrix that means your matrix multiplication by a can just be replaced by scale multiplication everything is going to simplify and so what will happen if you don't if these.
Birne Binegar: What what your county here if you sum up all the geometric multiplicity you're counting up the number of linearly independent eigenvectors.
Birne Binegar: If they don't add up the three that means you don't have enough independent directions in your space to reach every point in your face, because your variables.
Birne Binegar: This basically a problem is taking place in a three dimensional space, but the number of basis vectors i'm getting is only two, so the eigenvalues describes like the shape.
Birne Binegar: calculus three like if it's a quadratic scribe whether it's positive or negative or in determining.
Birne Binegar: You know it's not so much shaped dimension dimension is a better word to think of So these are representing the dimension of the course planning, I could say.
Birne Binegar: What you would like, in an ideal case is to have as many like in spaces many linearly independent eigenvectors, as you have eigenvalues.
Birne Binegar: Then, then what will happen is you'll have enough degrees of freedom to like every.
Birne Binegar: vector inside our three as a linear combination of your three eigenvectors, but if you only have two eigenvectors there's going to be a missing direction and what that's going to cause problems for us later on when we try to find good coordinates associated with a matrix tell.
Birne Binegar: yeah the the practical interpretation of these two numbers is going to come out.
Birne Binegar: it's going to be evident in lectures I do on Friday and Monday.
Birne Binegar: In fact, one of the reasons for being able to calculate these things i'll show you next week is that you can solve differential equations by diagonal eyes and matrices but in order to protect to do it, you have to make sure you have enough independent directions in order to.
Birne Binegar: achieve that goal, and so these numbers will become important but right now they're just a numbers that we can observe from our calculation to I point them out now, so that when we actually need them next week, we know what they are.
Birne Binegar: Okay.
Birne Binegar: So other questions.
Birne Binegar: If anyone has not picked up.
Birne Binegar: The first or second exam You see, I have quite a bit of unclaimed exams i'm no longer going to hold them myself because i'm not often in my office.
Birne Binegar: But I will leave them with a map office, so if you if you here today i'll give it to give them to you, but if you're not here if you're online somehow then.
Birne Binegar: You should come into the math office and pick up your two exams because I guarantee that some of the problems on the final exam are going to come directly off the midterm exams, so you should definitely.
Birne Binegar: pick up your exam so you figure out what your ears for me.