Birne Binegar: couple things.
Birne Binegar: First of all.
Birne Binegar: homework said eight what's due on Monday and courses, just to stay concurrent with lecture homework set nine is cover some material that will finish up today.
Birne Binegar: Finding eigenvectors and eigenvalues in their algebraic geometric multiplicity so all sign it today, I haven't yet posted it to can't do web assign but as soon as I get back to my office i'll do that.
Birne Binegar: And then, it looks like we'll have besides homework set nine it will be last problem said homework problems that can that all signed probably next week, but at any rate i've already posted all the solutions to the web assign problems well.
Birne Binegar: simple solutions, because of course the individual problems that you solve are going to be different numerically than the ones that.
Birne Binegar: That I do in.
Birne Binegar: In my solution, but you can see how each problem has to be approached by looking at the public solutions.
Birne Binegar: Okay, so.
Birne Binegar: What i'm going to do today is.
Birne Binegar: rehash real quickly a discussion of eigenvectors and eigenvalues i'm going to introduce the morn nomenclature.
Birne Binegar: So called algebraic and geometric multiplicity that are associated with and I get I can value and then the main purpose of today's lecture is just to work through some examples where all this stuff is calculated.
Birne Binegar: So.
Birne Binegar: Let me start.
Birne Binegar: by reviewing the so called I can vector eigenvalue problem which begins with a stated matrix called a let's make it specifically in n by n matrix.
Birne Binegar: And the idea is for that given matrix to find vectors D in numbers Lambda so that, at least on these vectors matrix multiplication is equivalent to scale and.
multiplication.
Birne Binegar: Now one thing you should notice here is that.
Birne Binegar: For a given number Lambda.
Birne Binegar: up well forgiven Dexter D there's only going to be one possible choice of a number Lambda because you can't satisfy to.
Birne Binegar: Two equations of this form, where the right hand side changes the left hand side is staying in the same so you can have two different eigenvalues for the same same eigenvector.
Birne Binegar: But the same equation also shows us that if you have one eigenvector and you multiply it by a scale or you know say you multiply by two.
Birne Binegar: To be will still satisfy the same equation, because you end up multiplying both sides of the equation, by the same number, so you have shot of this is for.
Birne Binegar: For any particular eigenvector there's only one eigenvalue but for any particular eigenvalue there may be multiple eigenvectors.
Birne Binegar: So that would suggest that maybe we should find the eigenvectors first and then find the eigenvalues.
Birne Binegar: Because the eigenvalues would be determined by the vectors button, in fact, we do it in the exact opposite way.
Birne Binegar: first thing we do is, we find the allowable I can value and then for each eigenvalue we find the corresponding eigenvector and what will happen is.
Birne Binegar: The solutions of this equation for a given eigenvalue Lambda well that's going to be a substrate so it's going to be an infinite set of vectors in general it's going to be a no space of a particular matrix feedback.
Birne Binegar: So the nomenclature, is the numbers that work in such an equation are called the eigenvalues and the vectors that work are called the corresponding eigenvectors.
Birne Binegar: One thing keep in the back of your mind is that you can always satisfy this equation by substituting the zero vector on both sides.
Birne Binegar: But that's the one solution that we're least interested in, and so the structure of the eigenvector problem is to find non zero vectors that satisfying to see an equation like that.
Birne Binegar: Alright, so the way we solved, it is first of all, we took the condition, where CG and we wrote it in more linear algebraic terms when I phrase, the same condition.
Birne Binegar: But a single matrix times the vector is equal to zero well so here X is playing the role of D this displays the eigenvectors as the solutions of a homogeneous linear system, a certain coefficient matrix times the vector is equal to zero.
Birne Binegar: Until.
Birne Binegar: What we do is.
Birne Binegar: We want more than just a solution of this equation, we want to find non zero solutions, and we have a theorem and then.
Birne Binegar: Actually, the fundamental theorem is linked up in a vertical matrix is said that the only way, you can have multiple solutions of this equation is when the determinant of the coefficient matrix is equal to zero.
Birne Binegar: And so defined non zero factors that satisfy this condition or equivalent equation to you need this determined to vanish and that's going to give us a condition that determines her eigenvalues.
Birne Binegar: Okay, so.
Birne Binegar: Because the the determinant this actually determines whether or not Lambda is an eigenvalue or not having that equal to zero, which.
Birne Binegar: And the fact that a determinant is always a polynomial in the entries at the course by matrix we use the determinant of a minus Lambda is to create a polynomial Lambda that's going to be called the characteristic polynomial of the matrix a.
Birne Binegar: And it's going to depend on a parameter Lambda actually I should call that an indeterminate Lambda.
Birne Binegar: So by indeterminate.
Birne Binegar: Language is such that I don't really care what it is it's just a symbol, and in fact the big surprise is that if.
Birne Binegar: So usual thing you do when you evaluate a polynomial is you take the variable and you substitute a number for it, and then you evaluate it.
Birne Binegar: Actually it's a theorem that instead of substituting in a number for Lambda instead I substitute in the original matrix it satisfies the characteristic equation.
Birne Binegar: So what i'm saying is is that.
Birne Binegar: Right in out this way.
Birne Binegar: If key a of Lambda is defined as the determinant of a minus Lambda times I that's going to equal some polynomial.
Birne Binegar: The Lambda so it's going to be an expression of the form athletes, going to be a money polynomial be Lambda C n plus some coefficient a minus one land the N minus one etc what's a one Lambda.
Birne Binegar: A zero tip that's typical in order ponyo meal a statement is is that.
Birne Binegar: it's called the COPA.
Birne Binegar: I guess it's kaylee serum but I won't name it i'm not sure now the surprising facts is sure yeah you can substitute numbers for this Lambda and then evaluate the polynomial at a point a corresponding that number but what's also true.
Birne Binegar: Is that if you replace Lambda I the matrix these powers are still going to be defined, you can take a square matrix and multiply it times itself and you'll get another square matrix at the same size.
Birne Binegar: The foursquare major see so and make n by n times n by n it's always going to be an n by n matrix so these powers of the matrix will all be defined and what you'll find.
Birne Binegar: Is that this is.
Birne Binegar: The zero.
Birne Binegar: n by n matrix.
Birne Binegar: So that's surprising and said the matrix itself saw it satisfies the same.
Birne Binegar: polynomial equation, as its eigenvalues do.
Birne Binegar: right but don't worry about this, this is just sort of a remarkable fact.
Birne Binegar: That you don't really need unless you go on in linear algebra.
Birne Binegar: Rather, what we'll do is we'll think of beings characteristic polynomials is always depending on a parameter Lambda and that the.
Birne Binegar: The solutions of this polynomial equation, are going to give us the eigenvalues so because it's important we call this equation that characteristic equation for the matrix day.
Birne Binegar: and
Birne Binegar: One thing to note about it is because.
Birne Binegar: it's computed like taking determinant of an n by n matrix it's going to be a polynomial degree in in.
Birne Binegar: Its variable X or Lambda and then.
Birne Binegar: Just some facts about polynomial equations if you have a polynomial equation every polynomial can be factor is over the complex numbers in the sense that.
Birne Binegar: Once you have a polynomial degree in it could be fact drives it's a product of an overall constant and then a bunch of linear.
Birne Binegar: functions of the underlying variable X don't hear one linear function times now they're learning your function times, etc, since this isn't polynomial degree, and you can have exactly in factors here and for each factor.
Birne Binegar: you'll have a corresponding offset from the variable those.
Birne Binegar: offset they're actually defined the solutions, a P of X is equal to zero, so that X equals R is the solution of the equation P of X equals zero only if well if, and only if.
Birne Binegar: X minus R is a factor here and so that's, in fact, how you solve I order polynomial equations you do so by doing a complete factorization and then identifying these offset as a solution.
Birne Binegar: Right so that's why I stay here the strategy for solving.
Birne Binegar: The characteristic equation, which is going to be an example of a poem when your equation is the factor is a polynomial and then grab the root.
Birne Binegar: Alright, so let's do a quick example here's a.
Birne Binegar: Three by three matrix.
Birne Binegar: yeah your three by three matrix.
Birne Binegar: will look for the eigenvalues of a that's going to be the solutions of the characteristic equation.
Birne Binegar: This shifted matrix a minus Lambda times the identity matrix it's just a itself, where we subtracted a parameter Lambda along the diagonal from the upper left to the lower right.
Birne Binegar: And so, for this matrix it's already upper triangular so that determine it's really easy to compute right for upper triangular matrix like a matrix in row Echelon form they determine and it's just the product along that main diagonal.
Birne Binegar: So what do we see well, we have a route Lambda equals one you have another route Lambda equal to two and we have a third route well it's not really a different route.
Birne Binegar: If the Lambda minus one equals zero exactly when Lambda minus one equals zero so, in fact, we have two distinct eigenvalues in this case.
Birne Binegar: For this polynomial.
Birne Binegar: The equals zero we either have to have Lambda equal to one and Lambda equal to remember that this condition what were the where it came from is by demanding that we had non zero vectors for which.
Birne Binegar: Non zero factors that satisfied the eigenvector eigenvalue Paul.
Birne Binegar: Alright, so now i'm going to give you a little bit more detail here in in some nomenclature.
Birne Binegar: So.
Birne Binegar: In algebra so forget about.
Birne Binegar: eigenvalue eigenvalues right now, you see, simply have a polynomial the fundamental theorem of algebra guarantees as it can be factor eyes as a product of linear polynomials so long as you allow yourself to use complex numbers, then you'll always be able to achieve this factorization.
Birne Binegar: And then.
Birne Binegar: In algebra we usually call the solutions of P of X equal to zero, we usually call the root at the polynomial.
Birne Binegar: and
Birne Binegar: We call the number of factors.
Birne Binegar: So here what i've done is i've taken all the factors Lambda minus or one and i've collected them together.
Birne Binegar: And so, our our one in our two aren't in that all these arcades are assumed to be distinct numbers and then what I done in writing it in this fashion, like collected all the similar factors in terms, where I rage Lambda minus a given.
Birne Binegar: route I just select them together in the same set of factors that way I can ascribe to a given eigenvector a certain multiplicity and that's an algebra they're called exactly that this this number in one is called the multiplicity of the solution or one.
Birne Binegar: Or the equation P of X is equal to zero.
Birne Binegar: However, and linear algebra we're going to use slightly different terminology will still be looking at polynomial equations but instead of calling.
Birne Binegar: The solutions of this polynomial equation route well they play the role of the eigenvalues so we'll just call them directly the eigenvalues of the matrix a.
Birne Binegar: And then.
Birne Binegar: Instead of calling them, the number of factors.
Birne Binegar: The number of factors that ever occur for a given eigenvalue rather than calling it the multiplicity of that route, or the multiplicity that eigenvalue will have a.
Birne Binegar: more specific name algebraic multiplicity because it's going to turn out there's a different kind of multiplicity that we can associate to the.
Birne Binegar: Two and given eigenvalue and so here the adjective algebraic Lee is just, meaning that the multiplicity we're concerned with here is coming from this algebraic point of view of what this quick equation.
Birne Binegar: outbreak characterization of this polynomial how many factors of Lambda miners are occur in this polynomial that's going to be exactly, then the algebraic multiplicity.
Birne Binegar: Alright, so i'm going to talk in a couple minutes about different kinds of multiplicity and what's that's going to be it's going to be correspond to the number of linearly independent I can bet you have corresponding to the given eigenvalue.
Birne Binegar: So.
Birne Binegar: let's go back to the the original matrix net example way back here so.
Birne Binegar: It was a matrix we started off with and We calculated it's characteristic polynomial and it turned out to be.
Birne Binegar: Expressed as a product of three linear polynomials you can see that.
Birne Binegar: The eigenvalue Lambda equals, one that factor occurs twice, and so the eigenvalue Lambda is equal to one it's going to have algebraic multiplicity to because there's two factors here the.
Birne Binegar: eigenvalue Lambda equal to only occurs once in the factorization at the characteristic polynomial so we'll say has algebraic multiplicity one.
Birne Binegar: So.
Birne Binegar: Here I just restate, but I just set given this factorization of the characteristic polynomial.
Birne Binegar: We count the number of times for similar factors occur, and then we can relate those to those multiplicity these two particular eigenvalues.
Birne Binegar: Alright, so.
Birne Binegar: Having factor is the characteristic polynomial we identify the eigenvalues of that me drinks what remains to solve the eigenvector eigenvalue problems to find the corresponding eigenvectors.
Birne Binegar: So what do we need ah well.
Birne Binegar: yeah I can vectors corresponding to a given eigenvalue are are going to be all the factors, such that the eigenvector eigenvalue condition is maintained well those factors.
Birne Binegar: In when we collect them together for not only a subspace of our end or subset of art and they're actually a sub space and we call that the audience space a corresponding to the eigenvalue are are more succinctly, the art, I give space, the matrix a.
Birne Binegar: and
Birne Binegar: that's how it should be fine, but this is how we compute it, it turns out that this condition.
Birne Binegar: is equivalent to saying that this coefficient matrix time to D is equal to zero, and so what we do is we solve the linear system with this coefficient matrix in order to identify all the eigenvectors with eigenvalue are.
Birne Binegar: All right, though.
Birne Binegar: there's a little bit of.
Birne Binegar: Problems with my nomenclature, I mean this is this DEMO engineers, this is not a problem, but we tend to be really pedantic in in mathematics.
Birne Binegar: So.
Birne Binegar: When we saw the eigenvector eigenvalue problem.
Birne Binegar: Our we do want to find this this so called Eigen space.
Birne Binegar: But generally speaking, it, this is going to be an infinite number of vectors and we don't want to have to write down an infinite number of vectors so rather the way we solve find the Eigen states we find a basis for and.
Birne Binegar: Health how hard is that well it just means we saw this homogeneous linear system.
Birne Binegar: expand the solution in terms of three variables and then we grab those constant vectors that are being multiplied by the three variables.
Birne Binegar: There are going to be a basis factor, the basis factors for this no space we use that finite set of basis vectors as we think of those as being the eigenvectors of a.
Birne Binegar: When, in fact, what we really need this Eigen space is going to be the set of all possible linear combinations of the basis vectors for this space.
Birne Binegar: Alright, so.
Birne Binegar: that's what i'm saying here.
Birne Binegar: When you if you want to find all the eigenvectors of a matrix a that's kind of it's impossible to write it down because there's infinitely many.
Birne Binegar: But there'll be a finite basis for this space is generated by any linearly independent set of.
Birne Binegar: actors and that basis will tend to call the eigenvectors of a when, in fact, not only will these guys be eigenvectors a but any linear combination of them would also be an eigenvector a with the same eigenvalue.
Birne Binegar: So we abusive language, a little bit these guys are more these guys are basis for all the space of all eigenvectors of a but I shorten that into saying that these vectors are simply the eigenvectors.
Birne Binegar: Alright, so.
Birne Binegar: We have a special name, the number of eigenvectors you have number of.
Birne Binegar: Independent eigenvectors you have, so the basis for this this number of basis factors, you have for the space we call that the geometric multiplicity of the eigenvalues are.
Birne Binegar: So many reviews and the nomenclature.
Birne Binegar: Thus far.
Birne Binegar: Given an n by n matrix a we have a polynomial we can attach to a what it is, is determinant of a minus Lambda times the identity matrix that the characteristic Paul.
Birne Binegar: Paul know meals can be factor is when we do do that to the characteristic polynomial we can collect similar factors together.
Birne Binegar: The number of times the given factor occurs well these offsets from the variable Lambda those are the exactly the eigenvalues and the number of factors that occur of Lambda minus R I that occur in the characters polynomial that's the corresponding.
Birne Binegar: algebraic multiplicity So for me, I can value are to the algebraic multiplicity is the number empty.
Birne Binegar: distinct from this algebraic multiplicity well, once we have an eigenvalue.
Birne Binegar: This is now a specific number, not a variable we have a no space that we can compute by solving.
Birne Binegar: The equation a minus R I times the identity matrix X is equal to zero, will get a bunch of solution vectors.
Birne Binegar: The basis for that Well, this is exactly the ri I can space in a in any basis for this, they are going to be the the what we refer to as the eigenvectors of a went but more properly I could really should call these.
Birne Binegar: vectors a basis for the Roi in space, but we tend to call them simply eigenvectors days.
Birne Binegar: And then.
Birne Binegar: The number of basis vectors you have so i'm not assuming that are is one dimensional could be multi dimensional so if it's multi dimensional having.
Birne Binegar: More than one I get more than one basis sector and we call the geometric multiplicity of a given eigenvalue the dimension of the space.
Birne Binegar: Okay, so that's the nomenclature, and when you do.
Birne Binegar: The web assign problems at nine most of those problems will ask you to find the eigenvalues.
Birne Binegar: determine their.
Birne Binegar: algebraic multiplicity their geometric multiplicity and the corresponding eigenvectors so what i'm going to do now, yes that's basically the substance of all the problems and the problems that nine so.
Birne Binegar: Might notice page, as you work on those problems yeah because that's this page contains all the relevant definitions and all you have to do is compute.
Birne Binegar: Alright, so.
Birne Binegar: i'll see.
Birne Binegar: Before I actually do some examples, let me introduce a little bit some two basic properties of these algebraic multiplicity and geometric multiplicity.
Birne Binegar: So.
Birne Binegar: Remember the geometric multiplicity there's the number of times you have a factor of Lambda minus are eyeing the characteristic polynomial.
Birne Binegar: Now let's go always happen is that the sum of your algebraic multiplicity is always going to add up to the size of up essentially the size of your matrix why.
Birne Binegar: Because it's a is n by n matrix this determinant it's going to be a polynomial degree in so it's going to be a polynomial that look like this, but.
Birne Binegar: it's also going to have a factorization and.
Birne Binegar: When we factor is it.
Birne Binegar: suppose I write it as.
Birne Binegar: Lambda minus 412 some power and one.
Birne Binegar: land or minus arcade or some power mk that we have collected together all the common factors and assign them, thereby assigned multiple algebraic multiplicity to each eigenvalue what's going to happen, though, is that.
Birne Binegar: This polynomial degree in has been completely fact derived by linear polynomials you can see that the degree of the highest term here is just going to be you'll get a.
Birne Binegar: Just use a binomial formula or something to expand it power, the leading term is going to be Lambda to the m.
Birne Binegar: m one leading term of this polynomial is going to be Lambda to the power mk and when I multiply the leading terms of all these polynomials together i'm going to get a total degree in So in fact.
Birne Binegar: These number of these exponent in one through mk they have to add up.
Birne Binegar: They have to add up exactly the end because both of these have to be.
Birne Binegar: degree and polynomials right so that's all this is saying that your algebraic multiplicity always Sunday and here's a fact that's.
Birne Binegar: Not.
Birne Binegar: immediately obvious and it's APP it's a little bit tedious to prove it turns out that the geometric multiplicity have are now.
Birne Binegar: not strictly less than but they're less than or equal to the corresponding algebraic multiplicity.
Birne Binegar: And what what will find any examples that we do today that well the algebraic multiplicity is have to add up through effectively the size of your matrix.
Birne Binegar: The geometric multiplicity do not, they could know when add up the geometric multiplicity these the dimensions of the corresponding I in spaces, they might not add up to the end and that condition is going to be.
Birne Binegar: relevant to the next topic we consider which will be dying realization of matrices.
Birne Binegar: Alright, so let's continue with this same example I started off with.
Birne Binegar: This three by three matrix for which we've already computed this characteristic polynomial and saw that it factor is this Lambda minus one I think squared times Lambda minus two.
Birne Binegar: And what we'll do for each of these eigenvalues will now determine the corresponding spaces right, so he subscript one by mean be I can space of a with eigenvalue one.
Birne Binegar: So what it is it's a no space with a certain coefficient matrix that I create by subtracting a given eigenvalue from a diagonal entries of the original matrix.
Birne Binegar: Right so here, it is now not again now, I have an explicit matrix notice all the variables have disappeared, and we can compute this.
Birne Binegar: This.
Birne Binegar: This no space fairly simply what we do is we.
Birne Binegar: rub reduce the coefficient matrix.
Birne Binegar: To reduce rotational and form, and then we try to solve the linear system associated with this.
Birne Binegar: matrix and we do throw Echelon form but what's going to happen.
Birne Binegar: This first row it's going to give us an equation zero times X one plus one times X two plus zero times X three has to equal zero or simply X equals zero second row with the.
Birne Binegar: We do show national on form it's going to give us the condition that X three has to equal zero and this last row simply that's redundant or trivial it's just a zero is equal to zero.
Birne Binegar: In this first column there's No Entry so there's no pivot in this first column, that means our first variable X one is actually going to be a free variable in the solution.
Birne Binegar: So we can now write down the solution we know that X two has equals zero X three x equals zero and X one is arbitrary, so my solution vector.
Birne Binegar: has to have.
Birne Binegar: has to have this form.
Birne Binegar: And then expand in terms of the free parameter here's my basis sector and so i'm going to call this constant vector the Eigen that vector of the original matrix a corresponding to eigenvalue one.
Birne Binegar: Because I only have.
Birne Binegar: One free parameter my organ space is actually one dimensional and so.
Birne Binegar: i'll say that the geometric multiplicity of that eigenvector, which is actually the dimension of the corresponding Eigen space that's going to be one.
Birne Binegar: Alright, so that was one worked out in detail for one of the two eigenvalues have a we had a second eigenvalue which was London equal to two.
Birne Binegar: And we'll do a similar calculation, we first of all need to figure out this island space it's going to be exactly the no space of the coefficient matrix we get by subtracting to from the diagonal elements of a.
Birne Binegar: Oh there it is it's really simplified and then after doing row reduction of this matrix to reduce row Echelon form right here's a pivot pivot.
Birne Binegar: Above and below pivots I have nothing but zeros and the pivots or equal to one oh, this is the reduce row Echelon form of effectively that matrix and.
Birne Binegar: We now go back to equations to write down typical solution vectors first row is telling us X one minus X two X equals zero second row is telling us that extra half that you put zero.
Birne Binegar: Well we'll do is typical X two corresponds to a call with add a pivot so that's going to be my free variable and the solution and X three and X one are going to be fixed variables.
Birne Binegar: that's going to lead me.
Birne Binegar: When I express all the components in terms of free variables leave me with this expression until I have any eigenvector.
Birne Binegar: Have a with eigenvalue to has to be a multiple of this vector 110 here's the corresponding eigenvector.
Birne Binegar: And in fact notice that, since I only have one free parameter my Eigen space is going to be one dimensional and so again, for the second eigenvalue landing equal the to the geometric multiplicity the dimension of the Eigen space is going to be wanting again.
Birne Binegar: But.
Birne Binegar: The constant vector that's being multiplied by the free parameter we now interpret that as the.
Birne Binegar: I can vector of the original matrix a corresponding to the eigenvalue Lambda equal to two.
Birne Binegar: Alright, so for this particular matrix you calculated the characteristic polynomial and as soon as we factor eyes it, we can identify the eigenvalues and their algebraic multiplicity.
Birne Binegar: That becomes.
Birne Binegar: Those two items to come immediately apparent once you talk like the characters the polynomial in factorization.
Birne Binegar: But then.
Birne Binegar: It took more work to actually figure out what the corresponding eigenvectors word that's when we actually had to sell some some linear systems with specific coefficient matrix but what way it worked out is.
Birne Binegar: corresponding to this eigenvalue Lambda equals one it had algebraic multiplicity to, and then the dimension of the corresponding Eigen space turned out to be one dimensional it had a single basis director and so.
Birne Binegar: The first row tells us what we've learned about.
Birne Binegar: properties of the value Lambda equal to one there that one minus Lambda occurs twice in the in the characteristic polynomial but the solution set the course of corresponding to the corresponding Eigen space is only one dimensional until that's why.
Birne Binegar: The geometric multiplicity is only one, so what this number is here it's the number of vectors you you found when you found the basis for the corresponding audience base.
Birne Binegar: For the eigenvalue to well it's algebraic multiplicity was one because there's only one factor to minus Lambda in the in the characteristic polynomial when we.
Birne Binegar: found that the.
Birne Binegar: And we actually found the eigenvectors corresponding to the eigenvalue land equal to, we only found one linearly dependent on this one, and so the geometric multiplicity the eigenvalue to is one.
Birne Binegar: notice that those any qualities that I stated.
Birne Binegar: That.
Birne Binegar: If I if the algebraic multiplicity.
Birne Binegar: of an icon value.
Birne Binegar: or I in meal I equals the dimension of the corresponding Eigen space.
Birne Binegar: called the geometric multiplicity.
Birne Binegar: Of the corresponding eigenvalue notice that.
Birne Binegar: For both these eigenvalues ui is less than or equal to me.
Birne Binegar: Because one is less than two and one is less than or equal to one so that that I.
Birne Binegar: property is been checked and then notice also that when I add the algebraic multiplicity two plus one equals three and that's after exactly equal to the size of the original matrix and that's the second identity, I gave you can you sum up.
Birne Binegar: All the multiplicity you get exactly effectively the size of the matrix you started off with this matrix a.
Birne Binegar: dope both those properties have been confirmed in this example.
Birne Binegar: Alright, so let's do a second example or will essentially do the same calculation on a different matrix and i'll be doing less exposition except for the fact showing you how the calculations flow.
Birne Binegar: So let's start off with this three by three matrix.
Birne Binegar: and
Birne Binegar: What we'll ask you to do is determine not only the eigenvalues in the audience.
Birne Binegar: But also the algebraic and GMO multi multiple geometric multiplicity of each eigenvalue.
Birne Binegar: So we begin by calculating the characteristic polynomial we shift the matrix a along its main diagonal by a parameter Lambda and calculated to determine.
Birne Binegar: Now here.
Birne Binegar: You want to keep Lambda as a young as a variable and so rubbery option is probably not the method you want to do to calculate this determinant better to use co factor expansions so that.
Birne Binegar: You don't have to worry about how do I divide by this so that I get rid of this this pivot here don't use row reduction in each calculation rather look for a Co factor expansion and, in this case I do a Co factor expansion along that first row.
Birne Binegar: Have minus Lambda minus four and then that then the the other two terms on that first row are not going to contribute.
Birne Binegar: And so what i'll need is minus four minus Lambda times the determinant of this minor but it's pretty clear that the terminal that minor.
Birne Binegar: This is 02 minus Lambda squared minus zero, and so, not in calculating this determinant using the Co factor expansion, not only did I get a polynomial but I got a polynomial that was actually pre factor is for me.
Birne Binegar: that's a good thing.
Birne Binegar: Soon, as I calculate this polynomial i've seen how it factor ISIS and I can identify the root i'm going to need Lambda equals minus four.
Birne Binegar: or Lambda equal to two if this polynomials can n equals zero so Those are our two eigenvalues.
Birne Binegar: We can also see that the eigenvalue negative for as algebraic multiplicity one if there's only one factor of minus four minus Lambda.
Birne Binegar: While the eigenvalue Lambda equal to two it occurs twice in the characteristic polynomial so the algebraic multiplicity of the eigenvalue to is going to be two.
Right.
Birne Binegar: you're right you're right that's absolutely true, should have been positive for.
Birne Binegar: No wait a minute.
Birne Binegar: We need is minus Lambda minus four equals zero minus Lambda is equal to four or Lambda is equal to negative for.
Birne Binegar: Do we need this factor to vanish so you said, the entire thing equal to zero, and then you solve for Lambda.
Birne Binegar: yeah so you said.
Birne Binegar: multiplication right So if I had like fraction multiplied the yes, yes yeah and what you'll find particularly with your web assign problem they might actually demand that you put in some standard form, so we can recognize it.
Birne Binegar: yeah and so yeah it's often the case, so that it takes easier, right down to if you if you have a common factor like one third, and each term and multiply the whole vector by three and then you only have integer entries.
Birne Binegar: yeah so you can you can.
Birne Binegar: freely adjust your patients, doctors, may not require them to have a specific link what we might like this, then be easy to write down.
Birne Binegar: Okay, so in fact yeah yeah that's what actually happens in this case here.
Birne Binegar: For this eigenvalue Lambda equal to negative four.
Birne Binegar: finalists corresponding Eigen space, I have to create the coefficient matrix.
Birne Binegar: By subtracting the eigenvalue from the diagonal elements of a that produces this matrix and then you got to do the rigmarole of solving the.
Birne Binegar: homogeneous linear system with this coefficient matrix and I don't do the recovery option, but this matrix row reduces to this matrix and reduce show Echelon for.
Birne Binegar: Making these pivots equal to one kind of forces you to take these guys to be fractions but as far as writing down the eigenvector corresponding to the eigenvalue negative for.
Birne Binegar: You can certainly use.
Birne Binegar: This this factor is an icon vector of a with eigenvalue negative four but so is um let's see I want to get rid of this denominator so i'll multiply the whole eigenvector by seven and i'll get minus six minus five and seven, this is also.
Birne Binegar: What we would call the.
Birne Binegar: I vector.
Birne Binegar: A with eigenvalue.
Birne Binegar: minus four, even though this definite article is really not appropriate.
Birne Binegar: Any scale or multiple either one of these doctors is still going to be in it, and I can back to eigenvalue.
Birne Binegar: Okay, and frequently when you're doing web assign problems you're probably most likely to get the answer that web assign.
Birne Binegar: Once by if you have.
Birne Binegar: fractional entries in your eigenvectors to just sort of find a common factor that will make all the components integers and then feed that in.
Birne Binegar: Okay.
Birne Binegar: Well that's the first.
Birne Binegar: that's.
Birne Binegar: The analysis of that first eigenvalue.
Birne Binegar: We had a second eigenvalue Lambda equal to two so let's calculate the corresponding Eigen space.
Birne Binegar: Oh, here it is in the calculation of sense it's the no space of the matrix we get by subtracting the number two from the diagonal elements of a which attract the eigenvalue along that diagonal we get this matrix and.
Birne Binegar: Once I reduce through ocean one form, I get this matrix, by the way this is.
Birne Binegar: Something I should point out.
Birne Binegar: When i'm calculating these I can spaces.
Birne Binegar: Once I do the row reduction to rely on form, I got a zero at the bottom you'd really need that zero row.
Birne Binegar: If you're trying to find.
Birne Binegar: Is no space or trying to find the corresponding Eigen space if, after the reduction of the coefficient matrix to reduce your national on form you're not getting into a row that means you've made a mistake you've made a calculation error, because.
Birne Binegar: The reason why we chose this specific vendor number minus four is because, for that particular number the determinant of this matrix is equal to zero, and that means that.
Birne Binegar: Look along this main diagonal the determinant this matrix is one times one time zero, we need that factor of zero here so that determine is this matrix.
Birne Binegar: is actually equal to zero, so that we have non trivial solutions with the corresponding linear system, so one check that you'll always find early on in these calculation of eigenvectors is.
Birne Binegar: You choose and eigenvalue you construct the matrix and then you try to calculate it no space, so you row reduce to reduce your left one for you always have to have at least 100 at the bottom otherwise you've made a mistake somewhere.
Birne Binegar: Alright, so let's see here here's.
Birne Binegar: The two Eigen space with the matrix a it's an old space of this matrix after we've reduced to.
Birne Binegar: reduce your national on form it's the no space of this matrix and from this, we can read off the eigenvectors.
Birne Binegar: What we need is when the first row it says that X, one has to equal zero the second row with this matrix when you do show Echelon form is telling me that X three x equals zero and the variable corresponding to the second column where there's no pivots that's left free and so.
Birne Binegar: No space of this is generated by this single vector.
Birne Binegar: That single factor, then I can vector of the matrix a with eigenvalue to.
Birne Binegar: Alright, so.
Birne Binegar: summarizing this calculation here's the original matrix We calculated characteristic polynomial and observed that we have to eigenvalues Lambda equal to negative for and Lambda equal to to.
Birne Binegar: The eigenvalues Lambda equal to two as algebraic multiplicity too because there's two factors of two minus Lambda in the characteristic polynomial.
Birne Binegar: The eigenvalue Lambda equals four negative for has only one factor and the corresponding factor in the characteristic polynomial so which algebraic multiplicity is one.
Birne Binegar: Once we calculated the eigenvalues We then proceeded to analyze the corresponding matrices and determine dimensions of their no space.
Birne Binegar: Those basis factors, so the know spatial the corresponding coefficient matrix disease are exactly these guys.
Birne Binegar: Since we only have one basis factor for the eigenvalue minus for y'all geometric multiplicity is one and again for the eigenvalue Lambda equal to two we only have one day inspector that's why we have geometric multiplicity of the argument to equal to one.
Birne Binegar: Let me do.
Birne Binegar: One last trivial example now I suppose I have.
Birne Binegar: The matrix.
Birne Binegar: The identity matrix.
Birne Binegar: And let's calculate the eigenvectors in eigenvalues of the identity matrix three by three identity major determinant one minus Lambda times I it's just going to be the determinant of one minus Lambda 0001 minus time to 0001 minus Lambda and what's going to happen here is.
Birne Binegar: The.
Birne Binegar: determinant it's going to be exactly one minus land, the third, and so this matrix only has one eigenvalue it's eigenvalue is going to be Lambda equal to one and it has.
Birne Binegar: algebraic multiplicity Joe.
Birne Binegar: eigenvalue.
Birne Binegar: equals one and it has algebraic multiplicity.
Birne Binegar: Three What about the dimension of the corresponding Eigen space, so the one I can space at the identity matrix it's going to be the no space.
Birne Binegar: Of.
Birne Binegar: One minus one times the identity matrix it's just going to be the null space.
Birne Binegar: Well, as soon as I subtract one from the diagonal here it's just going to be the zero matrix.
Birne Binegar: And the zero matrix time to any vector there's always equal to zero, so, in fact, every director is an injector of the identity matrix well, you could have seen that.
Birne Binegar: Simply from that identity.
Birne Binegar: The identity matrix time T is equal to be well think of this as one time Steve and now I displayed one is the eigenvalue.
Birne Binegar: Any vector V, with respect to the again, and she knows well, this is all sort of self consistent in this example, though.
Birne Binegar: I what I see is that the one I can space it's going to be three dimensional vector it's going to be a three dimensional vector space, because any vector in our three is going to satisfy this condition and so for the identity matrix.
Birne Binegar: The algebraic multiplicity is just going to be equal to the size of the matrix while the geometric multiplicity of the eigenvectors is also going to equal.
Birne Binegar: The dimension of the.
Birne Binegar: The size of the matrix also going to equal three in this case.
Birne Binegar: Okay, so.
Birne Binegar: I mean do some more examples on Friday, but.
Birne Binegar: If you can now be.
Birne Binegar: done homework said eight you can compute determination you should be able to work through all the problem in homework set nine as well, and if you do so before Friday, then I can, if you've run into any difficulties I can comment on that one last thing.
Birne Binegar: I still have a bunch of uncollected exams if you haven't haven't picked it up, yet you show now if otherwise what i'm gonna do is i'm gonna because i'm sort of sporadically on campus otherwise i'm getting to leave them in the math office, all you would have to do is go to.
Birne Binegar: math science computer science building right next to wipers go to the front desk and ask them for your exam and they'll they'll be able to give it to you.