Birne Binegar: yeah.
Birne Binegar: today.
Birne Binegar: I was revising it just before lecture today anyway so on i'll soon as I get back to my office all i'll post it yeah yeah sorry about that.
Birne Binegar: yeah I was actually working on it right up to the last minute.
Birne Binegar: Revising the presentation.
Birne Binegar: yeah.
Birne Binegar: Okay, so.
Birne Binegar: Okay, so um.
Birne Binegar: let's see.
Birne Binegar: Whether sign problem set eight was do today, I mean do, in a sense that i'm hoping you'd be able to compute determinants so you can follow.
Birne Binegar: computation of eigenvalues eigenvectors today's lecture but do try to do our problem said eight.
Birne Binegar: If not tonight relatively soon, so that at least the calculation of determinants is you know, several methods for doing so today what we're going to do is discuss an application that determinant.
Birne Binegar: Really, a principal application in this course, one of the main reasons you're in this course, presumably is because your APP your.
Birne Binegar: Relative disciplines.
Birne Binegar: need you to understand how to solve eigenvector eigenvalue problems i'm going to stay this issue mathematically but.
Birne Binegar: Towards the end i'll give applications on it and the applications are really quite different than what you might expect here for the first time we have an application of linear algebra which.
Birne Binegar: can only be solved using linear algebra but it's surprising in the context of the problem you're solving i'll give you an example, you know i'll show you how it's related to.
Birne Binegar: Like vectors and eigenvalues later, but if you have a three dimensional object and you toss it in the air it's been tried and, in fact, you can break that motion down into you know.
Birne Binegar: rotation one way rotation and other rotation, I went through our fb will be three separate axis of rotation that are natural for that object.
Birne Binegar: engineers need to do to understand that, so they can see how ports balance on a particular object well the way you can have a oddly shaped object like this desk or chair.
Birne Binegar: Finding those natural axes rotation, how did, how could you possibly do that it's going to turn out that those natural axis of rotation are determined by.
Birne Binegar: If I can vectors of a particular matrix and in fact the corresponding eigenvalues are going to be the moment of inertia around those particular axes of locations so.
Birne Binegar: there's a physical context in which the I vector eigenvalue problem mental state today has a direct application, but.
Birne Binegar: What will work with today is just the linear algebra aspects of it not try to do applications later, you also can use eigenvectors in eigenvalues to solve differential equations it's a.
Birne Binegar: pretty important concept in this course, so I want to do it properly so we'll start by just stating what the mathematical problem is and later in the week we'll get the applications.
Birne Binegar: So begins with a square matrix, and so let a B and N by n matrix what we're looking for our numbers Lambda and vectors.
Birne Binegar: That can be acted upon by that matrix O n by n matrix vector and N dimensional space and what we want to do is this peculiar identity, we want matrix multiplication of the vector to be equivalent to scale and multiplication of the vector by a number.
Birne Binegar: So right away there's sort of a contradiction you doing two different things to that vector one is matrix multiplication the other scale and multiplication and we're demanding that those those two actions on D are actually equivalent to one another.
Birne Binegar: When that happens.
Birne Binegar: This is only going to happen for special directions such directions are called I can vectors of the matrix and the corresponding numbers are going to be called eigenvalues.
Birne Binegar: So here's an example.
Birne Binegar: here's a two by two matrix here's a two dimensional vector when I multiply the vector by the matrix and I calculate I get a vector for four.
Birne Binegar: Which is just the scale of multiple by the number four of the original vector so here's an example of I can vector of the matrix a with eigenvalue for.
Birne Binegar: So that's the kind of phenomenon we're going to try to reproduce and find conditions for when that can happen.
Birne Binegar: So again, the nomenclature is.
Birne Binegar: vectors that satisfying equation like this are called eigenvectors it's only going to work for particular value or particular numbers Lambda.
Birne Binegar: Those particular numbers but which such equations have solutions those have going to be called eigenvalues, and here I want to stress here.
Birne Binegar: That we're only interested in solutions where the vector V is not equal to the zero vector.
Birne Binegar: Because the zero vector will work you just put the zero vector from both sides, you have an identity, but that identity is just zero vector is equal to zero vector once you calculate so it's not really.
Birne Binegar: much information you're going to extract if you have a zero vector here we're interested in the case when we have non trivial solutions of that equation.
Birne Binegar: Alright, so.
Birne Binegar: Some nomenclature here.
Birne Binegar: let's assume we have a number Lambda so that we do have vectors that satisfy this equation, then we can find all the factors were given Lambda that satisfies the equation, those are going to be called I can vectors and effectively what the Lambda I can stace.
Birne Binegar: it's going to be the set of all vectors for which Lambda is an eigenvalue of.
Birne Binegar: All vectors D that's that.
Birne Binegar: We say they have eigenvalue Lambda so if the collection of all vectors it once you fix Lambda this collection of all vectors that satisfy that condition.
Birne Binegar: We call that the land eigenvector stage and now what i'm going to do is massages a little bit so we get it into a more familiar setting more familiar setting inside linear algebra so here's the condition we want matrix multiplication to be equivalent scalable nation.
Birne Binegar: But what i'm gonna do is i'm going to take this term, to the opposite side of the equation, and then this combination of terms.
Birne Binegar: i'm going to try to get a common factor of be so when I do is I insert the identity matrix in before that schema multiplication.
Birne Binegar: And then, I have two matrices a Lambda times I times the I factor is that and here's the equivalent condition this condition is equivalent of saying that a minus Lambda times the identity matrix when it axon I can vector I get zero.
Birne Binegar: So that makes it a little bit clearer what we're trying to do linear algebra.
Birne Binegar: To solve this to find doctors D that satisfying equation like this well it's the equivalent of finding solutions up an equation like this, which is a homogeneous linear system.
Birne Binegar: A coefficient matrix times a vector of unknown variables is equal to the zero vector homogeneous just means we have the zero vector on the right hand side.
Birne Binegar: Okay, so what you should notice here is that that coefficient matrix it's not a itself it's such a kind of shifted version of the matrix a what we're doing here so effectively subtracting from a.
Birne Binegar: This matrix in its totality is the matrix a but we're along the main diagonal i've shifted the entries by an amount Lambda.
Birne Binegar: Lambda shifting matrix is going to end up being the principal calculations optics in the in the discussion will be more often than not, will be looking at this matrix rather than the original matrix a even when we're concerned with the eigenvectors and eigenvalues of a.
Birne Binegar: Alright, so in terms of the shifted matrix.
Birne Binegar: That the Lambda Eigen space.
Birne Binegar: Of a matrix a is exactly the no space of the shifted matrix.
Birne Binegar: So here we we can least see how to calculate all the eigenvectors would argue value a we create this matrix and we find it no space, in other words we solve this linear system.
Birne Binegar: In the usual fashion, take this matrix interpreted as a coefficient matrix reduce it to reduce your question one form.
Birne Binegar: separate the three variables from the fixed variables and write down the solution to expand in terms of the free variables, etc, identifying the Eigen spaces as a no space of the matrix Well, now we know how to do the calculation that will exhibit this explicitly.
Birne Binegar: Alright, so.
Birne Binegar: The main thing we need to do, though, is that for most values of Lambda it's going to turn out that they're, the only solution of an equation equation like this it's going to be the zero vector.
Birne Binegar: And so, the first thing I need to talk about a little bit is.
Birne Binegar: Well, first of all let me.
Birne Binegar: step back a little bit.
Birne Binegar: This.
Birne Binegar: Lambda Eigen space, the set of all possible vectors that have eigenvalue Lambda that of all possible vectors that satisfy.
Birne Binegar: eigenvector eigenvalue problem.
Birne Binegar: Having identified that as a no space of a particular matrix well that identifies it as a subspace right, the solution set of a homogeneous linear system is always the subspace but what i'm gonna do is i'm going to prove this lemma more directly.
Birne Binegar: Just so you see how this apparatus is working.
Birne Binegar: So the claim is.
Birne Binegar: Once you have an eigenvalue and you.
Birne Binegar: Look at the set of all possible solutions and a minus Lambda I X is equal to zero, creating the subspace that it's well creating that set of vector she's actually have subspace.
Birne Binegar: let's check that real quickly closure under scaling multiplication.
Birne Binegar: Suppose.
Birne Binegar: supposed to be lives in the Lambda Eigen space let's keep this story up here.
Birne Binegar: He Lambda equals.
Birne Binegar: keep everything up here same time.
Birne Binegar: So say is a square matrix.
Birne Binegar: and
Birne Binegar: Lambda.
Birne Binegar: it's going to be the set of vectors.
Birne Binegar: are in such that.
Birne Binegar: A times X.
Birne Binegar: I get vector of a eigenvalue Lambda, which is the same thing as saying.
Birne Binegar: The vectors.
Birne Binegar: are in such that a minus Lambda times the identity matrix times X is equal to zero.
Birne Binegar: And you know.
Birne Binegar: The show that this is actually a subspace is a set of vectors the show that, as the substation, we need to show which closed under scale or multiplication in fact tradition so suppose.
Birne Binegar: Suppose a B is a vector that lives inside this set.
Birne Binegar: The.
Birne Binegar: Lambda means that.
Birne Binegar: A times B is equal to Lambda times D.
Birne Binegar: Well, so that's what i'm going to assume about this vector V let's look at the scale or multiple of it so.
Birne Binegar: Because killer multiplication commutes with matrix condition, I can pull that factor through, but then sent a B is an eigenvalue of a with eigenvalue Lambda a B is equal to Lambda be.
Birne Binegar: And then I can just reverse the order of the scale or factors and you see now that a time scale or multiple the is equal to Lambda times the scale or multiple of the so K times T satisfies the same condition.
Birne Binegar: Acting on it my matrix multiplication is the same as scale multiplication by Lambda and so take time to be.
Birne Binegar: Long to the set so the set is closed under stealer multiplication.
Birne Binegar: Similarly, if you have two vectors that start off in a Eigen space you add them together well, you can distribute that matrix multiplication across the vector some.
Birne Binegar: And, in each case, since our hypothesis is that V1 and V2 or eigenvectors with eigenvalue Lambda we can replace a times, for example the one with Lambda times.
Birne Binegar: And then we can pull out the common factor of Lambda and we see that a times this factor is it exactly equal to Lambda times the same director, and so the one plus the two is also living in that space.
Birne Binegar: having shown that his coat on their skin location closed under a vector addition we confirm that the subset of our end is actually a subspace of.
Birne Binegar: All right now, so this.
Birne Binegar: This situation.
Birne Binegar: it's going to.
Birne Binegar: bring up.
Birne Binegar: Some some semantic problems, a little bit because.
Birne Binegar: The way people are used to speaking when they are looking for eigenvectors is well what's going to happen.
Birne Binegar: If you have an eigenvector.
Birne Binegar: it's going to live in a subspace and so you're always going to have well there's only one subspace it only has one vector that's the zero vector.
Birne Binegar: So that's the subspace that he says only up the zero vector that is a subspace but it's the one where least interested in more generally, if we have a subspace of a RN.
Birne Binegar: Well zero dimension is just that point at the origin, but all the other substances are one dimensional two dimensional three dimensional higher and so they contain an infinite number of vectors.
Birne Binegar: What we're really going to be looking forward, we saw when we speak of finding the eigenvectors of a matrix is.
Birne Binegar: Not a single eigenvector but generally will get if we find any non trivial solution we're going to have influence in the many other solutions.
Birne Binegar: But they're all going to lie in the same subspace and so actually what we try to find when we specify this I get space is we try to find a basis factors we try to find a basis for this subspace what you'll find when you leave this course.
Birne Binegar: Professor paper is talking about finding the eigenvector of this matrix what they really mean is the.
Birne Binegar: One I can vector and maybe from the others, you can generate the rest of the space so typically when you say you solve the eigenvector problem it's not that you find write down every Eigen factor for that you'd have to formulate it as a as an infinite set like this, but usually.
Birne Binegar: sort of slightly from a mathematical point of view, what you really want to do is find a basis for this subspace.
Birne Binegar: And the normal common nomenclature is called that basis the eigenvectors corresponding to that eigenvalue.
Birne Binegar: Okay, so let's read review the problem here the eigenvector eigenvalue problem is the problem, finding, first of all, the numbers, such that.
Birne Binegar: The spaces.
Birne Binegar: Have are more than just the zero vector we're looking for non trivial solutions, so our first condition is we're looking for, I can vectors with a vector here is not the zero vector that's too easy you know it doesn't do as much good we're looking for non trivial solutions of that equation.
Birne Binegar: And then.
Birne Binegar: We will proceed as well first figure out which numbers will work for us.
Birne Binegar: But what numbers Lambda do we have more than one solution of this condition or its equivalent condition here.
Birne Binegar: And then, once we find the eigenvalues well that's going to give us.
Birne Binegar: Particular numbers now they're going to be values it's not going to be a parameter that I did note by Lambda it will be a specific number, maybe the eigenvalues two, three or four.
Birne Binegar: matrix a generally always going to be a numerical matrix so once I replace Lambda here thought of initially as a parameter want to replace it by a number this becomes a concrete numerical matrix for which we can.
Birne Binegar: calculate with.
Birne Binegar: they'll.
Birne Binegar: A first step before we can actually solve find the eigenvectors to find the appropriate Eigen.
Birne Binegar: Then, once we find in eigenvalue then this becomes a concrete problem, this will be a particular number, and so this.
Birne Binegar: thing inside parentheses it's just going to be a pure numerical matrix, and so we can solve that linear system to find the corresponding eigenvalues notice what's important here is we don't solve the system, corresponding to a itself, but rather the shifted version of a.
Birne Binegar: Alright, so.
Birne Binegar: let's turn to the first part.
Birne Binegar: Of the eigenvector eigenvalue problem.
Birne Binegar: With a goal is to figure out what numbers Lambda.
Birne Binegar: Do we have Eigen spaces, which are more than one consist of more than just the zero vector we want to have non zero vectors that satisfy that condition.
Birne Binegar: Alright, so.
Birne Binegar: I take the eigenvalue problem I rephrase it into a form that's more familiar to us here's the eigenvector eigenvalue condition expressed as a homogeneous linear system.
Birne Binegar: And what we're looking for our solutions of this system for which there's a solution other than X equals zero.
Birne Binegar: Right So what does it mean.
Birne Binegar: How do we impose this condition, how do we make sure that the solutions of this linear system have solutions other than a trivial solution.
Birne Binegar: Well, we have to step back to just before the first midterm what we discussed the fundamental theorem of convertible matrices it told us that anytime we had a square matrix, then the following statements or equivalent.
Birne Binegar: That a has an inverse that every linear system, a X equals B has a unique solution that X equals zero so unique solution to eight X is equal to zero.
Birne Binegar: That the row reduce cholesterol on form at the matrix is the identity matrix and as we've seen in last week's lecture that the terminal a sorry that should have been not equal to zero.
Birne Binegar: turn a not equal to zero guarantees that a matrix has an inverse now those first 14 terms.
Birne Binegar: We prove this right before the the first midterm and what it had to do with the way we connected, the three statements, is it all had to do with essentially where the pivots live in a reduced row Echelon form of a matrix.
Birne Binegar: First of all, a matrix is only going to have an inverse when it can be reduced to the identity matrix that's effectively.
Birne Binegar: Connected here, and then, once you have an inverse of the matrix you can easily solve the system by multiplying both sides by the inverse and, in particular, if you have an inverse of the matrix.
Birne Binegar: You can multiply both sides of this system by a the minus version, you find it has to equal to zero vector so.
Birne Binegar: This first for statement we uncovered that as soon as we understood how to calculate the inverse of the matrix and relate it to, for example.
Birne Binegar: Solving linear systems.
Birne Binegar: But.
Birne Binegar: What we did last week, when our discussion of the terminus which we use determines to detect whether or not a matrix was convertible.
Birne Binegar: So.
Birne Binegar: If any one of these statements is true, then off for the other statements is also true.
Birne Binegar: But that also means that if any one of these statements is false, all the other statements are false now what we're looking for.
Birne Binegar: Our non trivial vectors to satisfy an equation like this well think of this as a coefficient matrix rather than just a itself if this linear system is going to have non trivial it's going to have only the trivial solution, then we're going to need a determinant of a.
Birne Binegar: Two not equal zero.
Birne Binegar: X equals zero is gonna.
Birne Binegar: If we're going to have solutions other than X equals zero so i'm negating statement three, then the occasion of statement five with the determinative a is equal to zero, and so that's how.
Birne Binegar: that's, how will determine the appropriate values of Lambda.
Birne Binegar: You only time this this equation here it's going to have non trivial solution is one to determine of this entire coefficient matrix when that is equal to zero.
Birne Binegar: All right, so here's the statement.
Birne Binegar: homogeneous linear system has a non trivial solution solutions other than that the zero vector only if only one the determinant of the matrix is equal to zero.
Birne Binegar: And now we can apply this to our I conducted rally problem what we're looking for.
Birne Binegar: Sorry got lots of typos here we need looking for a non trivial solution of that equation well that's kind of require that each permanent of this matrix equals zero.
Birne Binegar: Okay, how do we utilize that.
Birne Binegar: As we've seen and, hopefully, if you calculate it in your homework that the determinant of a matrix is always a polynomial in the entries of a matrix so in particular once I shift a.
Birne Binegar: So a is equal to 2113 then a minus Lambda times, why what that matrix is it just your original matrix where I shifted.
Birne Binegar: The diagonal elements by the parameter Lambda.
Birne Binegar: Now, when I take a determinant of this matrix it's going to be a polynomial and the Lambda is this to time three minus one times three minus two minus Lambda times three minus Lambda minus one times one it's going to end up being a polynomial.
Birne Binegar: But it's going to have to set up it's going to determine is going to have to equal zero so what happens is we end up with a polynomial equation, it will be a polynomial in the parameter Lambda.
Birne Binegar: A itself contains only numbers but we're adding to the problem is a new parameter Lambda when we calculate determine.
Birne Binegar: A minus Lambda is what it works out to be is to minus Lambda three minus Lambda minus one I E a polynomial in Lambda.
Birne Binegar: and setting that determining equal to zero means we have to solve a polynomial equation don't attack we effectively leave linear algebra for a minute, we have to consider ordinary algebra over.
Birne Binegar: Considering how we solve polynomial equations.
Birne Binegar: Alright, so.
Birne Binegar: This is the most important equation right now.
Birne Binegar: to determine the numbers Lambda which will have non trivial eigenvectors.
Birne Binegar: that's going to require us to solve a certain polynomial equation, we call that polynomial equation a characteristic equation for a.
Birne Binegar: And the solutions of this equation, the numbers Lambda, which is actually true those are going to be exactly the eigenvalues of a.
Birne Binegar: Okay, so.
Birne Binegar: yeah i'm repeating that let's do an explicit example.
Birne Binegar: Okay, so here's a two by two matrix I don't know anything about it, yet what i'm going to try to do is figure out.
Birne Binegar: The numbers Lambda such as i'll have non trivial eigenvectors.
Birne Binegar: Alright, so virtual calculate this so called characteristic polynomial.
Birne Binegar: The the this expression here is called the characteristic polynomial.
Birne Binegar: Of the matrix a it depends on a through and but it's a polynomial in this offsetting parameter Lambda.
Birne Binegar: don't we calculate it.
Birne Binegar: That take the determinant of a minus Lambda times the identity matrix the addition of this term, all it does to the original matrix is it shifts the elements along the main diagonal from the upper left to lower right it shifts it by an amount Lambda.
Birne Binegar: But now we can not calculate determinants in the usual fashion.
Birne Binegar: Here all use the difference of diagonal products that can keep the determinant at that two by two matrix and it works out to be certain polynomial and Lambda and now.
Birne Binegar: Let me remind you that to solve polynomial equations the typical procedure is to try to factor ISIS polynomial in terms of linear polynomials.
Birne Binegar: The Lambda squared minus two Lambda minus eight you should know from just high school algebra that this thing will factor, it says Lambda minus four times Lambda plus two.
Birne Binegar: And so.
Birne Binegar: The solution set a determinant a minus Lambda is equal to zero well here's that.
Birne Binegar: That determinant explicitly calculated and it's only going to equal zero when one of these factors is equal to zero So either we need Lambda equals four or Lambda to equal to.
Birne Binegar: So that's the eigenvalues have a are exactly the numbers for and negative two.
Birne Binegar: Alright, so.
Birne Binegar: If you have a polynomial that's quadratic.
Birne Binegar: You actually have a couple ways of solving the corresponding polynomial equations for quadratic polynomials, you have the quadratic formula that will always give you solutions, but more generally.
Birne Binegar: What you try to do for higher order polynomials for which there is no analog of the quadratic equation, the way you find solutions to a polynomial equation is you achieve that by fact driving the polynomial.
Birne Binegar: And in fact the fundamental theorem of algebra is the statement that if you have a polynomial degree in then there was always a complete factorization in terms of linear polynomials so that.
Birne Binegar: If you have some arbitrary polynomial.
Birne Binegar: A an x to the N, plus dot dot dot plus a zero if that's your polynomial, then the fundamental theme of algebra says that this arbitrary polynomial.
Birne Binegar: is always going to have a complete factorization in terms of linear polynomials where the degree of each factor is just degree one and the underlying variable.
Birne Binegar: And what happens once you achieve this factorization is that the solutions, the corresponding polynomial equation P of X is equal to zero they're going to be exactly these offsets of your variable.
Birne Binegar: so effectively you solve polynomial equations I.
Birne Binegar: guessing or constructing a factorization of the polynomial Oh, I should point out, though.
Birne Binegar: This is the one point in the course where complex numbers are sometimes unavoidable well I didn't stay explicitly in this theorem.
Birne Binegar: Is that the only way you guaranteed that you can always get this factorization is, you have to be able to admit complex numbers, for example, the polynomial equation X squared.
Birne Binegar: plus one is equal to zero, it has no real solutions, but X equals plus or minus I equals plus or minus the square root of minus one, this will satisfy the equation, and in fact this polynomial factor is is this Lambda minus I.
Birne Binegar: Am the plus.
Birne Binegar: So in fact this polynomial indeed that does have a factorization but it doesn't have a factorization necessarily over the real numbers now in this course.
Birne Binegar: which I try to avoid complex numbers because it makes computation much more strenuous but.
Birne Binegar: In fact, everything you've learned in linear algebra if you replace the real numbers by the complex numbers, they would still work, you can still.
Birne Binegar: write down a matrix which, instead of having real numbers has entries, you have complex numbers of entries, you can solve linear systems where the coefficients are complex numbers and the.
Birne Binegar: And what will happen, though it's generally when you saw the system of equations system of linear equations with complex coefficients and set your solutions end up being complex numbers as well.
Birne Binegar: But what's really happening here is the overall linear algebra is exactly the same, even if I was solving a linear system with complex numbers in it that method of reduction.
Birne Binegar: Using real reduction to solve those in your system is exactly the same, except for the point where you actually get down to the computations and you start multiplying numbers are adding numbers together that's where.
Birne Binegar: The complexity of the complex numbers come in, but the formula ISM is exactly the same it's only at the base level computations were using a risk to take that you see a difference.
Birne Binegar: But as you know, if you've ever dealt with complex numbers it's more tedious to deal with multiplication and addition of complex numbers because you have the real part and imaginary part to constantly keep separate and keep track up.
Birne Binegar: In this course we try to avoid complications of complex numbers and we keep things simple.
Birne Binegar: By we're keeping things real and usually you'll notice in homework problems and exam problems I even usually specialized to integers because they're even easier to add.
Birne Binegar: So the reason why we avoid complex numbers in this course is simply because the calculation become more tedious but the theory is exactly the same.
Birne Binegar: Okay, so so just a blade a warning.
Birne Binegar: I will a little bit later in the week show you that for some matrices.
Birne Binegar: Complex eigenvalues are unavoidable that i'm going to sidestep that issue for as long as I can just to show you how the apparatus works.
Birne Binegar: Alright, so let's start with.
Birne Binegar: Another example.
Birne Binegar: will try to find the good guys in the matrix this time it's going to be a three by three matrix.
Birne Binegar: Now, remember that the determinant up a polynomial is the determinant of a matrix is always a polynomial the same degree as the matrix size of I have a three by three matrix what I should expect when I calculated determinant it's going to be a polynomial degree three.
Birne Binegar: And so let's take this matrix will shift it along the diagonal by a parameter negative Lambda and will calculate us determine this polynomial that will result, which is the characteristic polynomial of the original matrix a.
Birne Binegar: Well, so here.
Birne Binegar: we'll do a Co factor expansion, I guess you a Co factor expansion along the second row, so we have.
Birne Binegar: This would be.
Birne Binegar: All right, more detail here, this would be.
Birne Binegar: The zero entry times minus one to the in this slot it's in the second row first column, and then I have the determinant the corresponding minor but i'm not going to compute it because it's being multiplied by zero.
Birne Binegar: Then I have the second term which is zero minus Lambda time minus one to the two plus two now, I have a non zero trivial term so let's actually write down that matrix it's exactly here.
Birne Binegar: One minus Lambda so what i'm doing here is i'm ignoring the rowing column here one minus Lambda one one.
Birne Binegar: minus Lambda.
Birne Binegar: And then I have pick up.
Birne Binegar: The term and the co factor expansion, corresponding to the third element in the second row, and that will be exactly this.
Birne Binegar: I get a minus sign here.
Birne Binegar: Because i'll have one times minus one this guy here is in the second row third column will be minus one to an odd power that's where that minus sign come from and then this is the minor we get by ignoring removing the.
Birne Binegar: column and row that contains the item of the entry of interest.
Birne Binegar: Alright, so that's.
Birne Binegar: A review with that determinant calculation out set up and then, once we.
Birne Binegar: simplify.
Birne Binegar: Here we do the direct calculation, so these two by two matrix these i'm just using the difference of the diagonal products to calculate the determinant and then, once I simplify I get as predicted polynomial degree three polynomial degree three in the parameter Lambda.
Birne Binegar: Alright, so now we have a problem.
Birne Binegar: For polynomials of degree three or ponyo music degree, for we don't have anything like the quadratic formula.
Birne Binegar: it's a simple solution for immediately writing down all possible solutions of the polynomial gration rather we have some polynomial that we're going to.
Birne Binegar: You only way we can figure out how to solve this find the values of Lambda for which PA of Lambda is equal to zero so we're going to have to factor ISIS.
Birne Binegar: For two weeks.
Birne Binegar: it's usually not so.
Birne Binegar: Not like quadratic equation towards, generally speaking, you can, if you see a nice pattern of the coefficient you can factor is it.
Birne Binegar: Rather, the case at hand.
Birne Binegar: work will have to do is first of all identify one easy factor all terms here proportional to Lambda so that means two things that means Lambda has to blend the equals zero has to be a route because if Lambda equals zero.
Birne Binegar: it's pretty obvious that the right hand side is going to evaluate to zero and Lambda is equal to zero, and so Lambda will be a solution.
Birne Binegar: But the find the other solutions.
Birne Binegar: What we'll do is we'll pull out the common factor of Lambda and now we'll try to factor ISIS quadratic polynomial and that's that's easier to do.
Birne Binegar: I can recognize this as Lambda minus two times Lambda plus one until the full factorization of our characteristic polynomial is going to be minus Lambda Lambda minus two Lambda plus one.
Birne Binegar: And from this factorization we can read off the eigenvalues for this polynomial equals zero in and the Lambda is equal to zero Lambda equal to two or Lambda equal to negative one.
Birne Binegar: Until those are the Eigen values.
Birne Binegar: that's the only new part you know, in the eye that you'll have to learn for the eigenvector eigenvalue problem and it's pretty simple.
Birne Binegar: Given a matrix a.
Birne Binegar: So, to find the eigenvalues.
Birne Binegar: The first thing you do is you form.
Birne Binegar: The shifted matrix a, which is a minus Lambda I you just subtract lamp the parameter Lambda from the diagonal entries at bay and then you compute.
Birne Binegar: A polynomial P, a Lambda which is going to be defined as determinant of the shifted matrix.
Birne Binegar: And then, what you do is you factor eyes.
Birne Binegar: PA a Lambda.
Birne Binegar: to identify.
Birne Binegar: See, I can bags.
Birne Binegar: Okay, but I can bet your eigenvalue problem has two parts.
Birne Binegar: we've only sold the first part, we found the allowable I can values for a given matrix following this procedure.
Birne Binegar: define the eigenvalues well.
Birne Binegar: What we do is we take one of the solutions that we found here.
Birne Binegar: Now Lambda is no longer a parameter we're interpreting it as a number one of the root of the characteristics equation.
Birne Binegar: And then we get when Lambda is a number this becomes a concrete numerical matrix and we can thereby use row reduction in the like to pitch blissfully find a solution and then we've got the corresponding eigenvectors.
Birne Binegar: So we finished the eigenvector eigenvalue problem.
Birne Binegar: Like once we've got the eigenvalues.
Birne Binegar: We substitute them in one by one, so in the i'm now interpreting Lambda as a particular number a particular route of the characteristic equation.
Birne Binegar: there'll be a numerical matrix or reduced to rational and form once I get that I can look at the columns that have pivots and the columns that don't have pivots.
Birne Binegar: to distinguish between fixed variables and pre variables, I can write down the equations corresponding the radio show ashland form.
Birne Binegar: express the fixed variables in terms of the free variable write down as typical solution vector expand that solution in terms of the free parameters and then.
Birne Binegar: The constant vectors that are being multiplied by the three parameters, those are going to be your basis vectors for the solution set, they will be a basis for your Eigen space, and those are what will call the eigenvectors to pay will actually.
Birne Binegar: be looking for these basis factors.
Birne Binegar: Okay, so.
Birne Binegar: yeah.
Birne Binegar: basis vector is that the basis for the other I.
Birne Binegar: know no show for each eigenvalue gonna have to do a separate you're gonna have to solve a separate linear system, every time you change this number, the solution is that equation change, and you have to resolve the system.
Birne Binegar: yeah but typically what will happen, though, is.
Birne Binegar: yeah it typically happens here you'll have a unique basis factor, because the solution set will be one dimensional.
Birne Binegar: And so there's only one direction in that entire audience base you usually just want to find that one direction.
Birne Binegar: That kind of fits with the the physical interpretation of our solutions or I described that problem of finding the natural axes rotations.
Birne Binegar: Those axes rotations they're really determined by a direction, not by the length of the victor but just the direction it's going in space that's usually what one means by the eigenvectors that you're looking for those directions, not necessarily unique vectors but a unique directions.
Birne Binegar: Okay, so yeah I guess i'll stop here today what we'll do on Wednesday is we'll explicitly find some eigenvectors and begin solving.
Birne Binegar: several examples eigenvector eigenvalue problems.