Birne Binegar: Today we're going to continue to discuss.
Birne Binegar: diagonal ization of square matrices but today, the emphasis is going to be on one we can actually dial is a matrix so I showed you how to but there's a caveat involved in that.
Birne Binegar: Or we do that stuff.
Birne Binegar: start with and then we'll look at specific ways, you can tell.
Birne Binegar: Whether or not a matrix is dying likeable.
Birne Binegar: oftentimes before you even proceed to find it eigenvectors noggin values.
Birne Binegar: Right so.
Birne Binegar: The basic idea that what we're trying to do, has giving us square matrix a we want to find a matrix see that will convert it into a simpler matrix one that's diagonal so that it's only non zero entries occur on that main diagonal.
Birne Binegar: so effectively what this matrix see is doing it's converting a into a diagonal matrix and i'll show you on Wednesday some applications of that, but the way this works is because.
Birne Binegar: If we write a matrix in terms of it's called there's two multiplication identities at play so let's take a matrix arbitrary n by n matrix, let me express it.
Birne Binegar: As a bunch of column vectors.
Birne Binegar: That i've inserted next to one another inside of matrix what these identities, have to do with and they follow just from the rules that matrix multiplication.
Birne Binegar: is how matrix multiplication X on another matrix so, in particular, if you multiply your matrix see from the left by an arbitrary matrix a.
Birne Binegar: Well, the effective, that is why you can computer directly by taking dot products Rosa a with a column to see, but the end result is interpretive all is simply the first column of a product is just a times the first call monthly second factor see second call Mike cetera.
Birne Binegar: et, by the way, there's also another identity, where instead of.
Birne Binegar: multiplying of a matrix from the left by another matrix a I can also think of multiplying see from the right by matrix but then they don't have such a nice identity.
Birne Binegar: What the analog of this formula is is that a X from the right on a matrix by acting on it rose rather than it's called.
Birne Binegar: So there is an analogous identity from multiplying see from the right, but instead of writing the result in terms of the columns you write it in terms of the row but it's analogous identity.
Birne Binegar: there's also a second identity here I am multiplying see from the right by a matrix but i'm restricting attention to when D is diagonal.
Birne Binegar: And when he is diagonal if I multiply a matrix as specified in terms of columns from the right by diagonal matrix the effect of that is just multiplying the original columns of.
Birne Binegar: see my a corresponding scale or factor this Lambda one is going to be the first diagonal entry of D, the one in this upper upper right hand corner and the two would be the second one, etc.
Birne Binegar: All right now, these are just identities that fall by specifying a matrix in terms of its columns and.
Birne Binegar: When we then argue, then, if we're looking for a matrix see so that we can convert a into a diagonal matrix well there's a consequence of those.
Birne Binegar: Two identities, first of all, taking the identity, we want and multiplying both sides by the matrix see.
Birne Binegar: And so what's going to happen is that CC the miners first just going to produce a factor of the identity matrix and what we get is the identity that he has to equal CD so anytime we have a matrix that diagnosis, a given matrix a there's got to be this relationship as well.
Birne Binegar: And now we use these matrix multiplication identities, a time some matrix see consisting of column C, want to see n that's exactly this.
Birne Binegar: And on the other hand, see times diagonal matrix on the right well that's going to produce this, and now you compare the two that the quality column by column, and you see that each column.
Birne Binegar: On the right on sorry each column on the left is.
Birne Binegar: scale or multiple of them well, a time see one is some skill or multiple see one, which means that see one is an eye conductor of a with eigenvalue Lambda one.
Birne Binegar: And in fact that's true of each of these calls, so if you're going to have an identity, like this one thing that has to happen for this matrix see it's it's columns have to be eigenvectors have a.
Birne Binegar: and, moreover, for this diagonal matrix D it's it's diagonal entries have to be eigenvalues of the matrix today and, in fact.
Birne Binegar: The eigenvalues are paired with these eigenvectors in the natural way Lambda one is exactly the eigenvalue corresponding to the eigenvector see one similarly Lambda to be the eigenvector corresponding to the second.
Birne Binegar: I can vector.
Birne Binegar: Alright, so.
Birne Binegar: This is so the way you might construct a matrix see the diagonal is a if you take a bunch of eigenvectors have a.
Birne Binegar: you arrange those as the columns at the matrix similarly you take.
Birne Binegar: The corresponding eigenvalue so Lambda one is going to be the eigenvalue of a were be one is the corresponding eigenvector and similarly you order your columns on the right hand side and you order the corresponding eigenvalues and exactly the same order, then this identity is guaranteed.
Birne Binegar: And it would seem that we've been sold the eigenvector eigenvalue problem but there's one caveat.
Birne Binegar: The identity that we want does imply that AC is equal to DC However, the opposite implication is not true AC equal DC does not necessarily mean that CD minus first AC equals D, because that inverse might not exist.
Birne Binegar: And so, you may be able to make a matrix of eigenvectors that satisfy this relation, but you won't be able to do the desired relation unless.
Birne Binegar: That matrix see is in fact convertible.
Birne Binegar: So what I began to talk about on Friday last week where conditions that we can that would guarantee that those.
Birne Binegar: conjugating mater CC are in vertical.
Birne Binegar: Right so.
Birne Binegar: Let me give you a simple convertibility criteria.
Birne Binegar: I stay here this last time, but I want to.
Birne Binegar: make it seem much more plausible to you.
Birne Binegar: Remember that the way you calculate an inverse of the matrix a let's call it C so it's even more explicit.
Birne Binegar: About you would find see the minus first is, you would reduce see a joined with it, and if you get the identity matrix here, then what happens on sorry on what happens on the right hand side.
Birne Binegar: ends up being inverse or you get something else.
Birne Binegar: Something else.
Birne Binegar: and see the minus first does not exist.
Birne Binegar: that's the calculation, by which we originally learned how to apply an inverse of the matrix these, I also want to remind you of.
Birne Binegar: How to find a basis.
Birne Binegar: A basis for a column space.
Birne Binegar: Of a matrix see well what you would do is you would take see you would really do too it's a radio show I show on or.
Birne Binegar: or even.
Birne Binegar: Further to reduce real Echelon for.
Birne Binegar: let's call that a well let's call it let's call it C double prime well the way you can find the basis for the calm state of this matrix if you look at where the pit is live.
Birne Binegar: And then.
Birne Binegar: show the pivot.
Birne Binegar: Sorry, these columns.
Birne Binegar: Of the original matrix see corresponding to.
Birne Binegar: The columns.
Birne Binegar: Of this, reduce show Echelon form.
Birne Binegar: So let's see double prime that have pit it.
Birne Binegar: Those are your basis vectors for the column space of the original matrix so that that calculation has been known for a while and we're going to use these two facts to prove.
Birne Binegar: This lemma.
Birne Binegar: So if Asian season vertical then it can be reduced the identity matrix that's this calculation, essentially, but then.
Birne Binegar: If the identity matrix is a relational on format see, then that means the row so on format see as a pivot and every call right because that's the identity matrix you always have a pivot on that main diagonal.
Birne Binegar: Every column is going to pivot well, that means that when we go back if you want to find a basis for the column space.
Birne Binegar: Every column of the original matrix is going to be a basis doctor that's, this is a rule by which we find basis factor for the columns days well that just means that every column vector of the original matrix see since it's the basis sector, they must all be linearly independent.
Birne Binegar: A basis is a set of vectors that's lit one linearly independent and to capable reaching every point in the space they generate well.
Birne Binegar: We find a basis for seen it turns out, it see if reducible to that again entry basic that every column of see can be used as a basis, Director for the column states that means that all the columns to see are linearly independent and then that.
Birne Binegar: So what we need in order for our matrix that converts a given matrix a into a diagonal matrix we need an inverted matrix C is kind of be inverted if the if it's columns are linearly independent.
Birne Binegar: On the previous slide we saw that.
Birne Binegar: See also have that consistent eigenvectors of a.
Birne Binegar: And so, putting this all together if you have a square matrix a it's going to be diagnosable.
Birne Binegar: If, and only if you have enough linearly independent eigenvectors since every column of the matrix see that that set of columns has to be literally independent since you're going to make the call them some eigenvectors of.
Birne Binegar: matrix a you're going to have to have an linearly independent eigenvectors in order to construct a vertical matrix.
Birne Binegar: Right so that's essentially our current criteria, but now and that much I covered last week, what i'm going to focus on today is waste of spotting whether or not a given matrix has enough linearly independent eigenvectors to construct that diagonal ice matrix see.
Birne Binegar: Okay, so.
Birne Binegar: The first way i'll try to answer this question is by reviewing a little bit more of our eigenvector eigenvalue calculations.
Birne Binegar: Remember, if you have a square matrix a.
Birne Binegar: You can calculate is eigenvalues always by looking at the characteristic polynomial and which is just the determinant of a minus a parameter Lambda times the identity matrix.
Birne Binegar: And because of the fundamental theorem a calculus any polynomial of degree N, such as this is always going to be factor I level in terms of.
Birne Binegar: linear polynomials don't potentially these offset from Atlanta in each factor, they could be complex numbers but there anyway, going to be the eigenvectors of the matrix a.
Birne Binegar: Because they will be root of the condition that this characteristic polynomials equal to zero.
Birne Binegar: Right, so they gave us a certain multiplicity that we can attach to a given eigenvalue this number one, the number of powers of the factor Lambda miners are one in the characteristic polynomial that's known as the algebraic multiplicity of the given.
Birne Binegar: The given eigenvalues and one thing to note about this, because this is for it n by n matrix the characteristic polynomial is always a polynomial degree and well, that means that these factors actually these multiplicity are going to have to sum up to.
Birne Binegar: To the size of the matrix because both have to be polynomials of the same degree.
Birne Binegar: Right they'll those are the algebraic multiplicity they're related to the factorization of the characteristic polynomial and that factorization you want anyway, because you want to find the actual I get down.
Birne Binegar: But the state from.
Birne Binegar: These algebraic multiplicity you had another kind of multiplicity and that was how many effectively how many linearly independent I can vectors do you have was a given eigenvalue what we do is we.
Birne Binegar: The set here at the set of all vectors inside or an such that the vectors or eigenvalues of the matrix a with the given, I can tell you are.
Birne Binegar: So it's a number of linearly independent basis vectors.
Birne Binegar: or equivalent need the dimension of this no space or equivalent Lee the number of free parameters, when you reduce this matrix.
Birne Binegar: To reduce electron form and you can't number of columns without pivots so that's how you calculate these numbers i'm sorry this should be subsequent there because i'm.
Birne Binegar: I get one geometric multiplicity.
Birne Binegar: or each eigenvalue.
Birne Binegar: All right now, I think I want to point out here, I didn't really remark.
Birne Binegar: or make them strong comment about this at the time we were discussing geometric multiplicity but what these geometric multiplicity are doing is they're counting.
Birne Binegar: The number of linearly independent eigenvectors in each Eigen space right once you have a basis, then you have.
Birne Binegar: A fixed set of linearly independent vectors a number of just a dimension of your subspace So in fact what these geometric multiplicity are doing.
Birne Binegar: they're counting the dimension of the Eigen spaces associated with eigenvectors or equivalent only they're counting the number of linearly independent like in vectors in each Eigen space so to diagonal is a matrix.
Birne Binegar: Well, to get a.
Birne Binegar: matrix see that's a diagonal that will die analyze a it needs to have an inverse in order to have an inverse, we have to have enough linearly independence vectors to us as columns so that matrix will reduce the the identity matrix well.
Birne Binegar: Well, what this geometric multiplicity doing this it's doing a sort of a piecemeal approach to the the general problem.
Birne Binegar: In order to diagnose a I need a total as n by n amit a total of n linearly independent eigenvectors so I can create that matrix see.
Birne Binegar: Well, these geometric multiplicity count the number of linearly independent eigenvectors in each Eigen space so to get the total number of linearly independent eigenvectors what you have to do is add up all these geometric multiplicity you, you cannot, I can space by I can say.
Birne Binegar: All right now.
Birne Binegar: Those algebraic and geometric multiplicity is also had some common editorial identities first one was that.
Birne Binegar: The algebraic multiplicity a number of factors of Lambda minus R I that occurs in the characteristic polynomial that has to be at least one, otherwise you wouldn't have an eigenvalue.
Birne Binegar: But, in fact, it turns out that the geometric multiplicity is bounded between this lower bound and it possibly could be as large as the algebraic multiplicity but, in general, live somewhere in between.
Birne Binegar: The other thing and I already pointed this out on the last slide.
Birne Binegar: When you write down this factorization both sides have to be polynomials of degree in which means these numbers one through mk they have to add up to end.
Birne Binegar: So what this is going to mean these two identities if.
Birne Binegar: The algebraic multiplicity always add up and but notice if just one of these guys is less than the corresponding algebraic multiplicity.
Birne Binegar: Some of these guys will not add up to in anymore, I mean if the quality is whole the quality is true, they algebraic multiplicity are all equal to the.
Birne Binegar: joy, the geometric multiplicity are all equal to the algebraic multiplicity, then I can just substitute in, and I see that I have no.
Birne Binegar: linearly independent eigenvectors these guys are counting the number of linearly independent eigenvectors in each Eigen space when we sum over all possible I didn't spaces that's when we get the total number of linearly independent eigenvectors but.
Birne Binegar: I can't just substitute new for em here because.
Birne Binegar: it's not necessarily equal they could be less than and if any one of these numbers is less than the corresponding algebraic multiplicity that means whatever these add up to is going to be less than in.
Birne Binegar: Alright, so.
Birne Binegar: The statement then there's.
Birne Binegar: An n by n matrix with a given list of eigenvalues is going to be diagnosed I usable only if we have a total of the end linearly independent I can vectors.
Birne Binegar: For that to happen, it well for that not to happen it suffices that one of these geometric multiplicity is less than the corresponding algebraic multiplicity.
Birne Binegar: Alright, so same Oregon it's just because.
Birne Binegar: Right I add up all these geometric multiplicity I get the maximum numbers linearly independent eigenvectors but, if any one of these numbers is less than the corresponding.
Birne Binegar: algebraic multiplicity these numbers will no longer add up and and I won't have enough linearly independent like in vectors to construct an inverted matrix see.
Birne Binegar: Right now there's.
Birne Binegar: there's one special case it's fairly common and easy to handle suppose we have K.
Birne Binegar: Suppose the eigenvectors of.
Birne Binegar: The eigenvalues that they are all distinct.
Birne Binegar: So that means that to eigenvalues for different indices one through K, we have a different index here than the eigenvalues also different.
Birne Binegar: Then the corresponding eigenvectors are always going to be linearly independent.
Birne Binegar: Alright, so here i'm.
Birne Binegar: considering any set of eigenvectors for which the corresponding eigenvalues are different from one another.
Birne Binegar: i'm going to only prove this to the case when I only have two eigenvectors having two different eigenvalues and more general case we have an arbitrary number of.
Birne Binegar: eigenvectors corresponding to distinct eigenvalues that's done by induction but you'll see the basic problem here suppose V1 and V2.
Birne Binegar: or eigenvectors of the matrix a.
Birne Binegar: That means we'll have a one some multiple the one and a two is going to equal some multiple be too.
Birne Binegar: All right.
Birne Binegar: i'm gonna.
Birne Binegar: See if what conditions, these guys were linearly independent it there, see if they're linearly independent or not.
Birne Binegar: But let's take a new combination of these two factors said it equal to the zero vector and see whether or not these coefficients X one X two whether we can make have a non solution of this equation or X one X two are not equal to zero.
Birne Binegar: Well, in fact, i'm going to assume that i'm going to assume that say X one is not equal to zero, and then this equation will allow me to express one of the eigenvectors in terms of the other liking vector.
Birne Binegar: And now, since the wants to be an eigenvector a.
Birne Binegar: With eigenvalue Lambda one will have at one is equal to Lambda be one.
Birne Binegar: But now, if D one, two or literally.
Birne Binegar: dependent that I can replace the one by the scale a multiple of the two.
Birne Binegar: And then, this is just a scale of multiple so I can bring the factor, the matrix factor pass the scale of multiplication let it attack D to directly when a hits be to be to is an eigenvalue eigenvector have a with eigenvalue Lambda to don't get a kick out Lambda to V2.
Birne Binegar: And now i'll just reverse the order factors this guy here can be again identified as the one.
Birne Binegar: And so, now the extreme side say that Lambda times the one has to equal Lambda two times the one, but I know I have offices that B one and B to correspond to different eigenvalues so my hypothesis here is that Atlanta one.
Birne Binegar: Does not equal Lambda to.
Birne Binegar: So this can happen.
Birne Binegar: Three I mean if this were three initial, for it would say that three times the one is equal to four times the one that can't happen unless be one itself a zero, but I can vectors.
Birne Binegar: We always exclude the trivial eigenvector corresponding the zero vector I can vectors are non zero vectors so In fact this is contradiction.
Birne Binegar: Alright, so that shows you if you have two different eigenvectors but they have different eigenvalues they're necessarily literally independent of one another.
Birne Binegar: Okay, so.
Birne Binegar: Now we got a theorem suppose we have a n by n matrix and it has a total of n distinct eigenvalues.
Birne Binegar: well.
Birne Binegar: that's going to mean.
Birne Binegar: that's going to mean that all these multiplicity.
Birne Binegar: or equal to one, so, in fact, your eigenvalues you're going to be one through land to end the corresponding characteristic polynomial.
Birne Binegar: it's going to be exactly all, let me okay i'll write it this way Lambda minus Lambda one that that Lambda minus.
Birne Binegar: Lambda n so i'm sorry for this sort of.
Birne Binegar: funny notation here.
Birne Binegar: But i'm putting sub scripts on Lambda i'm meeting i've got specific numbers that are going to be, I can I can values of the matrix a.
Birne Binegar: Here Lambda is a parameter it's a variable in this polynomial, and these are offsets that I previously labeled.
Birne Binegar: As our rather than confusing you with the variable with the sort of corresponding room.
Birne Binegar: But what's going to happen here is that all these eyes are going to equal equal one that we know.
Birne Binegar: The new eyes.
Birne Binegar: Are bounded between one and am I now is the am I, are all equal one that's going to mean that these geometric multiplicity to also all equal to one, and so we have.
Birne Binegar: into some of the new eyes.
Birne Binegar: I equals one to end it's going to equal summation I equals one the end are all equal to one and we're going to get in we are going to have enough linearly independent eigenvectors to create that conjugating matrix see that will die guys a.
Birne Binegar: But one thing is not clear or wasn't clear up to this point was whether or not all the corresponding I conductors we're actually linearly dependent one another.
Birne Binegar: But now we've just proved that if you have different eigenvalues the corresponding eigenvectors are linearly independent, so in this situation.
Birne Binegar: Where we have and distinct eigenvalues we're also going to have an linearly independent eigenvectors and the diagonal ization construction or see goes to.
Birne Binegar: First eigenvector as first call second eigenvectors and second call last eigenvectors the last call these are all going to be literally independent so.
Birne Binegar: The one up in.
Birne Binegar: Our linearly independent and so C is in vertical.
Birne Binegar: And then C minus first AC is going to in fact died realizing the matrix a.
Birne Binegar: Alright, so now.
Birne Binegar: One quick way.
Birne Binegar: of determining or seeing that a matrix can be diagonal line is simply to calculate its eigenvalues and verify that you get a total of end state eigenvalues.
Birne Binegar: Okay, let me do some examples.
Birne Binegar: In this first example.
Birne Binegar: Well let's work through it we're gonna.
Birne Binegar: The question I want to pose for this two by two matrix is, can I diagonal is it.
Birne Binegar: Oh there it is so what we'll do is we'll look at the corresponding characteristic told them first.
Birne Binegar: The zeros of this polynomial we're going to course be the eigenvalues of the original matrix day and we calculate the determinant of a minus Lambda times I turns out to be a perfect Square and has unique route Lambda has to equal one so.
Birne Binegar: You only have one eigenvector we, we need to linearly independent eigenvectors so diagonal likability is not guaranteed, because all of our multiplicity are not equal to one rather, we have a double room land to equal to one, so the non.
Birne Binegar: Think of there being two factors here, because the algebraic multiplicity are not equal to one we have non distinct eigenvalues and proceeding theorem this term here does not imply because.
Birne Binegar: Effectively, are two eigenvalues corresponding to the two factors are the same Lambda equals one.
Birne Binegar: Lambda equals one time to Lambda equals one.
Birne Binegar: Alright, so we only have one eigenvalue because of that diagonal eyes ability is not guaranteed.
Birne Binegar: let's go ahead and calculate.
Birne Binegar: The Lambda equals, one I can space, because it could be that that Lambda equals, one I can stay says two dimensional, in which case.
Birne Binegar: Just that one I can space would give us enough eigenvectors.
Birne Binegar: But.
Birne Binegar: If we examine that movie calculate from down a little bit.
Birne Binegar: If we actually go ahead and calculate that Lambda equals one Eigen space.
Birne Binegar: well.
Birne Binegar: exactly like this matrix and the I got ideas to calculate it's no space, so we reduced to reduce your rational and form we look at this matrix it's the.
Birne Binegar: Reduce rational and form up a coefficient matrix and the equations correspond to it are going to be zero times X one plus one times X is equal to zero and zero is equal to zero.
Birne Binegar: So the solution said corresponding the homogeneous linear system with this coefficient matrix it's just the span of a single vector.
Birne Binegar: In other words, I only have one like in space, because I only have one, I can tell you that when I get space only has one linearly independent.
Birne Binegar: vector inside it, and so I only going to get one linearly independent eigenvector for the matrix a I need to to construct a diagnostic matrix see.
Birne Binegar: And so the conclusion is that there is no matrix see that actually is going to be neutral diagonal is a because I can't make an inverted matrix out of the using only the eigenvectors of Bay I don't have enough, I can directors to get it in vertical matrix.
Birne Binegar: Okay, OK, so the conclusion here is the original matrix a is not dying logical, because it doesn't have enough linearly independent eigenvectors but we seen that by that explicitly finding all the linear independent eigenvectors we could find.
Birne Binegar: Alright, so let me do a more complicated example sort of at the other extreme here a look at a three by three matrix and i'll ask the same question is this matrix diagonal icicles.
Birne Binegar: So we look as characteristic polynomial I calculated it works out to be this polynomial I guess you'd have to do something like a Co factor expansion along that second row in order to.
Birne Binegar: complete this calculation, but it ends up to being a cubic polynomial that.
Birne Binegar: it's factor eyes of all you at least the that common factor of Lambda in each term but then they want to remove the power of Lambda use the quadratic.
Birne Binegar: factor that remains factor I just like that, and you see that we have three distinct eigenvalues Lambda equals zero or Lambda equals negative one or Lambda equal to.
Birne Binegar: All right now we are in the situation we can apply the theorem that says, if you have.
Birne Binegar: For an n by n matrix if you have an distinct eigenvalues then necessarily you're going to have enough linearly independent eigenvectors.
Birne Binegar: Because for each of these distinct eigenvalues you're going to have a literally independent eigenvector Adam up and you get three linearly independent eigenvectors that's enough eigenvectors to create an invisible matrix see.
Birne Binegar: Alright, so.
Birne Binegar: that's how quickly, I can determine whether or not a matrix.
Birne Binegar: can be diagonal I so let's go ahead and dive realize it.
Birne Binegar: That just will allow me to review the computation first thing you need to do is fine, the three linearly independent eigenvectors.
Birne Binegar: will do that by extracting from each organ space a corresponding basis vector so we first look at the zero I get space.
Birne Binegar: that's going to correspond to the no space and a minus the eigenvalue times the identity matrix or this matrix The next step is to take this coefficient matrix and reduce it to reduce row Echelon for that here what happened between what reduction.
Birne Binegar: And what i've got here is to reduce row Echelon for.
Birne Binegar: All right, then, once you get the.
Birne Binegar: Reduce relational and form of a coefficient matrix of a homogeneous linear system actively a X is equal to zero.
Birne Binegar: Then what you do is you go back and you convert back to equations and try to write down the solution, and the solution set it's a hyper plane.
Birne Binegar: And what one finds in this particular example is you're going to have to have X one plus x is equal to zero.
Birne Binegar: that's gonna actually produce this vector this one is just saying the third component has to equal zero so I basically have to condition tana three dimensional vector i'm left with a one dimensional Stan alright, so I have exactly one basis vector for the Lambda equals zero I can stace.
Birne Binegar: Alright, so and there it is that vector here that generates the entire no space that's what we use as the eigenvector corresponding to the given eigenvalue.
Birne Binegar: We have tomorrow eigenvalues, and so what we want to do now is use those eigenvalues to find corresponding basis vectors.
Birne Binegar: So similar for the preceding case we subtract the eigenvalue from the diagonal entries a yielding this matrix then rewrote reduce.
Birne Binegar: The coefficient matrix to reduce relational on form and we try to write down the solutions actors.
Birne Binegar: In this case, what's going to happen is the first component has to equal zero making that line and it X two is going to have to equal to negative X Ray that's.
Birne Binegar: Why those components pop out, we have only one basis vector for the entire solution set and so that one basis that one vector is actually going to be our basis.
Birne Binegar: So now we've got.
Birne Binegar: And I can vector corresponding to the eigenvalue minus one.
Birne Binegar: We have one last eigenvalue was Lambda equals to say calculation you subtract two from the diagonal elements hey you got a matrix that you then row reduce to reduce show Echelon form and you then endeavor to write down.
Birne Binegar: The solution vector, let me remind you of this calculation what you do once you get to the matrix and reduce your national i'm for.
Birne Binegar: you write down the corresponding equation, so I have X one plus zero times X two minus three half X three is equal to zero second row is telling me that X two minus one half X three is equal to zero, and last row is telling me zeros equals zero.
Birne Binegar: We then.
Birne Binegar: Look at these equations and decide, some of these variables are fixed and some variables are free, we use.
Birne Binegar: A column that doesn't contain the pivot to be the free correspond to the free parameter until these equations become X one is equal to three half X three x is equal to one half.
Birne Binegar: X Ray that expresses two variables in terms of the three variables we write down a solution vector.
Birne Binegar: Three half X three one half X three x three and then we expand in terms of the free parameter to get that basis sector so that would be three halves.
Birne Binegar: One half in one right so that's why that's where that vector popped out of the full calculation and what we see is again we have a single basis factor for the entire Eigen space.
Birne Binegar: What you can often do once you have a basis factors, you can scale or multiply and you'll get another basis factor so here are all done.
Birne Binegar: Between here is i've gotten rid of these fractions by taking this basis vector multiplying it by two to get rid of the fractions and get rid of the integers and the denominators to get a simpler vector.
Birne Binegar: And i'll call the basis vector for the Lambda equals to Eigen space i'll use this spectrum three one and two.
Birne Binegar: Alright, so.
Birne Binegar: In this example of a three by three matrix we first of all found all this eigenvalues and we found that we there were three distinct eigenvalues if we have a three by three matrix that was already enough to tell us at the matrix will be dying or sizable then when I did.
Birne Binegar: As I proceeded to diagonal is the matrix see well we're about to wait, I saw the eigenvector eigenvalue problem for the given matrix.
Birne Binegar: I had for the eigenvalue zero I had an Eigen factor for the eigenvalue minus one I have an Eigen factor and for an eigenvalue to I have yet another eigenvector because these correspond to distinct eigenvalues the corresponding eigenvectors are linearly independent.
Birne Binegar: That means when I use these factors as columns, not only will they be I can vectors to the matrix a they're going to be linearly independent columns as well, so this matrix see is going to be in vertical.
Birne Binegar: That makes it possible to actually take the natural identity or a time see is equal to.
Birne Binegar: See times D.
Birne Binegar: That.
Birne Binegar: That identity can be inverted well, we simply multiply it by the inverse of seeing we get the identity, we want, we get a matrix that converts the given matrix into a diagonal matrix.
Birne Binegar: How did I know without actually doing this computation computing the inverse of see multiplying it by a and then multiplying by see again how do I know that the resulting diagonal matrix is exactly this matrix well that follows from the fact that.
Birne Binegar: Now, when you do guidelines and matrix see what's going to correlate between your diagonal matrix see and the diagonal matrix that result is that the.
Birne Binegar: entries along this diagonal matrix along the main diagonal they have to be exactly the eigenvalues.
Birne Binegar: of corresponding to the corresponding columns of the matrix see so since I use.
Birne Binegar: And I can value zero or spawning this call or write down that call, but then in the for the eigenvalue all use zero.
Birne Binegar: In the second column of us this MIT this column vector it's it's the eigenvector a with eigenvalue minus one.
Birne Binegar: So i'll put it in the second slot but simultaneously when I have to do is put the right eigenvalue in the matrix D so be second slot second diagonal elements.
Birne Binegar: And then the third column I grabbed the third eigenvalue and use that as last diagonal elements of the diagonal matrix D, so that that's how it works.
Birne Binegar: Alright, so.
Birne Binegar: Let me close today's lecture with.
Birne Binegar: An example that i'm not going to prove but it's pretty common and physical applications.
Birne Binegar: A matrix.
Birne Binegar: A is said to be symmetric first of all, it has to be a square matrix it's going to be symmetric if it's equal to X transpose.
Birne Binegar: So another way of saying that is that the entry in the ice rowan jade column of the matrix a is exactly the same thing as the entry in the Eighth row, and I call you're going to have a certain reflection cemetery.
Birne Binegar: If you have a let's put it neat I throw here and jace calm so there's to be calm I.
Birne Binegar: row column J ro I well somewhere down here i'm Nike row J and there might be a corresponding column, I hear.
Birne Binegar: What what has to happen.
Birne Binegar: At these two entries they're actually going to be on opposite sides of that main diagonal they have to be equal in order to be a.
Birne Binegar: trend in order for the transpose speaking see be the same, or in other words, if I take the rows of the matrix a and I.
Birne Binegar: form a new matrix using the Rosa as columns i'll get exactly the transpose matrix and what i'm saying is then the transpose is the same as the original matrix.
Birne Binegar: Anyway, these certain matrix occur quite frequently and physical applications and what's nice about them is first of all they're always diagnosable.
Birne Binegar: And second of all this is even nicer if they're eigenvalues are always real numbers what's going to happen in this case is that when you calculate.
Birne Binegar: The characteristic polynomial.
Birne Binegar: of US symmetric matrix you get a polynomial that.
Birne Binegar: You get you get a polynomial that only has real solutions when you said it equal to zero you're not going to get something like.
Birne Binegar: PA a bland is not going to equal something like X squared plus one which never equal sorry, maybe Lambda squared.
Birne Binegar: it's not going to equal something like this, because this is only has roots when Lambda is equal to the square root of negative one or plus or minus I the pure imaginary number.
Birne Binegar: In the case of a symmetric make multiplication this complication of having the leap up into complex numbers never occurs.
Birne Binegar: that's really turns out to be really important for a number of physical locations.
Birne Binegar: That means, for example, when you calculate the moments of inertia or rotational body those moments of inertia, are they actually turn out to the eigenvalues of repeating their matrix.
Birne Binegar: Those momentum and inertia turn out to be real numbers not complex numbers that will be vital to the physical interpretation but also.
Birne Binegar: and
Birne Binegar: You ever go far to do things like quantum mechanics and like.
Birne Binegar: Finding the energy levels of the hydrogen atom is tantamount to dying analyzing a certain matrix and what's important is that matrix turns out to be symmetric, and so the eigenvalues up your energy operator all real numbers, which means your energy to real they're not complex numbers.
Birne Binegar: Okay, so.
Birne Binegar: I think i'll stop here today what i'm going to do on Wednesday, so I want to show you a nice application of.
Birne Binegar: Having the ability to diagonal is a matrix what i'll show you is that in fact we can solve differential equations systems a differential equations pretty readily if we can diagnose a certain matrix and i'll be able to do so without doing any.
Birne Binegar: Much at all about the theory of differential equations I just want you to know how to solve the differential equation for the exponential function.
Birne Binegar: And then from that will be able to solve systems of differential equations that actually govern a large large number of applications and finance and biology.
Birne Binegar: Chemistry etc OK so i'll let you guys go today and tomorrow we'll do an application of diarization.
Birne Binegar: yeah.
Birne Binegar: Whenever.
Birne Binegar: I vectors you ended up deciding this entry in that entry, why did that happen.
Birne Binegar: Well Okay, so when you write out the equation you'll have X one plus x two X equal to zero right, but now you want to move the free parameters, the opposite side that's where that minus sign on.
Birne Binegar: So X one minus X two X one plus x two X equal to zero or X one is equal to minus X two so in the solution vector X one is going to be minus X two X two is going to equal X two yeah so on yours you got to this part and then you had one minus one zero and then your final one.
Birne Binegar: You ended up whenever you're creating see the values were changed right here, oh did I.
Birne Binegar: yeah I should have used that one.
Birne Binegar: will be the same result.
Birne Binegar: yeah it would be the I mean it but it's kind of a coincidence Okay, it would be, it would be the same result because.
Birne Binegar: At any rate, this is even though I switched the sign is still an Eigen vector with eigenvalue zero.
Birne Binegar: And so the idea is you taken I, and I connect you with the correct correct eigenvalue and you correlate eigenvectors with her eigenvalues and and so any one of those columns could be multiplied by any number, and it would still, it would still work okay all right, thank you.
Birne Binegar: Put another way, there's lots of eigenvectors but what we've been doing is grabbing basis vectors and so every other I can vector is just going to be a scale or multiple of that same factor, so you just just working one basis back to rather than the entire set of eigenvectors.