Birne Binegar: I posted the your scores over the weekend on canvas i'll pass back the papers now, but before I do that, let me also point out that solutions have also been posted on canvas So if you have questions on on how they were graded.
Birne Binegar: I suggest you look at the solutions first and then you can talk to me.
Birne Binegar: During office hours or.
Birne Binegar: After immediately after class whatever you can email me if you have questions on how the exam was graded but the average score was like 90% and so.
Birne Binegar: Things continue the way you're going to continue to do 90% and every time the exam that's going to be more than half the class get a's so Keep it up, I will say that the.
Birne Binegar: Subsequently, on the second exam and on the final exam will be doing more.
Birne Binegar: More sort of theoretical stuff, the main purpose of the first exam was to make you familiar with a row reduction calculation, which is underlying all the calculations in this course.
Birne Binegar: you'll see another application of reduction today when I discuss how we actually call the ties sub spaces, but let me pass these exams back and.
Birne Binegar: I see Paul aldrete here.
Birne Binegar: Joseph.
Birne Binegar: De de Leon.
Birne Binegar: Fisher.
Birne Binegar: Kristen.
Birne Binegar: carly.
Birne Binegar: Matthew Griffin.
Birne Binegar: or.
Birne Binegar: Killing her.
Birne Binegar: jocelyn homeless home.
Birne Binegar: grace POPs.
Birne Binegar: out.
Birne Binegar: Robert.
Birne Binegar: Dawson Walker.
Birne Binegar: Okay, if I put your someone's name anyone not get an exam back.
Birne Binegar: Okay, so now what I suggest you do is look over your exams, if you do have questions on out which graded please check the solutions first.
Birne Binegar: And then, if there is.
Birne Binegar: I sometimes don't see things everything on the page until their issues you can you can talk to me about that, but.
Birne Binegar: Always for you to come to me.
Birne Binegar: You can talk to me immediately after class today, if you want or have office hours it to be a zoom.
Birne Binegar: But otherwise you can you can you can talk to me send me an email or.
Birne Binegar: talk to me after class.
Birne Binegar: Okay, so what I want to do today is, I want to pick up.
Birne Binegar: The discussion we.
Birne Binegar: Were conducting right before the exam.
Birne Binegar: We talked about sub spaces, a little bit before the exam these are.
Birne Binegar: Effectively little sub spaces sitting inside larger vector spaces little vector space is sitting inside larger vector spaces geometrically that's certainly what they are.
Birne Binegar: with one exception there they're not just hyper planes they're hyper planes They include the origin.
Birne Binegar: You have to include the origin because one of the properties of the subspace is it's got to be closed with respect to our fundamental vector space operations scale and multiplication and vector addition and so.
Birne Binegar: We say a subspace of US order list of and numbers it's a subset of that so every vector inside the subspace is going to be an n dimensional vector.
Birne Binegar: But what we require in addition is that if we start with a vector that starts off in that subset and we multiply it by number, we should not leave that sunset nor if we add two vectors inside the subset we stay inside the subset so the counterexample to keep in mind here.
Birne Binegar: Is the familiar circles.
Birne Binegar: drop points that are equally equidistant from the origin, then all these vectors who's.
Birne Binegar: That line the circle, they don't respect these two properties, if I multiply affected by a number of, say, I multiply by two by get a vector that's twice as long and it takes me off the circle, or if I add two vectors together.
Birne Binegar: You can see what's going to happen here if I add this vector this backdoor i'm going to get a vector out here, which is again off the circle.
Birne Binegar: And in fact this is typical was most subsets of an n dimensional space there's not going to behave well respect to our two fundamental operations, but what do I do.
Birne Binegar: Then we have what we call a subspace.
Birne Binegar: Right and then these are the most.
Birne Binegar: well.
Birne Binegar: These subsets or and really dominate linear algebra that we're more concerned with sub spaces in general than we are with individual vectors.
Birne Binegar: And there's two ways that these arise that we've already encountered.
Birne Binegar: The first one way, you can create a sub state is take a given list of backers that live inside your vector space and consider all possible linear combinations.
Birne Binegar: This construction actually uses both scale and multiplication and vector addition, and in fact this rule for generating this set.
Birne Binegar: is effectively to take all possible linear combinations of the vectors to create the set.
Birne Binegar: Because you're taking every possibility it's always going to be closed under the two fundamental operations.
Birne Binegar: So this set is naturally a subspace you can create a sub space anytime you have a list of vectors you just take all possible linear combinations you get a bigger set that that could be infinite number of factors, because these coefficients range o'byrne infinite set.
Birne Binegar: Well that's one way of getting a subspace and.
Birne Binegar: i'm point out that geometrically when you create a sub state this way you're basically making a hyper plane that passes to the origin.
Birne Binegar: So.
Birne Binegar: In general, a general hydroplane.
Birne Binegar: recall that these are sets that you formed by starting off at a given point.
Birne Binegar: And then straying from that point on a fixed set of directions.
Birne Binegar: Such that these parameters.
Birne Binegar: are allowed to range of the material.
Birne Binegar: We aligned and up here subspace.
Birne Binegar: where you can get a subspace from this construction, if you take.
Birne Binegar: That offset here take it equal to the zero vector and then in the special case when you just when you this term drops out, then you get exactly the span of a set of actors.
Birne Binegar: All right, but that's one way of.
Birne Binegar: Getting a subspace look at a hyper planed it passes through the origin but another way of doing it is.
Birne Binegar: For linear systems.
Birne Binegar: A solution set up what's called a homogeneous linear system that's where the right hand side is zero.
Birne Binegar: Though.
Birne Binegar: So let me put on this platform.
Birne Binegar: System of linear equations and the general case.
Birne Binegar: Is when you can write a linear system as a coefficient matrix time to vector variables is equal to another vector.
Birne Binegar: Now that's.
Birne Binegar: Of course, a general, the solution set of a linear system isn't always a hyper playing this special case when the hyper plane is actually a sub space, so when the offset here is actually the zero vector this corresponds to.
Birne Binegar: what's called the homogeneous linear system.
Birne Binegar: And it's where instead of having an arbitrary vector on the right hand side only zero vectors allowed for a homogeneous linear system.
Birne Binegar: And what we'll see is this course progresses is that.
Birne Binegar: they're going to be many other sub spaces, that we consider but they're all going to basically when they get down to the nuts and bolts of it constructed by either taking a solution set of a homogeneous linear system or by looking at.
Birne Binegar: A span of a given list of backers this typical way that you make us some space.
Birne Binegar: All right now what we're gonna be talking about this week is.
Birne Binegar: Not so much about what sub spaces are, but how do we compare them.
Birne Binegar: Because.
Birne Binegar: I might have a.
Birne Binegar: A hyper planes solution of a homogeneous linear system, how do I identify the vectors V1 through bk so I can see that solution, set up a linear system like this athlete has this form and vice versa.
Birne Binegar: Well, what we're gonna do is we're going to coordinate.
Birne Binegar: The vectors that live inside of some space and then, once we have good coordinates, then we can make a comparison to see whether or not.
Birne Binegar: To vectors actually live in the same subspace.
Birne Binegar: Okay, so here's the basic idea of how we're going to get good coordinate projectors that live inside a given subspace problem is is that.
Birne Binegar: This is a larger space and the subspace is generally going to be leaf geometrically smaller set of vectors inside a larger set of vectors.
Birne Binegar: But if you're starting off at a 10 dimensional space, you have a subspace of that 10 dimensional state it's still going to consist of pen dimensional vectors.
Birne Binegar: But they're just going to be fewer them you're not going to get like totality of alternate dimension of that fish you'll pick out a.
Birne Binegar: Certain subset of that and that's what we're dealing with today is well, if I have a five dimensional subspace sitting inside a tenement stage how do I describe pointing at subspace I don't want to use 10 dimensional coordinates because the space on concerned about is only five dimensional.
Birne Binegar: And so.
Birne Binegar: The basic idea, though, is that if we create a subspace like at all possible linear combinations of instead of vectors.
Birne Binegar: didn't want idea for how to describe all the vectors that live in type of said is well light a factor out.
Birne Binegar: Oh, if it's the span every vector inside this set is a linear combination of the generators and so maybe the good thing to do would just use the numbers one through ck.
Birne Binegar: As coordinates for the given vector and then by changing these numbers, we could we could move anywhere, we wanted to inside that space and that's good that's certainly what we would hope to do.
Birne Binegar: However, unless we put some additional conditions on these generators, it may be may happen that.
Birne Binegar: Two different specifications of these coefficients could produce the same vector and so you'd have a single point in your subspace.
Birne Binegar: That would appear to have two different coordinates are two different ways of accounting for that point.
Birne Binegar: We don't want to over count, nor do we want to undercount so we want to, we need to get these coefficients what we like is that once I give you a list of coefficients I get a unique point inside this space not several points.
Birne Binegar: So here's the explicit demonstration of this problem here, I have a set of three factors.
Birne Binegar: The factors inside this span, by definition, are linear combinations of these three vectors until let me denote a vector V sitting inside the sub.
Birne Binegar: base so lets me denote it might coefficients that's how I make it I choose number see one time the first vector number see two types of second factor number two three times third vector and I add them up.
Birne Binegar: However.
Birne Binegar: In this particular example if I choose the numbers well one one and zero for my coefficient and I construct the vector it adds up to the vector to zero, but here we're different choice of quote unquote coordinates.
Birne Binegar: I take the linear combination, but I get the same vector so does this affect your to zero as an element of this.
Birne Binegar: up the space is generated by these three factors, does it have coordinates 110 or does it have coordinate 002.
Birne Binegar: Well, both work and that's the problem, because what we want for a coordinates is that you specify a given.
Birne Binegar: set of numerical values for your coordinates you should have unique point in your space, just like in three states if you have a point in three space once you specify it horizontal to horizontal component that vertical opponent, you get a unique point but.
Birne Binegar: Here we have multiple ways of writing down a vector and what we want is a unique way of writing it down.
Birne Binegar: Okay, so what we're going to do if we're going to define our way out of the problem at first so we'll say a basis for sub state it's going to be a set of generators.
Birne Binegar: Such that not only can we express every vector is a linear combination of these we're going to demand that there's only one way to do that.
Birne Binegar: And then, this is, by definition, if we have subspace is generated by basis, then these numbers see one two ck or by definition unique.
Birne Binegar: But I haven't really done any math I could just be find a way the problem i've said that the set of vectors.
Birne Binegar: are somehow such that every vector and be can only be written one way and so that would mean, for example here that these three vectors are not a basis for this space they generate because there's multiple ways of writing it.
Birne Binegar: All right, so.
Birne Binegar: But if this condition is true if there's only one way of writing a given vector inside such patience a linear combination of you've given list of generators.
Birne Binegar: Then this process of coordination works, because now, when, as soon as I have a vector I can expand it in terms of the generators but it's not ambiguous now.
Birne Binegar: Because I specified that these are unique then there's only one set of numbers that will produce a given set of vectors.
Birne Binegar: Right, but I haven't really done anything i've just.
Birne Binegar: hidden the problem into a property of the vectors rather than the subspace itself.
Birne Binegar: So what i'm going to do next, is giving you a simpler criteria to check.
Birne Binegar: With this definition it solves our problems by defining it away but it doesn't really tell me how to find the vectors V1 V2 MP3 such that every vector inside the subs things can be uniquely written in this fashion.
Birne Binegar: dope, in fact, to check it to check to see that a given list of vectors is actually a basis, according to this definition.
Birne Binegar: I have to check every vector is tidy and make sure that there's only one way of writing it now that's impossible because the sub base is going to contain an infinite vectors so what I do on the subsequent slides is I give you with simpler conditions.
Birne Binegar: This is involved a notion of what we call linearly independence, so we have a set of actors sitting in some vector space, if I can take a linear combination of them and produce the zero vector.
Birne Binegar: Then we'll say that the vectors are literally independent if the only way I can get them that sub zero it's like taking all the coefficients equal to zero.
Birne Binegar: So.
Birne Binegar: And then here's the relationships.
Birne Binegar: If I create a subspace like taking a given list of vectors and taking all possible linear combinations then that set of factors it's actually going to be a basis for the subspace if, and only if the factors are linearly independence alright so there's two things going on here.
Birne Binegar: So.
Birne Binegar: w.
Birne Binegar: Equal the span of what am I vectors.
Birne Binegar: w.
Birne Binegar: wk.
Birne Binegar: This means.
Birne Binegar: Every be w.
Birne Binegar: can be written.
Birne Binegar: As cold si.
Birne Binegar: si K ck wk all right, but it doesn't.
Birne Binegar: there's no uniquely in here and so to get uniqueness.
Birne Binegar: To get uniqueness.
Birne Binegar: that's where linear independence, it can come in.
Birne Binegar: And in fact and what's linear independence linear independence means the only solution.
Birne Binegar: See one.
Birne Binegar: w one.
Birne Binegar: Plus C K w K.
Birne Binegar: is equal to zero is the solution where all those coefficients separately equals zero.
Birne Binegar: Right now.
Birne Binegar: From the definition of a basis.
Birne Binegar: If every vector has unique.
Birne Binegar: representation as a linear combination of your given list of actors, then you have a basis, but the problem is, is when you write a backdrop here, no guarantee, yet that there's only one way of.
Birne Binegar: Producing a given vectors a linear combination of your prescribed set.
Birne Binegar: What the serum that I have an athlete stated completely there what it says is that you don't have to check every vector to make sure.
Birne Binegar: Every vector inside your subspace has a unique representation, you only have to check the zero vector and if that's true, then.
Birne Binegar: Then it's back to your original set of vectors is going to be a basis and every other better, besides the zero vector will also have a unique crafts representation as a linear combination of your generators.
Birne Binegar: Alright, so let me show you how this works it's just logic and definitions, so what i'll first show you that if you have.
Birne Binegar: subspace that generated by given lists of vectors i'm not supposed to get these guys are basis vectors yet, but if they're linearly independent.
Birne Binegar: Then they will be a basis Oh, I guess i'm showing the opposite i'm assuming that I have a basis, and the first thing i'll show you is that the only way.
Birne Binegar: If these are basis, and the only way.
Birne Binegar: w can be generated is if these guys are also linearly independence okay so let's assume that this these factors, provide a basis for w that means every vector inside w has a unique expression it's a linear combination of the given list.
Birne Binegar: In particular subspace always contains the zero vector why because you can it has to be closed under scale and multiplication.
Birne Binegar: scale multiplication by the number zero always produces the zero vector so whenever you have a subspace it's always got to contain the zero vector so let's write the zero vector as a linear combination of these guys well.
Birne Binegar: Since.
Birne Binegar: Since this is a basis and i've got a vector inside w written as a linear combination of my basis factors that makes these coefficients unique like definition.
Birne Binegar: Be is a basis if every expression of this this form produces a unique vector inside your subspace alright, but that then.
Birne Binegar: We need a unique well we're guaranteed a unique expression for w because zero belongs can w and w wk or basis.
Birne Binegar: But notice also that you can choose see want to see K equals zero and this equation is going to be automatically satisfied because zero cons any vectors just going to produce the zero vector.
Birne Binegar: So, because we have a basis, there should be only one way of writing this factor inside w it's a linear combination, but then this guy, this is the solution to that problem if you choose the coefficients all equal to zero, then this equation will be true.
Birne Binegar: And so.
Birne Binegar: If w wk is a basis, then there's only one way to write the zero vector and in fact it's the solution we're all these coefficients or separately equal to zero.
Birne Binegar: All right, so.
Birne Binegar: that's The first part is going one direction if you have a have a basis, then the generators in that basis have to be linearly independent.
Birne Binegar: What about the opposite way suppose we have a set of linearly independent vectors and I want to show that the vectors are going to be a basis.
Birne Binegar: So will create the space it generated by our set of linearly independent factors and what i'm going to suppose, this is, I have two vectors or a single vector that can be written two different ways, like in that early examples go back to it.
Birne Binegar: In this example.
Birne Binegar: I had a single vector that could be written two different ways as linear combinations my generators that's the situation i'm gonna i'm going to examine.
Birne Binegar: Your question.
Birne Binegar: Alright, so.
Birne Binegar: Yes, yes, so i'm going to suppose that I have a single vector living inside the span of one given set of generators, and that I have two different ways of writing it.
Birne Binegar: And i'm also assuming that might generators w wk are linearly independent.
Birne Binegar: Though I suppose I have two different ways of writing the back.
Birne Binegar: Then i'll subtract one equations from the others B minus D is going to be the zero vector and then you can select term so that see one w one minus P one w one is see one minus one times w.
Birne Binegar: and, similarly, you can group the terms the remaining terms on the right hand sides now now we impose a condition that.
Birne Binegar: The factors w want through wk are linearly independent That means that the only linear combination of these vectors that produces the zero vector is the one we're all the coefficients equal to zero.
Birne Binegar: So if I have zero is equal to a linear combination of w wk this coefficient has to equal zero second coefficient has equal zero on the last coefficient has to equal zero.
Birne Binegar: So.
Birne Binegar: If these coefficients have to equals zero That means that see one minus P, one has to equal to zero so see one has to equal the ones like just move this to the opposite side.
Birne Binegar: Similarly, this last condition the fact that this coefficient has to be equal to zero, if this linear combination of factors is going to equal zero.
Birne Binegar: That means that the last coefficient has to equal the Lascaux efficient So in fact see what has to equal he wants you to has to equal D to decay has to equal bk there's only one choice of coefficients.
Birne Binegar: Alright, so I proved it this approved if you have a basis, then the generators are linearly dependent and I proved that if you have a linear independence set of generators.
Birne Binegar: Then, now there are basis, and what that does, for me, it gives me an easier condition to check, rather than having to check that every vector inside my subspace has a unique.
Birne Binegar: is represented as a unique linear combination, I only have to check that the zero vector is unique and then, once I know that, then I know I have a basis and every other linear combination is going to be.
Birne Binegar: Okay, OK so.
Birne Binegar: The find a basis for subspace your essential task is to find a set of literally independent vectors to generate the subspace.
Birne Binegar: Put another way, you have a list of vectors it's going to be a basis for subspace if you create that subspace like taking all possible linear combinations of these factors, but.
Birne Binegar: If these factors are not linear independence, there could be multiple ways of writing a given vector in terms of these guys.
Birne Binegar: But if there are basis there's a unique way and that's equivalent to saying that these vectors have to be linearly dependent, meaning that the only way, you can satisfy this equation is by taking these coefficients equal to zero.
Birne Binegar: All right, so let me explain some more i'm.
Birne Binegar: Talking around this a little bit So you see how things are connected.
Birne Binegar: I guess, the first thing I want to clarify is some nomenclature.
Birne Binegar: If I have a set of vectors and they obey an equation like see one B one.
Birne Binegar: ck bk is equal to zero.
Birne Binegar: Be.
Birne Binegar: The only solution.
Birne Binegar: Is.
Birne Binegar: equal to zero C two is equal to zero etc last coefficient has equal zero if this is the only solution of this equation is the one where you set all the coefficient separately equal to zero, that means the set the vectors are linearly independent.
Birne Binegar: linearly independent.
Birne Binegar: Alright So what is this.
Birne Binegar: notion of independence really means let's look at that suppose Well, first of all let's look at that what the equation says, under those conditions it's basically says that zero.
Birne Binegar: is equal to zero because of all these coefficients are equal to zero, the entire right hand side is going to equal zero, and as soon as you do, the scale of multiplication and you get a trivial equation not putting any conditions on the vectors.
Birne Binegar: If this is true, on the other hand, if there's another solution.
Birne Binegar: Here we're talking about the solution we're all the coefficients equal to zero to one another solution would be another sequence of numbers, where at least one of the numbers it's not equal to zero right so yeah at least.
Birne Binegar: One quotations say see I it's not equal to zero, but if that's true I could take this equations and I could write VI is equal to one over see I.
Birne Binegar: See one do one see to be too, I guess, I need a minus on there.
Birne Binegar: And then we'll skip.
Birne Binegar: c plus dot dot dot plus C minus one.
Birne Binegar: minus one.
Birne Binegar: Plus C plus one plus one.
Birne Binegar: Plus other terms until you reach si K.
Birne Binegar: OK OK, so what have I done i've taken this equations.
Birne Binegar: Focusing on the term that involves the coefficient it's not equal to zero, and I basically solve for the corresponding vector by moving.
Birne Binegar: All the other terms to the opposite side and then dividing mice by car.
Birne Binegar: What i'm showing you here is that, if I have a non trivial solution something other than this easy solution if I have a non trivial solution.
Birne Binegar: It will allow me to express one of the vectors in terms of the other vectors and so this vector bi depends on the other, vectors that's where this language comes.
Birne Binegar: of dependence and independence.
Birne Binegar: If the only way I can solve this equation is by taking all the coefficients equal to zero, that means there's no relationship between these doctors upon evaluation.
Birne Binegar: For this solution just says zero is equal to zero, but if we have a different solution, it will allow me to solve, but what are the barriers in terms of the other until the vectors are dependent upon one another.
Birne Binegar: Okay, so now.
Birne Binegar: Some more nomenclature i'll refer to this as the trivial solutions, because if you simply replace these numbers these coefficients by zero this is identities, but for other solutions.
Birne Binegar: If there exists other solutions of the equation that's when we say that this equation is a dependent relation, because if we have other solutions, we can use the equation to express one of the vectors in terms of the other vectors.
Birne Binegar: alright.
Birne Binegar: So the language is kind of.
Birne Binegar: will say yes vectors heart literally dependent, if there is a dependence relation and if there is no dependent relation if there is no solution other than the obvious solution, then we say the vectors are linearly independent.
Birne Binegar: Alright, so.
Birne Binegar: What we're going to use.
Birne Binegar: to coordinate I sub spaces.
Birne Binegar: are going to be set to vectors that are linearly independent.
Birne Binegar: So what.
Birne Binegar: What i'm going to do next is show you an easy way to spot if a given list of vectors are literally independent.
Birne Binegar: I guess I should do this visually it do it on pros on this slide until I get down to the stage what i'm what I want to show you out now, if you look at these two vectors.
Birne Binegar: This vector V one has a component and the vector be to doesn't have a component that a non zero component in the same slot and similarly this vector be to as a non zero component in the third slot and the vector V one as zero this this is this is sort of going to lead us to.
Birne Binegar: That observations can lead us to conclude that B one and B to are linearly independent why because why take a linear combination of these two vectors and I tried to get the zero vector out of that well this one X one times one.
Birne Binegar: it's going to.
Birne Binegar: Do this vector some I get an x one here, but that can't equals zero but outside of X one is equal to zero, and similarly in this third component.
Birne Binegar: I have X one time zero, which is going to put zero but i'll add to it X two times, one which will be X two but that third component cannot equal zero without setting X two X equals zero.
Birne Binegar: So what i'm showing here is that if you have a set of vectors in which each one of the vector has us component that can't be cancelled out by the other vectors it's automatically going to be.
Birne Binegar: literally in the past, independent because.
Birne Binegar: Simply they get the cancellation and that first slot you have to take X one is equal to zero, and they get the cancellation of thirst lot, it cannot be.
Birne Binegar: It cannot be matched by contributions on the first factor until you have to take the second equal the zeros.
Birne Binegar: Yes.
Birne Binegar: Let me show you some things.
Birne Binegar: suppose I have.
Birne Binegar: A matrix.
Birne Binegar: In row Echelon forms.
Birne Binegar: My claim is that the rogue actors in this matrix are automatically linear independence, why, because if I take a linear combination of the three rows.
Birne Binegar: I would multiply the first row vector by some constant but I won't be able to cancel it would contribution from the lower rows.
Birne Binegar: And so, if I had see one times this robe actor see two times second row better see three times the third row backers night add up all these robot yours, let me write it this way instead.
Birne Binegar: You should think of them sitting inside a matrix.
Birne Binegar: But think of them as real vectors at these three rows by have a linear combination that's going to produce zero well that first slot I have one plus one plus zero Plus zero.
Birne Binegar: seconds slot I have zero to see two.
Birne Binegar: Zero the third slot I have.
Birne Binegar: See one plus C to.
Birne Binegar: c three I guess.
Birne Binegar: Okay now how could this vector how could this linear combination possibly zero I need see one day equals zero because I can't get these guys to cancel that guy so that has to equal zero us here we have another.
Birne Binegar: Another pivot low that pivot I have only zero and above the pivot to I only have zero there's no way I can actually take a linear combination of the three rows and get rid of the central term without taking see two equals zero.
Birne Binegar: But then, but then, if see one is equal to zero that's equal to zero, and now i'm forced also to have see three is equal to zero, if the some of these three robe actors is going to equal to zero vectors.
Birne Binegar: So.
Birne Binegar: What happens then is that.
Birne Binegar: If you have a list of vectors.
Birne Binegar: And you write them as the rows of the matrix and it turns out that matrix is in real life long form, then you know that those directors are also literally independent.
Birne Binegar: Alright, so.
Birne Binegar: I think I stayed as a theorem on the next slide.
Birne Binegar: suppose you have an n by N, make it real Echelon for.
Birne Binegar: Then the real vectors.
Birne Binegar: qualification there, the non zero row vectors of a are automatically linearly independent.
Birne Binegar: Let me explain why I need to specify zeros let's look at an example.
Birne Binegar: is equal to 111011000.
Birne Binegar: Okay, so suppose that a matrix, then, can I get.
Birne Binegar: See once.
Birne Binegar: How can I get see one times the first row.
Birne Binegar: I see two types of second row.
Birne Binegar: A plus C three times the third row of ace.
Birne Binegar: equal to zero, without taking all these coefficients equal to zero, well, I could set this equal to zero, of a concept that equals zero, and I could set a third coefficient equal to one.
Birne Binegar: Because it anyway it's multiplying zero vector and so which going to produce zeros so.
Birne Binegar: These three rows are not going to be literally independent because there's a way there's a linear combination of the three rows that produces the zero vector we're not all the coefficient or equal to zero.
Birne Binegar: But.
Birne Binegar: What if I get rid of the zero row then there's no way of adding these two rows together but taking linear combination of these two rows and produce zero without taking the coefficients zero.
Birne Binegar: So that's why I exclude the non zero So if I if I allow myself a non zero row it's easy to throw it in with the coefficient and you're still gonna it's not gonna affect its ability to proof the zero vector out of it.
Birne Binegar: Okay, so.
Birne Binegar: The way this works is.
Birne Binegar: If you have a matrix in real Echelon form, then the first row, you have the pivot that furthers to the lap and below it, you have nothing but zeros.
Birne Binegar: And so, this role is going to have a component that's can't be matched by the other vectors and so get this contribution i'd have to set X one is equal to zero.
Birne Binegar: But once I set X one is equal to zero effectively remove that term from the thumb and now i've got.
Birne Binegar: A smaller matrix under consideration, now I removed the first row, but I still have the problem pivot.
Birne Binegar: In this new matrix form the using the second to the last row with the matrix it's still going to have a pivot and know pivots below it.
Birne Binegar: And so there'll be no possible counts tentatively the pivot component of the second row with the other row to the matrix i'll have to set X two X equals zero and you just work your way down the matrix until you conclude.
Birne Binegar: That the matrix is in regression line for the only way, you can get these non zero rose to some to the zero vector is by taking all the coefficients separately equals zero, which means that are linearly dependent, and so the rows of a.
Birne Binegar: matrix in row Echelon form are always going to be a set of non zero rose always going to be a set of literally independent actors.
Birne Binegar: Right now i'm.
Birne Binegar: just making that argument again.
Birne Binegar: So maybe I.
Birne Binegar: get a little bit further here.
Birne Binegar: So far, but we have.
Birne Binegar: A is.
Birne Binegar: It a row Echelon for.
Birne Binegar: Its.
Birne Binegar: Non zero vectors.
Birne Binegar: are necessarily literally independent.
Birne Binegar: Now, define us a vector subspace attached to a matrix what i'm gonna do is i'm going to take the real vectors of the matrix and i'm going to take all possible linear combinations that should have been see I guess the end Oh, maybe this one X one sense, so.
Birne Binegar: So if I said.
Birne Binegar: row space.
Birne Binegar: Of a equal the span.
Birne Binegar: Of the robe actors.
Birne Binegar: Okay, let me be specific here if a is, and I am in rows.
Birne Binegar: And columns.
Birne Binegar: Then every row has gonna have exactly am entries so that puts the row space.
Birne Binegar: As a.
Birne Binegar: As a subspace.
Birne Binegar: Of.
Birne Binegar: The set of n dimensional vector each row vector is going to have him components, and so the span of those real vectors it's going to be some subspace of our EMS until.
Birne Binegar: The.
Birne Binegar: Non zero rose.
Birne Binegar: are going to generate.
Birne Binegar: space.
Birne Binegar: The Non zero roads, they not only generate the row space of a but they're also linearly independent.
Birne Binegar: That means there are a basis for the road of space so here's the first example of a subspace for which we can immediately write down a basis if we have a matrix and row Echelon form.
Birne Binegar: The basis for the subspace it generated by those road is always going to consist of a non zero or the non zero row vectors of the matrix general Echelon forms.
Birne Binegar: What we'll see next time.
Birne Binegar: is how to handle more general matrices.
Birne Binegar: From the first exam you know that not every matrix is irrational and form, you have to do some work to get it in that shape but.
Birne Binegar: So on Wednesdays.
Birne Binegar: What will show is that.
Birne Binegar: operations.
Birne Binegar: Do not change.
Birne Binegar: Aerospace of a matrix.
Birne Binegar: This should make a little bit of sense to you, if you remember where these elementary real operations came from.
Birne Binegar: We discovered elementary row operations when when you're trying to solve linear systems and the way we did, that is, we first showed you that you can solve linear system by.
Birne Binegar: manipulating the equations by doing operations that don't change the solution set.
Birne Binegar: Then what we did is we carry those operations over so that they acted on the augmented matrix rather than on the equations themselves.
Birne Binegar: But those elementary operations performed on the augmented matrix did not change the solution set.
Birne Binegar: And so what we'll see on on Wednesday is that the elementary operations you perform on a matrix.
Birne Binegar: they're going to change the individual rows of the matrix but the space it's generated from them it's not going to change, just like the solution set up the corresponding system of equation doesn't change under the rule operations.
Birne Binegar: What that's going to mean that we have now have a way of finding a basis we take an arbitrary matrix we perform elementary row operations until we reach real Echelon porn.
Birne Binegar: The real question a lot for having been obtained by operations, that is not changed the row space it's still going to.
Birne Binegar: The real issue on form is going to be this it's ro space it's going to be the same as the original matrix but.
Birne Binegar: For the role Echelon form the roads are linearly dependent and so we'll use the basis for the role Echelon that natural basis for the role is one form will use that as a basis for the road space of a mill be able to coordinate like a the space of a.
Birne Binegar: indirectly through which row epsilon form.
Birne Binegar: Okay, so i'll stop here.
Birne Binegar: i'll stick around if you guys have questions on your exam but otherwise we'll we'll pick up with this topic on Wednesday.