Birne Binegar: Okay, so your tests are graded.
Birne Binegar: I posted the your scores also on canvas and i've also posted solutions at the top of our canvas page, but you all did very well I, I think the average score was about 90%.
Birne Binegar: And so you know well over half the classes, at this point i'll be looking at an A in the course, but things are going to get more difficult as well, not so much more difficult, that you had you have to stay at it we're going to make a transition now into more theoretical material where.
Birne Binegar: relations that you did on the on the first exam are going to come into play, over and over again, however.
Birne Binegar: What i'm going to do is like, as I lecture now is i'm not going to step through the details of a row reduction calculation.
Birne Binegar: Maybe give you a matrix that you started off with and it's ro Echelon form, so I don't have to do those tedious calculations on screen so much, but let me pass the stack.
Birne Binegar: Yakubu you here.
Birne Binegar: sylvan.
Birne Binegar: Roger Bennett.
Birne Binegar: David caliber.
Birne Binegar: castello.
Birne Binegar: Lisa Fischer.
Birne Binegar: Eric.
Birne Binegar: Thomas.
Birne Binegar: Thomas kid.
Birne Binegar: pick it up.
Birne Binegar: paxton craft.
Birne Binegar: Thomas good to you here.
Birne Binegar: Okay paxson craft.
Birne Binegar: You like Lucas.
Birne Binegar: dancer.
Birne Binegar: Center here.
Birne Binegar: Jacob heron.
Birne Binegar: Taylor.
Birne Binegar: Jordan wheeler.
Birne Binegar: let's see.
Birne Binegar: Alexandra wilburn.
Birne Binegar: Madison Wilson.
Birne Binegar: Alright, so people wandered in.
Birne Binegar: Anyone not have an exam damn yeah.
Birne Binegar: Okay, so.
Birne Binegar: Mostly errors on the exam or.
Birne Binegar: trivial yeah there were some problems with rob reduction.
Birne Binegar: yeah.
Birne Binegar: Oh yeah.
Birne Binegar: Okay, so I can answer questions on the exam but, like, I suggest you you look at the solutions that are posted if you want to argue for more points in the like, please have a look at the solutions first and maybe maybe then you'll spot your inner but.
Birne Binegar: Given that the class average was 90 i'm not.
Birne Binegar: so inclined to step through the the examine class than that, and if you have any difficulties at all understanding the post it solutions you can come talk to me to sit me during office hours, etc.
Birne Binegar: Okay, but once again for those who came in, like the average was 90%, which means that you know, the vast majority are.
Birne Binegar: doing well in this portion you can probably continue to do so because i'm not going to change things up second exam we will review the same way, will give you a practicing damn and.
Birne Binegar: As long as you sort of follow my directions, you should do do well in this course alright so.
Birne Binegar: What I want to do today.
Birne Binegar: Is move forward in the material little bit just prior to the.
Birne Binegar: The exam I talked a little bit about sub spaces i'm going to review that real quickly and then.
Birne Binegar: Today, so today's lecture we're going to get into the meat of how to deal with spaces well.
Birne Binegar: Typically, your your basic problem is, you may might have a three dimensional subspace embedded in 10 dimensional states you don't want to use.
Birne Binegar: 10 different coordinates to specify the vectors that live in a three dimensional space.
Birne Binegar: And so we're going to address that problem as a coordination problem for such cases and will resolve that.
Birne Binegar: fairly quickly but you'll notice that the way I resolve it, you simply by changing my definitions do to give me something, it will be.
Birne Binegar: Allow me to coordinate ties sub spaces and that's going to be related to linear independence.
Birne Binegar: I choosing special vectors that are linear independence, then the coordination of substations is easy.
Birne Binegar: And then, at the end I get that far we'll look at the row space of a matrix, and this will be a subspace that we can attach to any matrix and will.
Birne Binegar: will describe have we actually coordinate ties those vectors and in fact this example here the row space of the matrix matrix is.
Birne Binegar: Actually, the prototype for a general calculation, even when we're dealing with sub spaces that are different than the row states of the matrix will tend to take.
Birne Binegar: Coordination problem is will rephrase it as a problem of finding the row space of a matrix so it's a it looks like a particular example, but in fact this one this example is crucial, the calculations.
Birne Binegar: Okay, so that remind you, but the subspace is basically the idea is to find a little vector space inside larger vector space, but the key point.
Birne Binegar: In linear algebra is also always those two fundamental operations at scale and multiplication in that tradition.
Birne Binegar: So what we require up a subspace as a subset to really be regarded as a vector space sitting inside a larger vector space, if we want to be able to maintain the two fundamental vector operations.
Birne Binegar: inside our little vector space, so what we require is that we not only have a subset so every doctor is going to be a vector in a larger vector space.
Birne Binegar: But what we require is, instead, if we take a vector that lives inside that space and we multiply it by any real number we don't leave that so space.
Birne Binegar: and, similarly, if we start off with two bags vectors that start off living into sub sub state, we have them together we don't want to leave that That said, we want the set to be.
Birne Binegar: Closed under scale or multiplication and degradation, so those two operations can be performed freely without leaving the sense.
Birne Binegar: Alright, so.
Birne Binegar: that's the idea, and of course the counter example you might keep in mind it's simple if you have a point on a circle, all these vectors here.
Birne Binegar: they're not going to be a subspace because, for example, if I multiply a vector by two I get a vector twice as long as it takes me off the circle.
Birne Binegar: If I add two vectors together, I get a point that in general is gonna lie off the circle so most of the sets that you know from playing geometry are not going to be sub spaces what they have to do is first of all, since I can multiply by any number.
Birne Binegar: The factors are serving can be as long or short as I want and they still have to stay inside this yet so any finite region in the plane it's not going to be a subspace because, like multiplying, by a large enough number, you could take it outside debts that's it.
Birne Binegar: Okay, but the point is, is that these are the very special kind to set that.
Birne Binegar: are really a dominant in in linear algebra.
Birne Binegar: Alright, so there's two ways of making a subspace that are pretty standard, the first way well, so what I want.
Birne Binegar: For sub spaces for to be close respect to scale or multiplication in Dec tradition, what if I take all possible scale and multiplication and all possible vector additions, well then, this these two conditions are automatically because the definition set this it like add two vectors.
Birne Binegar: Like so it's going to be inside the sacrifice scalable vector it's still going to be inside that set so this definition.
Birne Binegar: You can think of that it's a simple way of imposing these to get your ton of set you demand it every vector in the subset s is created by taking linear combination since and scale of multiples of a given list of vectors.
Birne Binegar: geometrically these guys are going to be hyper planes except their special hyper planes are hyper planes that actually pass through the origins, we call them more general hyper planes.
Birne Binegar: These are sets of the form, you have a starting point and then you stray from that starting point along it given list of directions say kp K and then you let parameters to you want to tk range over the entire real life, this is typical hydroplane the this formula here what's missing.
Birne Binegar: it's missing this offset P zero or you can think of it if I take this as defined to be zero, then I don't need to add to it and then you've got.
Birne Binegar: you've got that span displayed as a hyper plane so it's a it's a the spans are special cases and they correspond to black higher dimensional surfaces that passes through the origin.
Birne Binegar: A second way of.
Birne Binegar: Producing a subspace is a bicep by folding a linear system, but just like for.
Birne Binegar: For a general linear system.
Birne Binegar: Such as.
Birne Binegar: A times X is equal to be so this is some coefficient matrix times the vector a variable equal some vector.
Birne Binegar: constants here this general linear system.
Birne Binegar: Its solution said it's always going to be a hyper plane, but the special cases we're looking at today.
Birne Binegar: are going to be the span.
Birne Binegar: So substances so it's going to be where P zero is equal to zero and.
Birne Binegar: Also solution sets of equations but now the right hand side here it's not an arbitrary vector has to be the zero vector so the solution sets of linear equations of this form are going to be hyper planes of this form they're going to be exactly the span and that the special case.
Birne Binegar: Is what we're looking at today we have substances.
Birne Binegar: Okay don't problem we're going to address today is.
Birne Binegar: Given a set of vectors or a subset that generated by given list of factors like generated by taking linear combinations of the given list of actors.
Birne Binegar: How do we coordinate ISIS.
Birne Binegar: I might have five vectors here, but each vector itself, like the attendee dimensional vector So how do I coordinate is that five dimensional space when it lives inside of 10 dimensional space well here's the idea.
Birne Binegar: Every element in the span has to be a linear combination of the generators and so every vector inside w, it has to be capable of be written in this fashion, otherwise it would live in a span, it would not be a linear combination of those vectors So the idea would be to use these K coefficients.
Birne Binegar: put them in an ordered list and regard that as a coordinate vector for the elements in a w, however, unless we put some additional conditions on the vectors w wk.
Birne Binegar: There might be more than one way of writing a vector, and so we might have a single point with two different coordinates and that.
Birne Binegar: What you need for good coordinates it's a one to one pairing between the object you're trying to study and your numerical doctors that are describing point to.
Birne Binegar: It so let me give you an example of the issue here.
Birne Binegar: Well here's a set of three doctors.
Birne Binegar: And i'm i'm going to take the setup all possible linear combination of these two factors, so that means i'm considering vectors.
Birne Binegar: Of the form of some constant times the first factor some constants that conduct your some constant times the third sector.
Birne Binegar: Any vector of this form it's going to belong to, so I can think of coordinating the point of S by the number C one C two and tier three given those numbers.
Birne Binegar: I can construct a vector that's going to live inside that set well that's all good except here i'm going to show you two different choices of numbers that produce the same vector.
Birne Binegar: So here's one choice I choose the first coefficient of one second coefficient one, the third coefficient through some linear combination up, I get the vector to zero.
Birne Binegar: But for this same three generators, I can choose a different list of coefficients 00 and two and in this case I construct the linear combination.
Birne Binegar: Add them up, I get the same vector so is this point two zero which lives inside this subspace doesn't have coordinates 110 it doesn't have coordinate 002 it's ambiguous.
Birne Binegar: And so, this is the situation we're trying to avoid we want to have unique coordinates for every point in a subspace not a choice.
Birne Binegar: Alright, so the way will resolve this is effectively will define our way out of the problem.
Birne Binegar: you're relevant definitions.
Birne Binegar: A basis.
Birne Binegar: For sub spaces assertive vectors such that every vector can be written as a linear combination of these sectors, so this means in particular that F.
Birne Binegar: has to be the span.
Alex: Of these factors.
Birne Binegar: But in addition not only can we express them as a linear combination, but that with that adjective there.
Birne Binegar: We want a unique way of expressing every vector in terms of the vectors V1 through bk So if you if a set of vectors V1 bk is the basis, then not only can we reach.
Birne Binegar: Every vector in the set by taking linear combination, but in fact there's only one way to reach a given decter via such up a linear combination.
Birne Binegar: And then, what we could do.
Birne Binegar: Is now the choices of the number C one C two ck they're going to be unique that's hidden in the definition.
Birne Binegar: So if we have a basis and we see an expression for a vector inside of subspace written in terms of that basis, then we know the number one number see want to see K or unique, by definition, something we have to prove however ah, how do we actually check this that's.
Birne Binegar: harder given a given a.
Birne Binegar: subspace that were generated by list of vectors, how do we tell if attack the basis, do we have to check every vector inside the subspace to make sure that there's only one way to write it.
Birne Binegar: Well, that would that would be what we would need to do if we wanted to use the definition direct, but what i'll show you next is there's a simpler test to see whether or not a given set of factors is the basis.
Birne Binegar: So this has to do with a new notion hold linear independence.
Birne Binegar: So we get about substations for a second now suppose I have a list of factors and suppose they're related by an equation of this form scale and multiples of this set of vectors certain scale of multiple.
Birne Binegar: You add them up and you get the zero vector suppose that.
Birne Binegar: The only solution of this equation thinking of the coefficient X one X K is variable suppose the only solution to that is by taking all the coefficients separately equal to zero, then we'll say that the vectors w wk are literally independent.
Birne Binegar: Okay, so so that's.
Birne Binegar: A special conditions and.
Birne Binegar: What I want to show you.
Birne Binegar: Is that if you have a set of generators that are literally independent then they're automatically going to be a basis now that's an improvement over the definition, because to apply the definition directly, I have to show that every vector inside myself that has a unique.
Birne Binegar: representation as a linear combination of like given list of vectors but with this notion of linear independence.
Birne Binegar: The theorem about the show you says, I only have to check that the zero vector can be uniquely represented as a linear combination that all know that, in fact I generators are actually a basis for the subspace degenerate.
Birne Binegar: Okay, so let me prove this theorem me move it up statement up, so the middle.
Birne Binegar: screen suppose we have a subspace that's generated by given list of vectors.
Birne Binegar: Unless the question is is can every vector in w be represented by unique linear combination of these guys if that's the case, then the vectors are going to be a basis.
Birne Binegar: So the statement of theorem is that w wk is going to be a basis for the subspace it generated from them if, and only if the given list of vectors are linearly dependent meeting there's only one solution, only one linear combination of these factors that equals zero factors.
Birne Binegar: So let me show you how that works.
Birne Binegar: Suppose.
Birne Binegar: labeled by be the set of vectors and will suppose, this is if, and only if statement that means both the Ford statement and the converse statement or true i'll do the Ford statement first i'll assume that I have a basis and i'll show you that necessarily the vectors are linearly independent.
Birne Binegar: Alright, so.
Birne Binegar: If i'm.
Birne Binegar: To be as a basis, well, in particular, that means every vector inside w has a unique expression as a linear combination those vectors.
Birne Binegar: The zero back don't also have to live inside w because scale of multiplication by zero is always going to produce the zero and sub spaces are always closed under scale or multiplication.
Birne Binegar: So the zero vector is going to be an example of a vector that lives inside capital w what I want to show is that these guys are a basis that in fact the only solution of this equation is obtained by setting see one two ck separately equal to zero.
Birne Binegar: Well, because be as a basis.
Birne Binegar: Any factor w has a unique expansion, in particular the zero vector has a unique expansion.
Birne Binegar: Okay, so there's only gonna be one choice of the number see one two ck that satisfied this condition, given the hypothesis that these are places for the sub they say generate.
Birne Binegar: However, it's pretty awful pretty clear that see one is equal to 02 is equal to zero all coefficients equal to zero, if you make the substitution on the right hand side you're going to get zero.
Birne Binegar: So, in fact, this is a solution of this equation a hypothesis is that there's only one solution, and so, if this equation is true, then you can.
Birne Binegar: conclude then that all the coefficients all separately banish if an equation like that holds and all the coefficients have to banish That means that are linearly dependent and we're done.
Birne Binegar: Okay, so.
Birne Binegar: The object in this this for direction is to assume we have a basis and then use it that to conclude that the basis factors satisfied equation like this with only one solution.
Birne Binegar: The trivial solutions.
Birne Binegar: And then.
Birne Binegar: Go the opposite way suppose that I have a set of vectors that are linearly independent, so the only way I can add them up and producing zero vectors by taking all the coefficients equal to zero.
Birne Binegar: Well, and let's suppose also that we look at the space is generated by the factors, the claim is is that.
Birne Binegar: it's the factors are linearly independent, then these vectors are also going to generate w well that's given by the way, i've constructed w but now there's going to be unique way of writing any vector and type w so linear combination of these guys.
Birne Binegar: So.
Birne Binegar: here's what we'll do.
Birne Binegar: we're going to assume that our vectors are linearly independent and that they generate the given subspace capital w.
Birne Binegar: What I want to show if there's only one way of writing any vector inside w as a linear combination of the given list of vectors so i'll do a proof by contradiction suppose I had two different ways of writing a particular vector inside w.
Birne Binegar: Well, what I can do, then, is, I can take these two equations so I can subtract the second equation from the first will be minus D is going to be the zero vector and then.
Birne Binegar: See one w one minus D one w one is going to be this term, you can see how that is going to work out, but now.
Birne Binegar: My hypothesis is that the vectors w one through wk are linearly dependent well, that means that the only way, you can solve this equation is by setting all the coefficients equal to zero, so these guys are linearly independent each of these coefficients has to equal zero.
Birne Binegar: But if see one minus P one this first coefficient that's equal to zero, that means that see one is equal to one.
Birne Binegar: And if the second coefficient or last coefficient, for example, si K minus DK, the only way I can solve this equation is by taking all the coefficients equal to zero, well then i'll need ck minus DK to equal to zero and that's simply going to.
Birne Binegar: imply that seek a thing called a DK So in fact.
Birne Binegar: If the doctors were linearly independent there would be only one way of writing a given vector You can write it two different ways, but as soon as you apply linearly Leonard literal independence you able to conclude that that all the coefficients have to match anyway.
Birne Binegar: Okay, so.
Birne Binegar: What that tells us, then, is to be get a basis just get a good set of coordinates for vectors in a subspace, we need a set of linearly independent vectors to generate.
Birne Binegar: So let me put it another way, a given list of factors is going to be a basis for a subspace if, and only if, first of all.
Birne Binegar: You have to be able to reach every point in your subspace by taking linear combinations of your given set, and second, of all you want to be able to reach a given point inside this subspace like only one particular linear combination.
Birne Binegar: But i'm going to replace that uniqueness property, the uniqueness of an expansion of a vector and I w respect to that basis i'm going to replace it by the linear independence condition so.
Birne Binegar: set of factors is going to be a basis for w if you have enough factors to reach every point in w and, second, of all you want the vectors to be literally independent, so the only way they can.
Birne Binegar: satisfy an equation like this is like taking what an effect is trivial solution for all the coefficients or equal to zero.
Birne Binegar: This means.
Birne Binegar: Both directions, have to be true.
Birne Binegar: Regular.
Birne Binegar: Okay, so yeah It shows it's a good question because.
Birne Binegar: You.
Birne Binegar: Give me an opportunity to point out that you confusing terminology, we don't talk about hyper planes is being linear independence, we talk about a set of vectors to make the hyper plane.
Birne Binegar: Is the hype linearly independent or not so we're concentrating on the generators and dump and in fact it's not going to be a fighter plane in general.
Birne Binegar: What we wait it's gonna be a hydroplane that passes through the origin, in order to be a subspace so in today's lecture we're restricting your attention.
Birne Binegar: To hyper planes it passes through the origin, so called spans have a set of vectors are always sub spaces and our problem was to find good coordinates for them and.
Birne Binegar: Our solution, thus far, sorry this should have been I guess oh I changed the wrong letter here, there should be a linear combination of factors, do you want to bk and.
Birne Binegar: The language will use it anytime you have an equation, so I let me write the correct equation up there.
Birne Binegar: C one.
Birne Binegar: c K.
Birne Binegar: bk is equal to zero factor.
Birne Binegar: anytime you have an equation like this well, you always going to have one simple solution where you take all the coefficient see want to see Casey equal to zero, so that solution.
Birne Binegar: will call the trivial solution of one you get it easily simply by setting all the coefficient separately equal to zero.
Birne Binegar: If there are other solutions, besides this trivial one then we'll say that this equation implies dependence relation amongst the vectors.
Birne Binegar: The one through bk the reason why it's called a dependence relation.
Birne Binegar: Let me just give you an example suppose we have two X one plus three x to.
Birne Binegar: The one.
Birne Binegar: The two.
Birne Binegar: Plus.
Birne Binegar: Three the three is equal to the zero vector so this would be.
Birne Binegar: A dependent relation, if this is true, and all the coefficients are not equal to zero, if I if the coefficient or equal to zero, then i've just have any questions, it says zero is equal to zero, so this is really the only interesting case.
Birne Binegar: of an equation like this, the only one that really put restrictions on these variables or all these vectors but what you can do that as long as you have a non zero coefficient is you can use the equation itself.
Birne Binegar: To let's say minus 3D to minus three three you can use the equation to express one vector in terms of the other and that's.
Birne Binegar: that's the dependence we're talking about when we say we have a dependency ablation.
Birne Binegar: If we have a solution of an equation like this, where the coefficients are not all equal to zero, it means well we're really going to have.
Birne Binegar: An equation, other than zero is equal to zero, and we can use that equation to solve for one of the vectors in terms of the other, so in this presentation, the one depends on the vectors D to and D3.
Birne Binegar: So often talk not only about vectors be linearly independent, that means.
Birne Binegar: I cannot solve this equation, unless I take all the coefficients equal to zero, I can only solve this equation, by making it trivial zero is equal to zero.
Birne Binegar: But if I have, if I have solution other than that to trivial solution, then my doctors are linearly dependent, because I could use the equation to solve for one doctor in terms of the other.
Birne Binegar: If I don't if this is the only solution, then the equation is trivial and there's no relationship amongst the vectors do you want through bk, and so we say they're literally independent so that's where that language dependence and independence is coming from.
Birne Binegar: If there is an equation that it's not the trivial equation, where we set all the coefficients equal to zero, then we could use the equation to express.
Birne Binegar: One of the vectors in terms of the others, and so the vectors would be dependent upon one another.
Birne Binegar: But if the only way I can solve this equation is by separately, setting the coefficients equal to zero, then effectively I don't have any relationship between the vectors V1 through bk and they're independent.
Birne Binegar: Alright, so this is sometimes confusing logically because.
Birne Binegar: dependence and independence, they have sort indications of one another and sometimes we tend to.
Birne Binegar: We tend to focus on the case when their doctors are independent, which oftentimes you have to understand that is not dependence.
Birne Binegar: Okay, so.
Birne Binegar: let's see.
Birne Binegar: So we will say that.
Birne Binegar: If the only solution.
Birne Binegar: If there are non trivial solutions of this solution other than the trivial solution definitely say that the vectors heart literally dependent.
Birne Binegar: Okay, so.
Birne Binegar: That changes, our problem, finding good coordinates.
Birne Binegar: For a vector tennis subspace.
Birne Binegar: Originally, I phrases, the problem of finding a basis basis would guarantee that every vector has a unique expansion.
Birne Binegar: problem was they don't want to check every doctor in my space and what I found is that I can get a basis, as long as i'm working with sets of factors that are literally independent.
Birne Binegar: So let me give you one simple example where a set of vectors is obviously literally independent i'll show you what to look for.
Birne Binegar: I get it pros on the slide Let me try to explain.
Birne Binegar: More visually on the blackboard.
Birne Binegar: What i'm going to suppose here is that I haven't set a vectors.
Birne Binegar: And for which each of these vectors has one component that can't be cancelled out by the others, so let me give you example.
Birne Binegar: suppose I had the one equals 10012 and I had a second vector be two equals 01011 the same size five yeah and maybe three equals 00100 okay so here are three doctors I claim that I can just look at each factors and know that they're literally independent.
Birne Binegar: Why because suppose I multiply the things by constants.
Birne Binegar: And then I add them up see one the one I see two B to C three the three.
Birne Binegar: Well okay see one times here, but I won't get any contribution C to B to.
Birne Binegar: C one C to C three on both sides.
Birne Binegar: there's not gonna be any contribution that component or the third component so.
Birne Binegar: I get see one there and then here see one time zero is zero and death she to there but there'll be no contribution from the third factor.
Birne Binegar: Then i'll have C three and then, if last terms will be see one plus C two and then.
Birne Binegar: To see one plus C two alright so here's three vectors here's a general linear combination, the question is.
Birne Binegar: How can I make such a general linear combination equal to zero well to get this first component to vanish I need to take see one is equal to zero, because the zero vector.
Birne Binegar: Has all components equal to zero, and all I also would need see to is equal to zero, and I need to see three is equal to zero, and so, in fact, the only way a linear combination of these three five dimensional vectors.
Birne Binegar: can produce the zero vector is like separately, taking the number C one C two and see three all equal to zero that's the only linear combination of these three factors that will produce the zero vector.
Birne Binegar: Now, how did I spot it so quickly well because there's no way I can get rid of the contribution from the first coordinate by adding in.
Birne Binegar: multiples of the second and third factor, so this guy can't be canceled by any combination of the vectors D to MP3, so I would have to take see one is equal to zero.
Birne Binegar: Though see one is equal to zero, then this vector doesn't appear in the song that I look at this guy well in order for see two plus.
Birne Binegar: B to C three times the three in order for that to equal to zero vector i'll need no there's nothing in the.
Birne Binegar: Third vector I can cancel the contribution and the second component coming from the second vector so I have to take see to is equal to zero, and certainly like for C three so the statement i'm trying to still stay here is that if you have a list of actors.
Birne Binegar: And they have the property that.
Birne Binegar: For any given component, only one vector has a non zero off there you won't be able to add the vectors up to produce the zero vector without taking all the coefficients equal to zero.
Birne Binegar: This entry here can only be canceled by taking see one is equal to zero, because the contribution to C to B to C three D3 three.
Birne Binegar: it's not going to contribute to this first component, so I would have to take see one is equal to zero, and similarly in order to get the second component equal to zero in this total some.
Birne Binegar: Well that's nothing I can use in the first factor, or the third vector that he cancel out the contribution from this guy, so I would have to take see two equals zero.
Birne Binegar: Okay, so now we have a sort of a visual chat as to whether or not.
Birne Binegar: vectors can be literally independent and, in fact.
Birne Binegar: The hint it's what's coming.
Birne Binegar: notice if I take these vectors and I think of them as the rows of the matrix i've got this matrix in row Echelon for.
Birne Binegar: And what happens in row Echelon for is you know.
Birne Binegar: Any Paul any.
Birne Binegar: pivot in a call it's going to have nothing but zeros below it until you won't be able to do any cancellation.
Birne Binegar: will show you maybe next five minutes is that if you have a matrix that to enroll Echelon for that it's non zero rose will automatically literally dependent because they'll have this.
Birne Binegar: This problematic cancellation is that.
Birne Binegar: below the pivot I won't be able to use linear combinations of the Rose below a pivot in order to get rid of the contribution of the pivot to the total sum.
Birne Binegar: Okay.
Birne Binegar: let's see um yeah well here's here's the I guess the visual demonstration what i'm talking about before you have said two vectors.
Birne Binegar: Then they're not going to be linearly independent way I can spot, that is, that this component cannot be cancelled with that component.
Birne Binegar: And this component cannot be cancelled with that component itself in fact you'll.
Birne Binegar: get a linear combination of these two particular vectors equals zero if the separately takes the coefficient at one of the first factor and the coefficient X, to the second vector they separately, have to vanish.
Birne Binegar: Alright, and so i'm applying that observation to the rows of a matrix well, if you take a linear combinations of the road vectors of a matrix and the matrix is in row Echelon form that's going to be that there's going to be.
Birne Binegar: The first row has a pivot for this to the left and the Rose below it don't have any entries below that first pivot, and so I won't be able to can't.
Birne Binegar: Have these vectors some zero without taking excellent is equal to zero, because it has a entry this way off the left it can't be cancelled I the lower rose so that would tell me that X one and half equals zero but X one is equal to zero, and this term vanishes, then you look at.
Birne Binegar: The matrix from row to on we're down to that the case row, and then, if it's in row Echelon form there no lo the pivot in the second row, you would only have zero so there'll be no way to cancel.
Birne Binegar: The contribution from the pivot in the second row with remaining terms in this expansion, because the other vectors are not going to have components in that second slot and so you'd have to take X is equal to zero.
Birne Binegar: And so.
Birne Binegar: yeah so the argument would then go is that.
Birne Binegar: You first show that the pivot they.
Birne Binegar: Because there's nothing below the pivot in the first row normally get.
Birne Binegar: Zero vector by taking linear combination of the robe actors you'd have to take the first coefficient equal to zero.
Birne Binegar: But then you just have a smaller matrix it's still in real life form and the pivots coming from here, cannot be cancelled with terms from the.
Birne Binegar: rest of the expansion, so you get X is equal to zero, and you keep doing that, until you finally conclude that you have to take all the coefficient separately equal to zero.
Birne Binegar: If the rows of a matrix a linear combination of the rows of the matrix it's in a row Echelon form if that's the producers zero vector all the coefficients have two equals zero, and so the non zero row vectors of a matrix in row Echelon form or always linearly independent.
Birne Binegar: Okay well so that gives us a check.
Birne Binegar: We have a list of vectors we can write a message Rosa the matrix and if that matrix happens to be in row Echelon form then we'll know that our directors are linearly dependent, however, now ah.
Birne Binegar: Well, here here's the conclusion that.
Birne Binegar: I wanted to draw from the discussion on the previous slide if you start off with the matrix and real Echelon form then it's non zero row vectors are always going to be a basis for the space that you generate like taking all possible linear combinations of the rogue actors.
Birne Binegar: However, I stipulate also that I only consider the.
Birne Binegar: Non zero robot years.
Birne Binegar: Let me give you a.
Birne Binegar: sort of a pedantic point but suppose I have a matrix 111000100000 all right so there's a matrix.
Birne Binegar: With three rows and you could ask if it's possible to solve this equation see 11110 plus C 20010 plus C three times 000 all right now.
Birne Binegar: The three rows of this matrix are not linearly independent why because I could set this equal to zero, I could set that equal to zero, I could set that equal to one.
Birne Binegar: And, but one time to zero vector is still the zero vector so I can find a choice of constants she wants she's 23 so that these vectors some to the zero vector without taking all the coefficients equal to zero, but.
Birne Binegar: If I.
Birne Binegar: threw out that that zero row, then these two vectors would be linearly dependent, because I can't cancel the contribution from this guy with something in the second vector know.
Birne Binegar: Until I have to take the first coefficient equal to zero, and then this some multiple of this would have to equals zero, but that would force me to take the zero multiple in order to get rid of the contribution in the third component.
Birne Binegar: Okay, so.
Birne Binegar: Make your season row so on for always provide a basis for.
Birne Binegar: For the road space that the matrix But what if you have a matrix it's not in row so on.
Birne Binegar: What can you say that well here's something we want true but it's true.
Birne Binegar: The road space at the matrix.
Birne Binegar: doesn't change when you do an elementary row operation.
Birne Binegar: Well, I guess i'm about out of time, but let me just see if I can clarify this point and then we'll pick up with this point on Wednesday.
Birne Binegar: You have a if you have a matrix.
Birne Binegar: hey don't whatever row is one row one row to row three think of these as robe actors, when I do an elementary row operation, what happens to these is this might go to say Lambda times 01 and the second row might go to.
Birne Binegar: row two plus some other both pole.
Birne Binegar: row three and maybe the third row.
Birne Binegar: goes to row three minus row one two boats, I do some elementary real operations on this matrix the statement here is that.
Birne Binegar: Although the matrix is changing under the row operations.
Birne Binegar: i'll call this our have a.
Birne Binegar: vm the span of BC will set up all possible linear combinations of the three rows it's going to be the same set at the span of all possible in your combinations of the three rows so even though we're changing the matrix the space is generated by the road vectors it's not changing.
Birne Binegar: In fact, you can understand that, because if you remember, where the elementary row operations came from they came from demanding that the equations with that the solution without change.
Birne Binegar: And in fact neither played equations were just manipulating rows of the matrix that the solution is not changing the fact the equivalent statement is the space is generated by these rogue actors it's not changing either.
Birne Binegar: Okay, so i'll pick up well with this idea on Wednesday.