Birne Binegar: Okay, so today is effectively the last lecture, of course.
Birne Binegar: In fact, hit the enter today's lecture i'm actually going to start reviewing a little bit what i'm going to talk about today.
Birne Binegar: Is hunger rehash what I discussed on Friday, how we decompose a vector first of all component, the vector that's running parallel to another vector.
Birne Binegar: And second and, second, of all how you decompose a vector finding a piece of a vector that actually lives inside of subspace so i'll redo that those constructions and then i'll do two examples of that So you see explicitly how you calculate and then what we'll do is we'll.
Birne Binegar: begin our review what i'm going to talk about at the end of the lecture today just row reduction and how many different ways we use real reduction in this course.
Birne Binegar: Okay, and then on Wednesday and Friday this week well i'll do a more substantial review probably go over a practice sample exam etc.
Birne Binegar: Okay, so um what I talked about on Friday last week it's the work.
Birne Binegar: There we go okay so last week, I was talking about.
Birne Binegar: given to vector say a vector a in a vector be how do I find the component of a vector aid that's running parallel to be in another piece of a that turning perpendicular to pee.
Birne Binegar: Now, in this picture it's kind of clear that, basically, what you need is some plain geometry some little trigonometry and you ought to be able to figure out these components, for example.
Birne Binegar: At least the length of this vector here it's going to be well you play a photographer's the serum.
Birne Binegar: In a little trigonometry the length of the adjacent side of a right triangle side adjacent to the angle is the length of the high pot news times the cosine of the angle.
Birne Binegar: Between the two vectors and what we actually use in linear algebra though it's not this geometric picture, where we think of.
Birne Binegar: vectors as being directed line segment rather what we use is the Doc product which allows us to extract geometric information from the vector sister typically presented in this course so.
Birne Binegar: In this course.
Birne Binegar: A vector inside an n dimensional vector space is just a list of numbers.
Birne Binegar: Say be one to the end.
Birne Binegar: and
Birne Binegar: The way we extract geometric information about these ordered list of numbers is we use a dot product, so a B.
Birne Binegar: is going to be the first component of a time, the first component of B, plus a second component, based on the second component of be etc until we've exhausted all the components, so we can.
Birne Binegar: compute this, but what makes it useful to our intuition, is the fact we have another formula for this product that's purely geometric you take the length of a.
Birne Binegar: You take the length of be.
Birne Binegar: And you multiply by that cosine of the angle between the vectors amd and you get a geometric version of the same dot product now what we'll do is we'll compute the product.
Birne Binegar: Using ordered list of numbers but we'll use on the left hand side here to interpret that product geometrically from this formula here, we see that the length of a.
Birne Binegar: is just equal to the square root of the Doc product of a with itself and that the coast time between two vectors is just equal to their product divided by their links, but the links can also be computed.
Birne Binegar: Using this formula for the product be that be, and so what I did at the start of friday's lecture.
Birne Binegar: Is I took this geometric picture and then I tried to.
Birne Binegar: compute.
Birne Binegar: The various sub vectors this component of a that lives in the direction of be in this perpendicular component I use your non tree to compute those.
Birne Binegar: But those angles and those links are not going to be computed by using ruler and protractor is that I would do for this figure rather i'll use the.
Birne Binegar: linear algebra dot product this this expansion to compute those but in the end, what happened, though, is that my final answer, while computer will using dot products for n dimensional vectors is it's fairly simple.
Birne Binegar: This formula for the.
Birne Binegar: component of a that's going in parallel, be that simply the product of a would be divided by effectively the magnitude square to be times the vector these are, this is a scale, this is a scale or a factor so just a number times the vector that will produce a vector.
Birne Binegar: All right, told that business was pretty easy, but basically.
Birne Binegar: The derivation of these formulas.
Birne Binegar: Basically, came from this, we just need to how to.
Birne Binegar: Get the geometric information from the dot product and then apply that to get these two components.
Birne Binegar: The main problem, though I wanted to discuss on Friday was more general instead of funding the part of a vector be that is running parallel to a given vector i'm going to find a part of a vector V that lives inside of subspace w.
Birne Binegar: So here's the picture.
Birne Binegar: it's more like this, so this.
Birne Binegar: flat surface here that's our subspace.
Birne Binegar: Think of this it's embedded in a three dimensional space and I have a two dimensional subspace and I have a one dimensional vector sitting inside as well the idea.
Birne Binegar: of an orthogonal decomposition respect to a subspace issue try to find the component of your vector V that actually lives inside the subspace and a separate component that's actually perpendicular to that subspace.
Birne Binegar: Okay now to be perpendicular to a subspace what does that actually mean you're not perpendicular to the individual vectors in that subspace you're perpendicular to the lines that are in that subspace and so that makes it.
Birne Binegar: A little bit difficult and then this the sort of considerations that we use on the preceding slide where we just looked for the components of a vector of respect, you have a vector, then the simple geometry was sufficient to solve this problem.
Birne Binegar: As I did on Friday really requires a lot of apparatus for more complicated.
Birne Binegar: linear algebra gap to introduce another subspace and we're gonna have to find bases for the subspace we're going to have to expand the vector V, with respect to a particular basis so i'll go through that again.
Birne Binegar: But no one thing you should know before I even start.
Birne Binegar: Is that all the technique I solve this problem, even though it appears to be just a higher dimensional generalization of the first problem.
Birne Binegar: Right, only one to decompose a vector with respect to another vector here when I decompose it with respect to a subspace it, we need more formal apparatus of the entire course not just.
Birne Binegar: This geometric interpretation of the Doc product.
Birne Binegar: Okay, but the idea is similar to what we did before we take an arbitrary vector be and we want to break it up into a piece that lives inside the subspace and a separate piece and.
Birne Binegar: We get back to the original vector by combining those two constituents.
Birne Binegar: Alright, and the way we do, that is, we introduce an auxiliary subspace.
Birne Binegar: The sub base w subscript perp is the set of all vectors in the entire vector space that's perpendicular to every vector inside w.
Birne Binegar: Alright, so.
Birne Binegar: here's how we do it, first of all i'm assuming that we've got the we found this subspace and we've even gone as far as calculating a basis for it.
Birne Binegar: Then, what I do is I create the secondary subspace it's going to consist of all the vectors in the entire vector space that are perpendicular to every vector inside the given subspace.
Birne Binegar: Now that condition.
Birne Binegar: I leave it this way it's kind of hard to figure out how i'm going to find all these factors, but in fact what I showed you on Friday was is that.
Birne Binegar: If the subscript w and let me just list a set of vectors.
Birne Binegar: bk if this is a basis.
Birne Binegar: For your subspace w then.
Birne Binegar: Every w.
Birne Binegar: Inside w has a unique expression.
Birne Binegar: As w equals C one C plus plus C K UK to have a basis means you have enough directions to reach every point in your subspace using a unique expression of this form.
Birne Binegar: Now I want.
Birne Binegar: w.
Birne Binegar: So I want so suppose he lives in this perpendicular subspace then, what I want it's got to be this be has got to be perpendicular to every vector including this guy.
Birne Binegar: should be perpendicular to every vector of this form, so I need this.
Birne Binegar: This purpose perpendicular to every director inside w, but I can achieve that by making sure that this vector V is perpendicular to each of these basis vectors, so this is.
Birne Binegar: Automatic.
Birne Binegar: v.bi.
Birne Binegar: equals zero for each basis factor.
Birne Binegar: So rather than impose this condition.
Birne Binegar: That the is perpendicular to every vector of its form its supply says to say that it's a B is perpendicular to just these basis vectors V1 to bk then this condition you write it out can be written as a matrix equation.
Birne Binegar: take these vectors right him as the rows of a matrix.
Birne Binegar: And then put this vector V as a column vector.
Birne Binegar: Then, according to the rules of matrix multiplication.
Birne Binegar: The first component of this matrix product it's going to be first row with the first factor.
Birne Binegar: it's got product with the first column of the In fact the only column is the second factor, so that just going to be the one.
Birne Binegar: That deed in that first component, the second component is going to be exactly the two that the last component is going to be bk dotted would be, what I want is that all these products of the with a basis vector to equals zero and so i'll have.
Birne Binegar: Zero down to zero, and now I see.
Birne Binegar: That, in fact, these vectors they have to be the solutions of a homogeneous linear system where you have a coefficient matrix times the vector is equal to the zero vector, in other words.
Birne Binegar: The solutions.
Birne Binegar: are going to be.
Birne Binegar: Effectively, the no space.
Birne Binegar: But what do I mean by the solutions, what I want me to say is that the set of all vectors that are perpendicular to every vector inside w this is going to be exactly.
Birne Binegar: The solution, set up a linear system where the coefficient matrix is just that matrix that we form a using the original basis as rose.
Birne Binegar: Okay, so that's the auxiliary subspace that we need if we're going to deep decompose an arbitrary vector find out what part of it lives inside the subspace and what our lives outside the subspace.
Birne Binegar: We need to construct this auxiliary subspace and but that's easy to do we have a particular coefficient matrix, we need to find it no space.
Birne Binegar: meaning, we need to solve a linear system, and we do that by reducing the make the coefficient matrix the row Echelon form.
Birne Binegar: Converting back to the equations expanding this solution vector in terms of free parameters and blah blah blah, in the end, you get a bunch of free parameters times.
Birne Binegar: column vectors and those column vectors are exactly your basis, so, in fact, we not only can calculate this subspace we can actually find a basis for it as well.
Birne Binegar: And indeed.
Birne Binegar: Just from this form here, I already know what the dimension of that that solution set the dimension of this null space, because each of these guys.
Birne Binegar: There basis vectors so they're linearly independent That means that the rows of this coefficient matrix are all linearly independent.
Birne Binegar: So it's rank exactly equal to the number of rows dimension up the road spaces these rows are all linearly independent it's just going to equal, the number of rows so.
Birne Binegar: Put another way, this matrix is going to reduce to some reduce row Echelon form.
Birne Binegar: What do I do so since the rows are all linearly independent I can't get a zero at the bottom.
Birne Binegar: That means basically i'm going to have well i've got to have a pivot in each row so i'm going to have K pivots but each of these factors is also living inside our end.
Birne Binegar: And so the number of entries and number of columns I have is going to equal in so upon row reduction, this is going to become a matrix or I have a non zero rows, but I have a total of n columns.
Birne Binegar: That means the number of free parameters is going to be N minus K right you count the number of columns that don't have tibbetts to get the dimension of the solution, set.
Birne Binegar: So what I know about this space is first of all, you know how to calculate it, and in fact I already know which dimension it's just going to be the number of rows I have minus the number of columns I have minus number of rows.
Birne Binegar: Right, so when we calculate this and we actually find the solution vectors of this linear system.
Birne Binegar: what's going to turn out is that expansion of a typical solution vector is going to involve N minus K basis factors.
Birne Binegar: What I did on Friday, and I do again today is i'm going to count those basis vectors in a peculiar way at first, instead of starting.
Birne Binegar: My basis vector and calling it basis factor number one i'm going to call it basis vector K plus one, because the next idea is to extend this basis.
Birne Binegar: To a basis of n linearly independent vectors by a joining the original K basis vectors my sub base w a joining to that set of vectors.
Birne Binegar: The basis vectors for the solution set up this linear system that will be N minus K vectors and so when I put them together i'll go from one the K K plus one to end and i'll get a total of in linearly independent vectors.
Birne Binegar: Alright, so.
Birne Binegar: Well that's what we do and now we're almost done surprisingly it doesn't look like we're very close to getting this be composition, but now we use the idea of what a basis does for us again.
Birne Binegar: Since like So the first key factors of this basis span original sub base w the next N minus K, but basic factors.
Birne Binegar: They span this perpendicular subspace the set of all vectors that are perpendicular to every doctor inside w when I put them together, I get a basis for the entire vector space, so what I do is I take my given vector V.
Birne Binegar: And I try to express express it.
Birne Binegar: In terms of this basis for our end so have an expansion si K plus one K plus one dot dot dot plus C and the end now what's going to happen.
Birne Binegar: Is.
Birne Binegar: Each of these basis factors, do you want to bk they live inside they live inside w so this lies.
Birne Binegar: In w.
Birne Binegar: The second half of the vector V it's being expanded in terms of basis vectors for the perpendicular subspace so that piece with in the perpendicular step space, but what I do is I simply call that the w and I call this piece the perp.
Birne Binegar: And that's the desire decomposition.
Birne Binegar: So what remains in is through explain how do I get these particular number C one C to up the CN.
Birne Binegar: Well that's an old calculation that we've done.
Birne Binegar: Let me recall it for you.
Birne Binegar: If you want, if i'm.
Birne Binegar: Let me put the be on this side and let's say we have see one B one dot dot dot plus C and the end so I know all these faces fact here's what i'm trying to do is figure out what coefficients I need to multiply that my basis vectors by to their liking.
Birne Binegar: it's expressed given vector in terms of my basis factors this problem, the right hand side.
Birne Binegar: can be thought it's a matrix multiplication where R Us has columns.
Birne Binegar: My basis vectors.
Birne Binegar: and
Birne Binegar: Then it's multiplying see one, you see, and so, when you have a.
Birne Binegar: matrix times a column vector that can always be expanded by using the components of the column vector having them multiplying the corresponding.
Birne Binegar: Sir yep the components here multiply the columns of The coefficient matrix and what will happen is you'll reproduce the right hand side and on the left hand side you'll have a component.
Birne Binegar: Of the vector V.
Birne Binegar: This is a linear system where this is the unknown, it has the form the coefficient matrix times, an unknown vector is equal to a known vector.
Birne Binegar: And so what do we do.
We.
Birne Binegar: solve that linear system by forming the augmented matrix and real reducing to reduce rational on on.
Birne Binegar: What something nice happens in this situation because we're dealing with basis factors these columns are all linearly independent so what's going to happen during this row reduction is in the end i've got to get a matrix that has a pivot in every column until what ones that will happen.
Birne Binegar: Is on during this role reduction, the left hand factor is going to reduce to the identity matrix and the right hand side.
Birne Binegar: it's going to be your solution, because the equations corresponding to this augmented matrix or X one is equal to see one X two is equal to see two X three is equal to see three etc, so, in fact, the way you determine these coefficients.
Birne Binegar: Is you right you're given vector sorry you're right you're given vectors as the columns of a matrix you take the vector that you're trying to express in terms of your basis vectors you write that, as the last column of an augmented matrix and your relatives and then.
Birne Binegar: Once you reach reduce relational on form you just look at this last call and it's going to be exactly the coefficient you want.
Birne Binegar: So you'll be able to compute all these numbers and then to get the orthogonal decomposition all you have to do is keep more or less the first half of that up to up to.
Birne Binegar: Last based inspector for w you keep those terms, together, and you call that bw and then the remaining terms and that expansion they're all going to live in the perpendicular subspace.
Birne Binegar: call that contribution to the total vector be called that the perp and you've got your desired the composition.
Birne Binegar: Okay, so let me now work out and explicit example.
Birne Binegar: So i'm going to avoid row reduction in this example, but I will go through all the steps with real numbers so.
Birne Binegar: yeah so i've just explained this on the blackboard.
Birne Binegar: let's start off with a subspace this generated by two vectors all possible linear combinations of the vectors 101 and 011.
Birne Binegar: These are three dimensional vector so what I create in that fashion, is a two dimensional subspace are three.
Birne Binegar: it's pretty obvious that these two vectors are linearly independent because I can't find any linear combination of these two vectors that equals zero without taking both coefficients equal to zero, you can see that, because.
Birne Binegar: If I wrote.
Birne Binegar: See one time fee one let's see two times, would be to.
Birne Binegar: utter abby abby see one times what what is it.
Birne Binegar: 101 plus the two times.
Birne Binegar: Second, one is 011.
Birne Binegar: By simply add those together i'm going to get C one C two and see one plus the two now linear independence when you check for that issue, try to.
Birne Binegar: solve this equation, where you have a linear combination of your vectors and it produces the zero vector and what you see.
Birne Binegar: As soon as you compute here if there's no way I can get a zero vector on the right hand side without separately taking.
Birne Binegar: The two coefficients see one and see two equals zero and then that will also make the third component equal to zero, so the only solution of this of this equation is the one where the coefficients are equal to zero, that means that these two guys are linearly independent.
Birne Binegar: So I saw that right away, simply because I saw there's no way I can get the first component to cancel without because there's nothing that.
Birne Binegar: If I even if I multiply this vector by a number I can't get this piece to cancel out without making see one is equal to zero.
Birne Binegar: Nor can I make this second component equal to zero by adding in some multiple the first director, so these guys are never going to add to a linear combination of them is never going to produce zero unless the coefficients or equal to zero.
Birne Binegar: All right, so that just arguing why this is the basis what we're going to do in this example is figure out the component of this particular vector one four minus four which will find that the component of this factor that lives inside w as well as the corresponding perpendicular component.
Birne Binegar: So the first step.
Birne Binegar: is finding a basis for w, but I just verified that i've got a basis for debbie the second step.
Birne Binegar: is to find this auxiliary basis a basis for the subspace of our three that's perpendicular to every vector inside this space well that's done like constructing a matrix using these guys's rows.
Birne Binegar: And then, reducing to reduce relational on warm well soon if I write this down it's already in reduce your actual on board so.
Birne Binegar: I can actually figure out the know space of this matrix it's going to be the solutions of X one plus zero X two X three is equal to zero second row is going to give me equation zero times X one one times X plus one times X three is equal to zero.
Birne Binegar: So you get these equations and from these equations you can extract the hyper planes solution like moving the three parameters, corresponding to.
Birne Binegar: The columns without a pivot you move them to the right hand side to express fix variables in terms of three parameters and then you write down a typical solution vector.
Birne Binegar: In the end, what you find is the subspace w PR, which is the set of vectors that are perpendicular to every doctor and w.
Birne Binegar: Every vector in this space has to be a scale or multiple this particular vector so that guy right there is our basis factor for w.
We have.
Birne Binegar: A basis for w which was.
Birne Binegar: yeah these two vectors 101 and 011.
Birne Binegar: And now we just calculated.
Birne Binegar: A basis for the perpendicular, which is actually the no spaces.
Birne Binegar: The matrix I create by using these two vectors as rose, and what that turned out to be is the specter minus one minus one one.
Birne Binegar: All right now, I have a total of three basis vectors and I can use it to.
Birne Binegar: To expand any three dimensional vector in terms of these three vectors.
Birne Binegar: The.
Birne Binegar: equals.
Birne Binegar: 101011 and minus one minus one one that's now a basis.
Birne Binegar: Not for these individual sub spaces, but for the entire vector station are three in which both the subspace it's live.
Birne Binegar: Right somewhere close to the end now.
Birne Binegar: we've got a basis for the entire vector space are three let's express this vector in terms of that basis.
Birne Binegar: Well, the way we do, that is, we take our basis we use those basis factors to form the columns.
Birne Binegar: Of the matrix and we use the given vector as the being on the right hand side because what we're trying to satisfy is that see one do one plus C three the three a national put in the middle one who is let's see to to.
Birne Binegar: What we're trying to satisfy.
Birne Binegar: Is this equation and effectively what we have to do is solve linear system where.
Birne Binegar: The matrix type form using the basis vectors as columns times, an arbitrary that your C one C to C three is equal to my given vector, which is what is 114 minus four.
Birne Binegar: OK so.
Birne Binegar: The augmented matrix corresponding to the linear system that i'm trying to solve.
Birne Binegar: is exactly this I just took these vectors road and vertically and then I hope.
Birne Binegar: The backdrop trying to express in terms of this basis I wrote it as the last call and I wrote reduce to reduce rational on form, and so I see that the solution is going to be X one is equal to minus two X is equal to one and X is equal to minus three.
Birne Binegar: So that's our our.
Birne Binegar: Our solution to the problem, and now, that means that be, can be written as minus two times the first basis factor one time the second basis vector negative three times the third basis factor.
Birne Binegar: But some of these two vectors well i'm using a basis for w so this partial sum is going to be a vector inside w.
Birne Binegar: This last term i'm using this last spacious vector which lives in the perpendicular stuff space so that's going to live outside of view and so by simply.
Birne Binegar: Group, the first two terms together, I get the subscript w the perpendicular subspace is just that last term.
Birne Binegar: Alright, so there's first example of how we.
Birne Binegar: Given a subspace how we actually decompose an arbitrary vector one four minus four into a piece that lives inside the subspace and a piece that's perpendicular to that subspace.
Birne Binegar: Alright, so let's do.
Birne Binegar: Another example.
Birne Binegar: Where the main distinction is that, unlike the first example i'm going to specify a subspace not by telling you it's generators but saying that it's the subspace correspond solution set up an equation.
Birne Binegar: we're still going to need a basis, so what happens in this example is the first thing I have to do is figure out a basis for the subspace under consideration, though it's defined as the solution set of a linear equation.
Birne Binegar: and actually from that.
Birne Binegar: let's go ahead and solve that that linear system using our standard pepe.
Birne Binegar: We only have one equation.
Birne Binegar: it's um.
Birne Binegar: X one plus x plus x three is equal to zero, that means the augmented matrix.
Birne Binegar: is well 111 the coefficient of the first variable is one quotation second variables one coefficient of third variables one the right hand side is equal to zero, and we only have one equations or augmented matrix only has one row.
Birne Binegar: Well then, we would it's already in reduce show Echelon form.
Birne Binegar: So, once you get the augmented matrix and reduce your actual inform you write down the equations well.
Birne Binegar: Using the fact that they variables corresponding the columns without pivots those are going to be regarded as free parameters, so this.
Birne Binegar: For one and only row with this reducible Echelon form of augmented matrix tells us that X, one has to equal minus X two minus X three which tells us that our solution vector.
Birne Binegar: X has to have the form of.
Birne Binegar: minus X two minus X three x two X three.
Birne Binegar: This is the fixed variable and I use the equation to coming from the matrix and reduce your national on board to really express the fixed variables in terms of the free variable and then, finally, we expand the terms of the three variables get minus 110 plus x two.
Birne Binegar: minus 101 and then.
Birne Binegar: So at X three So these are my basis vectors.
Birne Binegar: For the solution set.
Birne Binegar: Which.
Birne Binegar: is exactly my subspace so the distinction between the preceding example and the current example is I gave you a basis in the first example in this second example I give you the subspace.
Birne Binegar: In a different fashion, I say it's a solution set of a linear system, and so we have to do a little bit of work.
Birne Binegar: To find a basis for that, for the basis for w.
Birne Binegar: Alright, but then once we've got that faces.
Birne Binegar: The problem of decomposing the vector one to one respect to the subspace to get to proceed, just like we did before we now have w equals the span of the vectors.
Birne Binegar: minus 110.
Birne Binegar: and
Birne Binegar: minus 101 and, in fact, these are basis.
Birne Binegar: vectors for w.
Birne Binegar: So now we can.
Birne Binegar: next step is going to be defined the perpendicular subspace this is going to be the north space.
Birne Binegar: matrix we get by using these basis factors.
Birne Binegar: As rose.
Birne Binegar: To as row so it's going to be the north face.
Birne Binegar: Of a C minus one.
Birne Binegar: minus 110 and minus 101.
Birne Binegar: All right, unfortunately it's not quite in.
Birne Binegar: Real Echelon form but pretty pretty simple, I think I think we I end up.
Birne Binegar: yeah So this is the matrix.
Birne Binegar: These are just using these basis factors that's rose and then reducing all the way to reduce your actual on on now i've got.
Birne Binegar: The the coefficient matrix and reduce show Echelon form, I can read off the equation X one minus X three equals zero X two minus X three x equals zero.
Birne Binegar: What I see is both X one X two corresponding to these columns with pivots they all can be expressed in terms of X three and, in fact, you can see that it's just going to be.
Birne Binegar: 111 it's going to be the basis factor.
Birne Binegar: see if I can do that slowly I have.
Birne Binegar: One zero minus 101 minus one that's the reduce row Echelon form, I get equations X one equals X three x is equal X three what i've done is i've written down the equation X one times one X two times zero minus one times X three is equal to zero i've moved.
Birne Binegar: From those equations i've moved the variable corresponding through the call that pivot over to the right hand side, and then I write down solution vector.
Birne Binegar: X three x three x three and you see that it every solution vector.
Birne Binegar: Is the scale or multiple of this factor so that's our basis sector.
Birne Binegar: that's our basis vector for.
Birne Binegar: The perpendicular subspace.
Birne Binegar: And what else almost done.
Birne Binegar: we've got a basis for w and.
Birne Binegar: i've also got a basis for the perpendicular subspace so now i'll use this basis for our three and i'll try to figure out.
Birne Binegar: The components of the vector one to one when I expand this vector one to one in terms of these three basis factors like the columns of this form.
Birne Binegar: I do a row reduction to be do show Echelon form and what I find is I get a unique solution X, one has to equal two thirds X two half equal minus one third and X three has to equal four thirds.
Birne Binegar: That means I can.
Birne Binegar: Take the vector.
Birne Binegar: Worse, it one to one, and when I expand it in terms of the three factors.
Birne Binegar: The components along the first X factor is going to be two thirds component long The second factor is going to be minus one third, and the component along the third vector is going to be four thirds, I think the first two terms.
Birne Binegar: Since are constructed from basis vectors for w.
Birne Binegar: Some of the first two terms is going to be in w and.
Birne Binegar: I didn't complete this problem.
Birne Binegar: That last what was to happen on that last slide.
Birne Binegar: Just.
Birne Binegar: began the idea was to find content C one.
Birne Binegar: c to.
Birne Binegar: c three.
Birne Binegar: So that the given vector.
Birne Binegar: To be expressed in terms of the other basis we just found this basis, and what we found is that see one had the equal.
Birne Binegar: Two thirds.
Birne Binegar: That see two has the equal minus one third.
Birne Binegar: And the third has to equal this last component for third.
Birne Binegar: And so, our vector D is, in fact.
Birne Binegar: Two thirds times, our first basis factor which is minus 110.
Birne Binegar: Plus.
Birne Binegar: or minus.
Birne Binegar: Second coefficient times the second basis factor which is minus 101.
Birne Binegar: And then we have the force or the third term coefficient times third base it back 111.
Birne Binegar: Then this part is the w.
Birne Binegar: Because it's constructed by taking linear combinations of basis vectors for w and this part is the perpendicular component and then that are decomposition you can explicitly calculate with this vector is then and the scale multiply that vector to get deeper.
Birne Binegar: Okay, so that's how those those calculations go.
Birne Binegar: Once you find a basis for w you need that to find a basis for w PR but then, once you have a basis for w and w PR you get a basis for the entire vector space and the problem becomes reducing.
Birne Binegar: A given linear system to reduce your actual on form and then use these entries in the last call to tell you what that expansion is, and then the last step is just.
Birne Binegar: Take this expansion and break it into the part that lives inside w plus the part that lives in the perpendicular subspace.
Birne Binegar: Okay.
Birne Binegar: So.
Birne Binegar: at eight minutes left, let me pause here if there are any questions on this.
Birne Binegar: Okay, so when i'm going to do in the last eight minutes of today's lecture.
Birne Binegar: it's just.
Birne Binegar: remind you.
Birne Binegar: Of all the things we've used row reduction, for I think the hard part of the final exam.
Birne Binegar: Is as you'll be faced with a problem and then you'll see about two inches of space left for you to do some calculation.
Birne Binegar: Well, probably what you get both why I live so much space there's because you're gonna have to reduce a matrix somehow.
Birne Binegar: So what I want to do today is show you the different ways that row reduction comes in, as we try to solve problems and linear algebra we almost everything we've done can be expressed in terms of a row yucky calculation.
Birne Binegar: So.
Birne Binegar: To do row reduction first thing you need to know is what is the matrix enroll exelon form and what is that matrix and reduce your.
Birne Binegar: role is one form just means that the first non zero entries in every row.
Birne Binegar: goes pivot of current a downward staircase from the upper left to the lower right any matrix where you have a bunch of zeros and they.
Birne Binegar: terminate with some non zero entry you want these pivots do occur in a downward staircase that's the first stage of a general reaction you try to get a matrix and this form by using elementary real operations to clear out entries below pivots.
Birne Binegar: But then, if you're actually trying to solve a linear system, you have to go further, you have to use more elementary row operations to recast the matrix so that it's pivots are all equal to one and so that above and below pivots you only have zeros.
Birne Binegar: You to do these if you're to.
Birne Binegar: transform a matrix into this form usually just using these three elementary row operations so.
Birne Binegar: use this operation, and when you have pivots are sort of out of order for suddenly, then you can interchange your order so that you can get a downward staircase.
Birne Binegar: use this real operation typically so that you can make pivots all equal to one and then this is the operation use most often use it in the first stage of row operation to clear out entries below pivot and use this this operation in the final stage to clear out entries that.
Birne Binegar: entries above pivots.
Birne Binegar: So that's that's the what row reduction does its utility is first of all, we use row reduction to solve linear systems that's done by.
Birne Binegar: Forming a matrix where we take the coefficient matrix and use that as the left hand side of an augmented matrix and we use the column vector on the right, as the final column.
Birne Binegar: The solving a linear system means taking.
Birne Binegar: Taking the linear system and re expressing it as an augmented matrix and then row reducing this augmented matrix all the way to reduce row Echelon form.
Birne Binegar: that's not the end of the problem, though, because once you get it and reduce relies on form, you have to convert back to equations and then you have to really express those equations as corresponding to a hyper plane.
Birne Binegar: But today what remaining six minutes, or so I just want to talk about the rope reduction part.
Birne Binegar: But I do want to point out that oftentimes.
Birne Binegar: These linear systems one we're interested in, or were there on the right hand side, we have the zero vector and which case.
Birne Binegar: The solution set really only depends on the matrix we call such solution sets the no space of the corresponding coefficient matrix but effectively we solve equations like this exact well we solve equations like this exactly like we solve equations like and the more general type.
Birne Binegar: A second reason for doing real reduction is it so now we're sorted in the material that we began after the first exam if we have a subspace.
Birne Binegar: Expressed as a set of all possible linear combinations of a given list of actors and we want to find a basis for that subspace what we do is rewrite the vectors as a rose of a matrix.
Birne Binegar: reduce that matrix to row Echelon form, and then we grabbed a nonzero rose that will give us a basis for this simple type of subspace.
Birne Binegar: There are some spaces can be specified in a number of ways, in particular, if, once we give you a matrix, there are three sub spaces, I can attach to that and we use rubber ducky to find bases for each of these three sub spaces.
Birne Binegar: In fact, what you can do if you have a given matrix.
Birne Binegar: You can find basis for all three substations by simply reducing your matrix to reduce your actual.
Birne Binegar: Once you've done that you get a basis for the row space I just looking at the take grabbing the non zero rows of this reduce relational and form you get a basis for the column space by.
Birne Binegar: Taking the reduce your actual on phone with you, given matrix looking at where the pivots live in that role exelon form.
Birne Binegar: Then, going back to the original matrix and grabbing the corresponding columns.
Birne Binegar: And then, finally, you get a basis for the North face of this matrix by calculating it reduce role Echelon form and then converting back to equations and.
Birne Binegar: Writing down the hyper playing solution and grabbing those constant factors that are being multiplied by the free parameters so again it's real reduction that's this the essential part of the calculation, but what you have to worry about, though, is once you've done that row.
Birne Binegar: row row reduction, what you have to do with that reduce your national on form to get the correct answer you kind of depend on what I asked you for.
Birne Binegar: So have I run out of time.
Birne Binegar: I got one minute left.
Birne Binegar: i'll just go through this quickly you know i'll cover it again on Wednesday.
Birne Binegar: We have linear transformations we also so functions between vector spaces that are well behaved respective scale multiplication and vector addition.
Birne Binegar: We use row reduction again define basis for sub spaces, we attached to these functions turns out that the range of a linear transformation is calculated by.
Birne Binegar: attaching a matrix to your linear transformation and then calculating the columns face of this matrix which you'll do by using row reduction.
Birne Binegar: And that the kernel of the linear transformation, the set of vectors in the domain that get sent to zero that can also be calculated, you have to first calculate the matrix.
Birne Binegar: corresponding to the linear transformation and then you have to reduce this matrix to reduce your national on form to figure out exactly what a basis for that space will be.
Birne Binegar: You also use real reduction to calculate determinants here, we take a matrix we reduce it to a row Echelon form, and then we get an expression for the determinant.
Birne Binegar: Based on how we wrote reduced a to its rotational on born the determinant of the original matrix it's going to be minus one to the number of times we had to interchange Roche, as we roll reduced.
Birne Binegar: Then we have to divide out by if we did any scale or modifications and doing this role reduction we're gonna have to pull out those factors.
Birne Binegar: And then, finally, we have the determinant of the matrix in row Echelon form but that's going to be an upper triangular matrix who's determine and it's just the product of its diagonal elements and so real reduction, you can be used to calculate determines in this fashion.
Birne Binegar: And then last thing we use row reduction for this one we're looking for eigenvectors if you want to find the basis vectors for an Eigen space, what you have to do is calculate the north face of your given matrix minus your eigenvalue times the identity matrix.
Birne Binegar: So you see that real reduction calculation occurs all over the place.
Birne Binegar: what's going to be critical, on your final exam is to first of all get the right matrix second of all reduce correctly and then Third, and most importantly, is to figure out what you need to do with that matrix and reduce cholesterol and form to get the desired answer.
Birne Binegar: OK so i'll stop here and i'll see you again on Wednesday.