Birne Binegar: i'm not sure how that.
Birne Binegar: So I just sure another.
Birne Binegar: yeah.
Birne Binegar: they're doing their.
Birne Binegar: This is.
Birne Binegar: we're at the final lecture I think what what we'll do on Wednesday and Friday will just be tried to prepare you for the final exam.
Birne Binegar: So one thing I should point out i've got a number of uncollected first exams.
Birne Binegar: The cumulative part of the final exam is going to come directly off the first two midterms I may change numbers around but there'll be the same problems, and in particular I guarantee that you'll see those definitions again on the final so.
Birne Binegar: If you haven't picked up your damn I strongly advise you to do so, I used to keep them in the math office, but then i'm going to start carrying them to class too So hopefully people will show up and collect their exam.
Birne Binegar: let's see so what i'm going to do today is, I talked about orthogonal the compositions on Friday, but I never really gave you any calculation examples of that so that's, the main purpose of today's lecture is just to give you a couple examples of how you do the calculations.
Birne Binegar: That.
Birne Binegar: needed to be performed, if you want to find that component of a vector that lives in a certain subspace.
Birne Binegar: So all i'm going to review orthogonal decomposition i'm going to give you two examples and then i'm just going to stop with.
Birne Binegar: sort of formal material and that will actually begin to do a little bit of review, I think what i'll try to talk about at the end of today's lecture or the many ways that we use.
Birne Binegar: row reduction to answer questions in in linear algebra you'll see that.
Birne Binegar: throughout the course we've been using reduction what becomes confusing on the final exam if you have a question.
Birne Binegar: that's problem, it is going to involve a calculation probably a row reduction, what do you do with that role reduction, because what i'll show you is there's at least seven different things, you can do after you reduce the matrix.
Birne Binegar: And so it will be important for you to keep these things sort of separate and.
Birne Binegar: distinct from one another, so that when you do have the final exam you're given a problem, find a basis for such and such as subspace you know exactly once you've reduced the appropriate matrix what to do with that result.
Birne Binegar: Okay, so so let's start with them.
Birne Binegar: Just reviewing it what was covered on Friday.
Birne Binegar: friday's lecture which concerned with a commonality.
Birne Binegar: here's the basic picture, this is taking place in a two dimensional plane, but if I have two vectors A and B, a question now not that can be posed is what is the component of a heading in the direction of be you want to analyze do forces.
Birne Binegar: What component of one force actually adds to.
Birne Binegar: A second force or many applications of this, and in fact we solve this problem, not so much using linear algebra but more or less just using plain geometry and trigonometry what made it.
Birne Binegar: What made our answer look like linear algebra is the fact that we have two different ways of computing adopt product, we can we think about.
Birne Binegar: A vector the way we do, mostly in this course as sorry.
Birne Binegar: If we think of a vector as a as an ordered list of numbers.
Birne Binegar: One through eight and then a dot product say a dot B.
Birne Binegar: is well in terms of order list of numbers, it would be a one B one plus dot dot plus A and B end if these are n dimensional vectors but on the other hand.
Birne Binegar: We learned early on, is that that product also has a geometric interpretation as the length of first factor, the length of the second vector.
Birne Binegar: times the cosine of the angle between the two vectors and so effectively what we did at one started lecture on Friday is we took this geometric picture, you can see that the length of this.
Birne Binegar: component here is the Jason side of a right triangle with angle theta so this guy's link is basically the length of a iPod news times.
Birne Binegar: The cosine of that angle, and in fact we can compute that cosine like taking products of n dimensional vector so In fact this picture could be lifted up into an n dimensional space.
Birne Binegar: The things we need to compute these components of a the part that turning parallel to be in the part that perpendicular be we can use the dot product they extract that geometric information, when all is said and done.
Birne Binegar: What we found is that the part of a that's running parallel to be can be computed by.
Birne Binegar: Taking the dot product with a and D dividing it by magnitude square to be times the vector be itself, you need that vector here because.
Birne Binegar: We want the answer that's a vector not not simply a link, we wanted to actually have same direction is be and then the perpendicular component.
Birne Binegar: That can be extracted once I know what this guy is and i'm given a vector a I can solve for this purpose component alright so that's how you.
Birne Binegar: You have two vectors sitting in an n dimensional space, you can still decompose one of them into a part that's perfectly parallel to the second vector or perpendicular to the second backer.
Birne Binegar: Alright, so that was pretty easy that just.
Birne Binegar: meant manipulating this geometric interpretation that the dot product, so we can extract geometric information, but the harder problems that we post.
Birne Binegar: The second half of friday's lecture was to do a similar thing, but now, instead of finding component of vector that running parallel to give me Dexter be rather we're trying to find a component of that vector that lives inside of subspace and so.
Birne Binegar: You have this geometric picture and.
Birne Binegar: Here, although there.
Birne Binegar: I give you a nice geometric picture of what's going on.
Birne Binegar: The the simple geometric interpretation of that product is not that much help because.
Birne Binegar: yeah first of all, we need to coordinate ISIS the subspace to find a vector that the actually lives in it, and in fact the procedure for finding the component of a vector be that lives within a subspace.
Birne Binegar: doesn't rely on on this formula at all, rather it relies on us finding a basis for w and also finding a basis for.
Birne Binegar: Another subspace so called w perpendicular that consists of all vectors in the entire space that are perpendicular to every vector inside w.
Birne Binegar: Because that's where this guy is going to live this guy does not live in who he's back perpendicular to w so that that that component, the perp it's going to live in some space of vectors that perpendicular to every vector inside w.
Birne Binegar: Alright, so now, but the idea is.
Birne Binegar: Given a vector and a subspace, how can we decompose the vector into a vector that lives within the subspace plus some correction, the part of be that lives outside to give me subspace.
Birne Binegar: Alright, so do we identify those two things what we actually do is we identify.
Birne Binegar: With what we have to do is find a basis for subspace w but will also when need to find a basis for a second subspace w perp.
Birne Binegar: Alright, so here's the algorithm that we've developed on Friday first step, so we still have this picture.
Birne Binegar: Where the is a given vector w is a given subspace and I want to find the part of the vector V that actually lives inside that subspace so.
Birne Binegar: I start with a basis for myself space and what I do is I first define this perpendicular stuff based it's going to be.
Birne Binegar: In fact, this joint from this subspace except the both contain the origin but otherwise there'll be disjointed what w subscript perk actually is it's all the vectors in the entire.
Birne Binegar: vector states are n or everything is living it's going to be the vectors that are perpendicular to every vector inside the subspace.
Birne Binegar: Now turned out that.
Birne Binegar: The subspace w perk we can identify that calculation Lee what it works out to be is well, first of all, in order to be perpendicular to every vector inside your subspace w it suffices to be perpendicular to every basis factor.
Birne Binegar: Because if.
Birne Binegar: If the is perpendicular to each basis factor.
Birne Binegar: Bi for you, given subspace then.
Birne Binegar: That means D dot.
Birne Binegar: E I.
Birne Binegar: it's going to equal zero for each for each of these basis vectors and then this condition can be thought of as a matrix equation.
Birne Binegar: right basis sectors that rose.
Birne Binegar: And I take the vector.
Birne Binegar: A vector perpendicular to each of these guys it's going to satisfy this matrix equation because.
Birne Binegar: When I compute this matrix product, the first component on the right hand side it's going to be well you take a row with the first factor in you got it with a column vector of the second factor, so it just going to be the one that you want.
Birne Binegar: Upon not carrying out that matrix multiplication and similar down here you'll have he and I should say K, because I only have.
Birne Binegar: A basis sectors in general, the.
Birne Binegar: Well D D and we want these are all equals zero, and so what we have it, if you look at the extreme side is a homogeneous linear system.
Birne Binegar: which we can solve.
Birne Binegar: I take this as a coefficient matrix this guy is the coefficient matrix times and unknown variable is equal to zero variable that's a homogeneous linear system.
Birne Binegar: Its solution set it's a no space of the corresponding coefficient matrix this guy.
Birne Binegar: And so we saw that we know how to do that we reduce this to reduce your rational and form we get the odd committed matrix up the solutions we.
Birne Binegar: Take those equations we use them as separate three variables from India from fixed variables and we write out the solution as a hyper planes meeting and expansion in terms of.
Birne Binegar: Three variables time certain constant vectors those constant vectors are actually going to be.
Birne Binegar: A basis for the subspace so in the course.
Birne Binegar: In the course of solving this linear system we're going to end up with basis vectors for the solution set.
Birne Binegar: And in fact.
Birne Binegar: We can actually even before we do the road actually we know how many basis vectors we're going to get because these rows or they're there they're already basis factors, so they gotta be linearly independent.
Birne Binegar: And so they actually form a basis for the road space, but they thought about another way if I reduce this guy to reduce your lesson one form.
Birne Binegar: i'm going to have a pivot every column, because these are basis sector, so I had a call without a pivot that would mean i'd have zero row, and one of these guys would not be a basis sector.
Birne Binegar: So when this guy gets row reduced to rational and form i'm going to have it i'm going to have exactly K tickets.
Birne Binegar: And that, but then i'm going to have since these doctors are all living inside an n dimensional space.
Birne Binegar: I have, I have here K rows.
Birne Binegar: And back there linearly independent rows.
Birne Binegar: And I have an columns.
Birne Binegar: Now, each one of the key rose is going to have a pivot so that means the number of columns of the row Echelon form of this guy that do not have pivots it's going to be equal to N minus K you just subtract off the number of rows with pivots and.
Birne Binegar: that's going to be the dimension of the solution, set.
Birne Binegar: Of this linear system which guarantees that a vector exactly perpendicular to every vector inside w.
Birne Binegar: And so what's going to happen is when I tried to solve a linear system, the homogeneous linear system with this coefficient matrix i'm going to get a solution set that N minus K dimensional.
Birne Binegar: i'm going to write the basis vectors for those so absolution said, not from one to N minus K, but rather from K plus one to N, the reason for using the subscription instead of just counting from one up to N minus K is the idea now is to combine.
Birne Binegar: These basis vectors for the perpendicular subspace to combine those with a basis vectors for the original subspace w.
Birne Binegar: notice that I have K vectors and I have N minus K doctors, the total number of vectors i'm going to get it's going to be when I combine these two bases is going to be in.
Birne Binegar: And literally independent vectors.
Birne Binegar: So there's the basis for w here's a basis for the solution set of this linear system put together, I get a basis for our end and now having constructed this basis for the entire vector space now I can.
Birne Binegar: carry out the decomposition because all I have to do is take the given vector be and express it as a linear combination of these basis vectors then.
Birne Binegar: First part of it, when I just some the first K terms i'm going to get a vector lips inside w and it's going to be exactly this guy and then the contribution from the last N minus K terms it's going to be exactly that guy.
Birne Binegar: So I think I feel this a little bit more deliberately on the.
Birne Binegar: Next slide here we go so.
Birne Binegar: What we try to do, then, is now I constructed a basis.
Birne Binegar: I basis be.
Birne Binegar: it's going to be a basis for w and then I joined to it, the basis for that perpendicular subspace and here we had to be one bk and then the K plus one to be an.
Birne Binegar: Alright, so this is the basis for w and then this is the basis for w PR.
Birne Binegar: But when I combine the two bases together, I get a basis for the entire vector space and so every director, including vector V that we started off with can be written as a linear combination of these basis vectors.
Birne Binegar: How do we actually find the coefficient so.
Birne Binegar: Because this because I have a basis, certainly this expansion exists and it's unique, but how do I find those coefficients well listen turn this setting be equal to this linear combination amounts to solving a certain linear system.
Birne Binegar: Effectively, this right hand side here to be written as.
Birne Binegar: A vector with component see one if the end.
Birne Binegar: Times a matrix.
Birne Binegar: With columns you want to be in.
Birne Binegar: dope.
Birne Binegar: If you do this matrix product what you'll get is.
Birne Binegar: exactly the right hand so right here that's going to equal D.
Birne Binegar: And what you see when I express this condition in matrix form, you see that Well, I can figure out what these coefficients are by solving the corresponding homogeneous linear system, this has the form of a X is equal to a particular column vector so we have a.
Birne Binegar: coefficient matrix that we get from using the entire basis as columns set equal to a particular vector will set it equal to are given vector V one are trying to express.
Birne Binegar: One we're supposed to analyze.
Birne Binegar: What part of the actually lives inside the subspace w what part actually lives outside the subspace.
Birne Binegar: And so.
Birne Binegar: We solve this simply by reducing but you know what's going to happen in this case.
Birne Binegar: Because these guys these columns all linearly independent this coefficient matrix it's going to reduce to the identity matrix there's going to be no free parameters.
Birne Binegar: Because these columns are all linearly independent So what will happen is one let's call this matrix a let's call that X let's call that be to keep with this standard notation what happens is that this ad is going to reduce to the identity matrix and then.
Birne Binegar: Something is going to happen to this last column as well, what it's actually going to be is the coordinate vector.
Birne Binegar: The coordinate vector for the vector D, with respect to this basis, in other words it's going to be exactly.
Birne Binegar: I and are unknown numbers C one C two C n.
Birne Binegar: So.
Birne Binegar: Let me step back and explain what we've done so far i've got writing all the boys.
Birne Binegar: This problem starts.
Birne Binegar: It starts with a vector V and a subspace first step well thanks first thing we do is we find a basis for w.
Birne Binegar: Now that depends on how w was presented to us was a given us a span of vectors well then there's a technique for extracting from a span of vectors use row reduction defined a set of.
Birne Binegar: linearly independent generators for that stuff space or if it's given us the solution set of a homogeneous linear system and we end up solving linear system but.
Birne Binegar: Depending on how the subspace was specified, we have a way of finding a basis for it and that's the first step to solving this problem.
Birne Binegar: Problem of finding what part of an arbitrary vector V lives inside of giving subspace, the first thing we need to do is to coordinate is that subspace.
Birne Binegar: Alright, so that's the first step, then what we do is we create that second auxiliary subspace that consistent all the vectors in the entire vector space that are perpendicular to debbie that involves.
Birne Binegar: Solving and other linear system and getting a basis for the solution set.
Birne Binegar: Then I combine the two bases together, and what I get then it's a basis for the entire vector space or in.
Birne Binegar: Alright, so that's construction.
Birne Binegar: of a particular basis for the entire vector space are N, that will be really useful and breaking an arbitrary vector into to such components.
Birne Binegar: What we do is then the idea is to write the vector V.
Birne Binegar: As see one, the first base at vector up the ck time for last patient vector now these first K basis factors.
Birne Binegar: They all are basis vectors for w so this part.
Birne Binegar: lives in w.
Birne Binegar: But then we have we've extended this basis for w to a basis for our n will have K plus one be K plus one until we reach the last basis vector see and the end.
Birne Binegar: All right, and then these guys this collection up terms, their constant time basis vectors for w PR so that piece is going to live in the perpendicular subspace, and so what i've achieved here is this desire decomposition.
Birne Binegar: These first K terms, consolidate them into a single factor that will be.
Birne Binegar: A part of the vector D that lives inside the subspace and these remaining terms are going to be the part of the of the original vector V that's perpendicular to every vector inside that subspace and that's your desire decomposition.
Birne Binegar: And then distress again what the calculation that you have to perform is is that once you found.
Birne Binegar: A basis for w supplemented that with a basis for the perpendicular subspace.
Birne Binegar: you write them all, as columns you read.
Birne Binegar: All of these n basis vectors that's columns you put your given vector D as a column on the right hand side of this augmented matrix and you reduce that they need to reduce relational and form.
Birne Binegar: Because these columns are all linearly independent the left hand side it's going to always reduce to the identity matrix.
Birne Binegar: And the right hand side it's going to reduce the sun factor will call it db but what it is it's going to be exactly these numbers one through CN that allow us to express the vector V, in terms of the given basis.
Birne Binegar: All right, and so there you have it once you've got both bases you, you can create this augmented matrix you reduce it to reduce your national and form and then.
Birne Binegar: You grab the first key components of this use that to construct the part of the that lives inside this up space and then the remaining components they're going to live in the on the perpendicular substation and you're done.
Birne Binegar: Okay, so that's the calculation, at least for me formally Let me now do a particular example So you see exactly how we compute.
Birne Binegar: The first example.
Birne Binegar: i've arranged things like choosing special vectors and special subspecies i've arranged thing, so that I don't have to do any reduction, it turns out, every time I.
Birne Binegar: get to the next stage where I have to convert the matrix in the rotational on form the matrix is already going to be in row Echelon form so i'm just avoiding those row reduction calculations.
Birne Binegar: But let's start with a substation subspace that generated by two vectors sitting inside our three and then, when I want to do with this information so.
Birne Binegar: there's a subspace and now here's the fixed vector inside are three components, one for and minus four what I want to do is decompose that vector into a piece that lives inside the subspace plus another piece that lives outside this application stack it lives in the.
Birne Binegar: orthogonal complement of the original subspace.
Birne Binegar: Alright, so first thing to note is that these basis factors, you can tell that they're already linearly independent because there's no way.
Birne Binegar: A linear combination of these two vectors can produce zero without taking both coefficients equal to zero, because the cancel this.
Birne Binegar: A constant time see one there's nothing here that I can use to cancel it so i'd have to take see one is equal to zero.
Birne Binegar: If a linear combination of these two vectors is going to equal zero, and similarly there's no way i'm going to be able to cancel out the second component i'm taking a multiple of this and adding it to the two without taking.
Birne Binegar: The the scalable fly use this guy without setting that equal to zero, so, in other words anytime you have two vectors where.
Birne Binegar: There are components that are zero in one but not zero and the other they're going to be literally independent vectors and so.
Birne Binegar: What I need to do, then, is, I need to find the subspace i've got a basis for w it's exactly these two.
Birne Binegar: These two generators i'm going to use that to create a basis for the perpendicular subspace now that involves finding all the factors that are perpendicular to both of these guys.
Birne Binegar: And so what I do is I think of this as the coefficient matrix for linear system, I write it down and then I reduce to reduce show Echelon form define.
Birne Binegar: say more explicitly what we're doing here.
Birne Binegar: That perpendicular subspace it's gonna be.
Birne Binegar: All the vectors.
Birne Binegar: In w that are perpendicular sorry all the vectors in our n that are perpendicular to every w.
Birne Binegar: Inside the subspace and what we've seen is it corresponds to the no space of the matrix I get like taking the basis sectors for w.
Birne Binegar: writing a math as wrote him a matrix and reducing.
Birne Binegar: Because effectively when I want to do is sold the linear system or this matrix times X is equal to zero.
Birne Binegar: All right now, so to calculate the null space of this matrix I would reduce the row Echelon form and then write down equations and expand my solutions well in this case the road up front is easy because i've chosen.
Birne Binegar: A convenient basis already for w.
Birne Binegar: And what will happen then having converted the coefficient matrix into a traditional Echelon form, I then convert back to equations first row says X one plus x three is equal to zero second roses X plus x equal to zero, then I also use this matrix to decide.
Birne Binegar: Which variables are fixed and which variables are free well.
Birne Binegar: The third column doesn't have a pivot so that's going to be our free variable I moved the X three to the opposite side, and then I see I can express X one X two in terms of X Ray and my general solution vector is always going to have this form.
Birne Binegar: And here is.
Birne Binegar: Our desired.
Birne Binegar: basis factor.
Birne Binegar: For that perpendicular subspace.
Birne Binegar: This these two factors are a basis for w I had to do this road actually calculation to find a basis for the perpendicular subspace but now i've got three linearly independent vectors in a three dimensional space i've got a complete basis for our three.
Birne Binegar: So.
Birne Binegar: i'll take the two bases.
Birne Binegar: This was the original basis for w, this is the basis we just found for the perpendicular subspace we put those together.
Birne Binegar: And now, what i'll do.
Birne Binegar: Is i'll try to figure out.
Birne Binegar: A linear combination of these factors.
Birne Binegar: That will produce the getting the vector we were given at the start this vector V so here i've taken these three vectors i'm writing them vertically, because this requires me.
Birne Binegar: to write the vectors the basis vectors this calls.
Birne Binegar: And then I put.
Birne Binegar: The vector V go back to the statement of the problem i'm trying to express this director, one for my niche for i'm trying to break it up and do a part that lives in w and a part that doesn't line w.
Birne Binegar: So here are my three basis vectors for the entire three dimensional space here's the backdrop trying to express in terms of those basis vectors.
Birne Binegar: When I do the row reduction, when I wrote reduce this to reduce your last one form, I get this matrix and now you see X, one has to equal minus two one has to you X to has to equal one and X three has to equal minus three so there it is.
Birne Binegar: Now.
Birne Binegar: This is the coordinate vector of the given vector.
Birne Binegar: Sorry, this is written wrong, this should have been one four minus four not one to one when i've.
Birne Binegar: i've now got the.
Birne Binegar: The Court and i'm now able to express.
Birne Binegar: The vector one four minus four in terms of this basis I expand it and then.
Birne Binegar: these first two terms i'm using basis vectors for w so these some of these first two terms it's going to be the part of be that lives inside w.
Birne Binegar: And the last term in that expansion.
Birne Binegar: This be three it's not a basis for w it's a basis for the perpendicular so this last term here it's going to live inside the perpendicular subspace but there's.
Birne Binegar: The end i've got the decomposition the original vector one four minus four decomposes with respect to the subspace who follows the part that lives inside w minus two one minus one and the part that is perpendicular to w is three three minus three.
Birne Binegar: Okay, so that there's a calculation where.
Birne Binegar: We didn't do any reduction explicitly but i'm.
Birne Binegar: showing you effectively how the calculation is carried out module all the real reaction, you have to do.
Birne Binegar: Let me do one more example, this time.
Birne Binegar: i'm going to state my.
Birne Binegar: The nature of the subs space, a little bit differently, what i'm going to do in this example again take a vector one to one, but this time i'm going to.
Birne Binegar: Try to figure out what.
Birne Binegar: What component of this director lives inside the solution set of that homogeneous linear system.
Birne Binegar: Okay, so I subspace solution set of this equation, and so you're i'm writing it down pretty easy to see that.
Birne Binegar: The coefficient matrix I mean as soon as I write down the young.
Birne Binegar: The coefficient matrix through the system it's already going to be in real Echelon form is just going to be a single row 111.
Birne Binegar: And in fact.
Birne Binegar: So that coefficient matrix.
Birne Binegar: solution set.
Birne Binegar: of X one plus x plus x three is equal to zero, well, we look at that.
Birne Binegar: Exactly.
Birne Binegar: But, not a single equation till just be zero here it's a solution set of this linear system notice the coefficient matrix is already in reduce rotational and form it just a single row in the pivot is already equal to one.
Birne Binegar: And so they'll X two X three are going to be the free parameters and what will happen as soon as I write down the equations corresponding to the single row i'm going to be able to do from that that X, one has to equal minus X two minus X three.
Birne Binegar: And then, when I expand the solution vector in terms of these two three variables here are going to be the corresponding basis vectors.
Birne Binegar: So this tells me that.
Birne Binegar: The general solution is going to be minus X two minus X three x two X three or X two times minus 110 plus x three times minus 101 alright so.
Birne Binegar: in more detail that's how I uncovered these two basis vectors for w now that's all this calculation didn't occur in the first example because I already gave you a basis, but in this again.
Birne Binegar: I gave you a subspace but I didn't give you a basis, so the first thing we have to do is find a basis for w that's what this calculation deals.
Birne Binegar: And then.
Birne Binegar: So i've got a basis for w The next step is to find a basis for the perpendicular subspace what I did, then, is, I take these basis vectors that I normally write vertically.
Birne Binegar: I write them horizontal because the conditions to be a vector inside this perpendicular stuff space is that vectors in this space, have to be perpendicular to each of these basis factors.
Birne Binegar: So we set up.
Birne Binegar: here's the coefficient matrix that we need to reduce here's it's reduced form now will convert back to the equations and it turns out, in this case.
Birne Binegar: we're going to have a single free parameter because we only have one column, and they reduce cholesterol on form that doesn't have a pivot and, in fact, every vector inside this space is going to be a multiple this vector.
Birne Binegar: All right now i've got 123 basis vectors i've got a basis for the entire vector space or three.
Birne Binegar: And so.
Birne Binegar: We combine the two bases together.
Birne Binegar: And then we now try to expand our given vector one to one in terms of the basis V1 V2 v3 That means we write the basis vectors as columns we put the vector that we're trying to expand in terms of the basis as the.
Birne Binegar: column on the very far right and rewrote reduce to reduce relational and form.
Birne Binegar: Reduce relational and form and then these components to third minus one for third that's what happens to this last call which we robotics, and these are going to be the component of the expansion of the given vector one to one with respect to this basis.
Birne Binegar: And now to get the component of the that lives inside dollars subspace we just use the old the first two basis doctors.
Birne Binegar: are going to live inside w so I take two thirds times diverse basis vector minus one third time second basis sector, and that will give me the piece of be that lives inside w.
Birne Binegar: Oh sorry and.
Birne Binegar: I should have finished that oh sorry I didn't really finish this one.
Birne Binegar: The perpendicular part.
Birne Binegar: it's going to be well that last.
Birne Binegar: coefficient times the third basis factor which is gets going to be what four thirds times.
Birne Binegar: 111111 all right, and then you can scale them multiply to get the precise result.
Birne Binegar: Okay, so anyway.
Birne Binegar: that's how you do D competitions, with respect to a subspace.
Birne Binegar: You first need to find a basis for the subspace, then you need to find a basis, and you in fact you use that basis to find a basis for the perpendicular subspace.
Birne Binegar: And then, with base bases for w and is orthogonal compliment you combine those bases to get a basis for the entire vector space are and.
Birne Binegar: Then you take you given vector you expand it with respect to your given basis for our end and you simply split.
Birne Binegar: That linear combination in half, use the first K vectors you use the vectors using.
Birne Binegar: The part that uses the basis for w you collect those into a single vector and you add to that the part that that uses the basis for the.
Birne Binegar: orthogonal compliment w PR and that will be your second component, the two components will be a part of me that lives in w plus the part of the that does not line w.
Birne Binegar: Alright, so.
Birne Binegar: yeah so.
Birne Binegar: example the coefficient of.
Birne Binegar: The equation is that determine our vector that we're trying to.
Birne Binegar: Okay, so remember what a basis does it gives you a way of writing an arbitrary vector as a specific linear combination.
Birne Binegar: So, because we have a basis for the entire vector space that vector be that we were given can be written in terms of those faces factors so that's sort of a separate calculation, but it's one we've done many times, if you want us to to solve.
Birne Binegar: X one be one.
Birne Binegar: plus x and be an.
Birne Binegar: equal to a given vector.
Birne Binegar: V well, the first thing you do is you try to stray away from this simple simple equation it's actually a vector equation, but we want to rewrite it as a matrix equation, in which case it becomes.
Birne Binegar: The end times X one X n is equal to be one to be n.
Birne Binegar: now see our goal is to find these coalition's, and so we have to solve a linear system this condition that says, I can express be in terms of the spaces, that is converted into a matrix equation that's up the form a X.
Birne Binegar: is equal to be it's a Homer isn't non homogeneous linear system, so we solve it by.
Birne Binegar: Taking the coefficient matrix and the column vector on the right and rewrote reduce to reduce row Echelon form.
Birne Binegar: And then we tried write down the solution what's going to happen in this case, though, is because these columns are all linearly independent the right hand side is always going to reduce to the identity matrix and the.
Birne Binegar: The last column here it's going to read it's going to reduce to the answer because upon this production you'll get 1001 all the way down here to one and then here you go get some values what am I call alpha one through out the end.
Birne Binegar: And it's going to say that X, one has to equal alpha one X two, it has to equal altitude, etc, so this last column is actually giving you exactly the numbers, you want.
Birne Binegar: And then so that's what we do, once we get the entire basis and we wrote we do and we grabbed this call vector and then what we do.
Birne Binegar: If we take that expansion and we split it there's a part that uses only the basis vectors for w that's those terms we're going to call the subscript w and then there'll be a part that uses only basis vectors for the perpendicular subspace that's going to be up repeat w perpendicular.
Birne Binegar: So.
Birne Binegar: So, I guess, I should point out that it's kind of surprising that that that case of us.
Birne Binegar: I mean the orthogonal projection of one vector on another all we needed what's the Doc product, but when we try to generalize it to figure out.
Birne Binegar: The part of a vector that lives in the subspace all the apparatus of linear algebra has comes into play, just to solve that problem.
Birne Binegar: So in fact I don't know if we can even solve the problem without a long discussion of basis vectors and sub spaces, etc.
Birne Binegar: Okay, other questions.
Birne Binegar: Okay, so what I wanted to do.
Birne Binegar: To finish off, today's lecture is just to prep you a little bit.
Birne Binegar: Will all continue reviewing on Wednesday and Friday of this week.
Birne Binegar: But the one thing I think that's really confusing typically at the end of the course when we've done so much row reduction is to figure out.
Birne Binegar: Given a final exam problem it's a long calculation, you can detect with I left lots of space for you to work on it so you're probably going to have to do a row reduction of some sort.
Birne Binegar: What do you, what do you what matrix to you know, reduce and what do you do, with the result, so what am I do quickly just do.
Birne Binegar: store org organize things it's describe.
Birne Binegar: The many reduction calculations we've done in this course.
Birne Binegar: First of all.
Birne Binegar: When you reduce your goal is, first of all to reach a matrix in a row Echelon form and that simply means that the pivots.
Birne Binegar: The first non zero entries in any given row they're occurring in this downward staircase from the upper right to the lower left that's usually good if you just want to find basis sectors, but it doesn't it's not good enough if you actually want to solve a linear system.
Birne Binegar: A complete row reduction is taking an arbitrary matrix and putting it in the form that it's in a row Echelon form but also in a way, such as the pivots are all equal to one and that below the pivot and above the pivots we have nothing but zeros.
Birne Binegar: What rubbery option does is it converts to now arbitrary matrix into either row F1 formula, all the way to reduce relational unformed just by using three.
Birne Binegar: Elementary row operations, you can interchange rose, you can scale or multiply rose or you can replace a row by at some with a multiple of another row.
Birne Binegar: This is the operation you use most often use this, for example, to clear out the pivots throughout entries below pivots and you also you tend to use this to clear out the entries above pivots and the second stage or reduction.
Birne Binegar: Alright, so.
Birne Binegar: So the way we use it i've got i'll give you seven quick examples, first of all, if you have a homogeneous or you have a linear system, then the way we solve it is rewrote reduce that matrix to reduce show Echelon form.
Birne Binegar: Included in this case, but a special case that we use over and over again, is went on the right hand side of the equation, we just have the zero vector.
Birne Binegar: Because finding the solution, set up a X is equal to zero, that really only depends on the matrix a it doesn't there's no auxiliary vector in the problem and oftentimes we find that.
Birne Binegar: for other purposes, finding the no space of the matrix or the solution set of eight X equals zero is really what we want.
Birne Binegar: What what's nice about the second case and i've used it constantly throughout the course is that I don't really have the form the augmented matrix to do the appropriate role reaction this case because.
Birne Binegar: Nothing is going to happen in that last column because it's it's all zeros and if I even if I do real operations on.
The.
Birne Binegar: When my augmented matrix it's actually have this form.
Birne Binegar: With the zero vector here when I do the row reduction absolutely nothing at all happens in this last column because.
Birne Binegar: Whether i'm scale are multiplying or interchanging rose these entries always stayed zero so usually, when I talk about solving a homogeneous in your system, I guess row reduce this part.
Birne Binegar: So it's, not that this disappears it just it becomes unnecessary to calculate.
Birne Binegar: All right, and then.
Birne Binegar: A second utility for row reduction we saw that after the first exam when we started talking about some spaces, if you have a subspace that generated by a set of vectors.
Birne Binegar: Well there's no guarantee that these are actually basis vectors that could you may have more directions and you need to reach every point in this space.
Birne Binegar: To find the idea of a basis we find a minimal set of directions that capable of reaching every point in this space, the way you find such a minimal set of generators is you take your given vectors you write them as the rows of the matrix you reduce that to row Echelon form.
Birne Binegar: If these vectors are not linearly independent then you're going to get some zeros at the bottom, that you can just ignore you just grab the non zero Rosa that row Echelon form, and that will give you a basis for the subspace.
Birne Binegar: there's also some other stuff spaces, that we can calculate the ones that are attached to him matrix so.
Birne Binegar: But when you're calculating sub spaces that are attached to a matrix you can use a single row reduction to find bases for three substations I can attach to a matrix so there's three sub spaces are.
Birne Binegar: I can take the span of all the row vectors in the matrix that's called row space and the way I find a basis for well, I simply reduce a to reduce your actual on form and I grabbed the non to roads, the same thing thing is, we would do if we just had a list of vectors.
Birne Binegar: To get a basis for the column space, again, I wrote us a to reduce your salon form but instead of grabbing a column of the of the reduce your national and form, what I do is I figure out where the pivots live in this guy and then I grabbed the corresponding columns of the original matrix.
Birne Binegar: And then, a basis for the North face that's the solution set up the matrix times a variable vector is equal to zero that you do by calculating reduce relational on form of the coefficient matrix and then write down the equations and expanding in terms of free parameters, etc.
Birne Binegar: All right, we also use real reduction to figure out the inverse of the matrix that came about, by setting up this.
Birne Binegar: Double matrix a joined with the identity matrix and row reducing this guy to reduce your actual on form, then it turned out that the right hand side would end up being the inverse.
Birne Binegar: If we had a linear transformation well what we could do is, we can associate a matrix to this linear transformation.
Birne Binegar: And then there are sub spaces that are associated with this linear transformation all the points in the.
Birne Binegar: Co domain that can be reached by applying the linear transformation that's called the range that turned out to be the column space of this matrix and i've just described how you find a basis for the column space.
Birne Binegar: The set of vectors in a domain that gets sent to the zero vector here that's called the Colonel and again that's calculate essentially be a row reduction calculation, where you solve feticide equation a subscript g of X is equal to zero.
Birne Binegar: And then we also use real reduction to calculate determinants.
Birne Binegar: here for a row reduces two matrix a prime and row Echelon form then determine a it's going to be minus one to the number of row interchanges I used in the sequence of operations, I have to.
Birne Binegar: extract from the determine if a prime all the scale or factors i've used to multiply rows of pain in the course of this row reduction and then I have the determinant of the relational and form of a but that's going to be just going to be the diagonal product long that main diagonal.
Birne Binegar: And then yeah and then, finally, we also use reproduction extensively when we try to find eigenvectors the the basis for the.
Birne Binegar: Are Eigen space if ours and eigenvalue for matrix a is found by calculating and no space.
Birne Binegar: Okay, so.
Birne Binegar: what's always got it what's probably the hard part of the exam is given a problem where there's a lot of space to work is to figure out what makes you have to write down and then what you have to do with the result, once you get the matrix and reduce show Echelon form.
Birne Binegar: Alright, so i'll let you guys go now i'm sorry ran three minutes over but.
Birne Binegar: we'll continue the review on Wednesday.