Birne Binegar: Okay, so.
Birne Binegar: let's see so.
Birne Binegar: Two people here today that's that's depressing.
Birne Binegar: So what i'm going to do today is i'm going to finish up with the last topic for the course on orthogonal D compositions and what it has to do is.
Birne Binegar: Basically, just given given two vectors in the plane.
Birne Binegar: Say vector a inner vector B is to decompose one of these factors, with respect to the other, in the sense that.
Birne Binegar: I can.
Birne Binegar: find a component of a that's perpendicular to the given direction, and I can find a component of a this parallel to that direction, but the idea is to break up an arbitrary vector.
Birne Binegar: up into two components for specified respect to a given direction, so these sort of things are often used, for example when you're analyzing forces.
Birne Binegar: You have some wind getting a plane here the actual force exerted on this system is only really depends on.
Birne Binegar: The fourth the component of force it's perpendicular to the surface, so if you think of this as your direction, be the interesting part of the force as far as it exert force on the surface, would be the perpendicular component and so you need a fourth factor to be.
Birne Binegar: divided up into a part that perpendicular to the surface or parallel to the surface so.
Birne Binegar: Our interest, though, is not in this.
Birne Binegar: picture quite this symbol here we're looking at the component of a vector that's running parallel and perpendicular to a given direction.
Birne Binegar: What we're gonna do today replace this direction, be by subspace and what will demand is a.
Birne Binegar: Now, think of this sliding part of a blackboard is a hyper plane embedded in three dimensional space at the surface.
Birne Binegar: We have a vector and three dimension it's going to have a certain component that's perpendicular to the surface and also, you will have a short of a shadow in the surface, as well, so if I have a vector popping out.
Birne Binegar: they'll have a certain chatter that lives inside the subspace and it also have a certain component as well that's perpendicular that subspace that's the sort of the competition we're aiming for, but what i'll talk about first is how I decompose vector.
Birne Binegar: With respect to a particular direction.
Birne Binegar: So let me start with that.
Birne Binegar: Simpler condition here's two vectors in the plane what i'm hoping to do for the given situation is decompose one of the vectors a into two pieces apart that's running parallel to the given direction be and apart that's perpendicular to it.
Birne Binegar: From this diagram you know that the way you add vectors geometrically is you.
Birne Binegar: You would take draw the vector A, B and then you would transport the tail The second factor to the tip of the first factor, and then the result and vector respect yourself, is the fact that that starts from the tail of this one and goes to the tip this factor so there's.
Birne Binegar: This picture of athletes could have been drawn the first week of class you're interpreting vectors as directed line segments.
Birne Binegar: In this picture we call.
Birne Binegar: The vector a subscript be the orthogonal well we call this decomposition of ad orthogonal the composition of a with respect to the direction be.
Birne Binegar: This component here is called a component of a in the direction of be and, naturally, a subscript perp if the component of a that's perpendicular to be.
Birne Binegar: All right now.
Birne Binegar: Other language that Jews for this situation, this instead of talking about.
Birne Binegar: The component along a certain direction sometimes people will say the same vector ad that combined with the perpendicular.
Birne Binegar: component of a to produce a sometimes they refer to this at the mental projection of a in the direction of be.
Birne Binegar: Where that language comes from if you imagine shining a flashlight down perpendicular to this direction D and then the shadow that the vector a lead.
Birne Binegar: On this sort of surface be that's that would be the projection but it coincides with what we and the last slide called the part of a that's running parallel to be.
Birne Binegar: All right now.
Birne Binegar: The discussion, thus far i'm drawing vectors in a plane and drawing them as directed line segments, but of course in this course, we like to think of vectors primarily is.
Birne Binegar: list of numbers ordered list of numbers, I have endless numbers in that list it'd be a vector inside the vector space are in, so what i'm going to do next i'm going to.
Birne Binegar: Look at this analysis, again, but now use instead of a protractor in the ruler to draw this picture we're going to specify vectors in terms of their components.
Birne Binegar: So the set this up.
Birne Binegar: First of all, bringing a little high school geometry.
Birne Binegar: That tells me a bit about the length of these.
Birne Binegar: Factors a purpose, a B, you have the Pythagoras theorem that says the length of this hypothesis is equal to some of the squares of the site.
Birne Binegar: Or, more precise move a little trigonometry three you know the length of this site adjacent to the angle is the length of the high pot news times the cons the cosine of the angle.
Birne Binegar: In other words, i'm drawing this picture, when you have a high pot news, then in an angle feta than the length of the adjacent side is the length of the high pot news is the cosine of the angle and the length of the opposite side is length of the high pot news is the sign of the angle.
Birne Binegar: Now what i'm going to do.
Birne Binegar: To get away from this purely geometrical picture is remember that.
Birne Binegar: In the vector space our end we don't have procrastination rulers, but we do have a Doc product.
Birne Binegar: So a dot B, so you know, a higher dimensional space we can extract geometry like this, using the Doc product, because this is going to be, on the one hand.
Birne Binegar: Computer will in terms of the components of these sectors.
Birne Binegar: In the end, sorry, these are no longer factories components of vectors we have a way of computing dot products in an n dimensional space which can in turn the interpreted geometrically.
Birne Binegar: And the same way as they are in two dimensions.
Birne Binegar: The dot product to two vectors is the length of the first factor linked to the second factor times the cosine of the angle between vectors and, in fact.
Birne Binegar: Using this geometric interpretation of the dot product will be computed this way, but we can interpret it this way, in which case a length of pay it just the square root of a dotted with itself similar for the link to be.
Birne Binegar: equal to the square root of a beam dotted with itself.
Birne Binegar: And this angle here, well, we can compute it as cosine data, all I do is divide both sides, so I have a.be on the right hand side lt by by these length but i'm going to express them.
Birne Binegar: In terms of adapt Robert.
Birne Binegar: E dot p.
Birne Binegar: Alright, so the idea here, and this was.
Birne Binegar: First.
Birne Binegar: explained in the first week of class, the idea is that we can use this.
Birne Binegar: Doc product of n dimensional vectors even though they're just ordered list of numbers, we can interpret them geometrically.
Birne Binegar: We can get linked up vectors by taking the Doc product of the vector of itself and then taking it square root and we can get angles between vectors in an n dimensional space I taking their product and then dividing by the corresponding links which are also computer using this formula.
Birne Binegar: And in that way we can talk, we can look at this picture, but we no longer have to think of it being confined to a two dimensional plane, well, we think of it is actually occurring in an n dimensional states.
Birne Binegar: And this formula for the product.
Birne Binegar: makes these geometric quantities, it makes it a computer.
Birne Binegar: Okay, so let's um.
Birne Binegar: Let me again try to express a in terms of two components, but now I want you to visualize these guys, as if they were living in a higher dimensional space, but that we still have these these geometric interpretation of our product which is defined on ordered list of numbers.
Birne Binegar: So um.
Birne Binegar: So here i'm just.
Birne Binegar: reviewing these interpretations and now let's look at the picture again.
Birne Binegar: If we want the length of this vector well it's going to be the length of the high pot news times the cosine of the angle in between, but now we have.
Birne Binegar: dot products of n dimensional vectors that we can use to compute length and angles, so when we plug in.
Birne Binegar: I haven't yet replaced the length with that product, but this angle.
Birne Binegar: that's how you solve for it, you can increase the angle, in terms of dot products and.
Birne Binegar: linked up of a is going to cancel, and so the length of this vector a subscript, be it just can be the dot product that they would be divided by the direction of be or byte magazine be.
Birne Binegar: All right now.
Birne Binegar: Actually, what I want.
Birne Binegar: A does not equal.
Birne Binegar: The length of the link to this it's not equal, the length of that plus link at that.
Birne Binegar: But as vectors they are true a does equal this specter plus this factor, so what i'm going to do is not rather than work with the length of these factors, I want to work with the vectors themselves.
Birne Binegar: This calculation tells me how long the spectrum is but it doesn't prescribe a direction to it, because this is just the number.
Birne Binegar: To get a direction to it well it's going in the direction of be so all you have to do is take a vector of this link and make sure it's going in the correct direction.
Birne Binegar: And so.
Birne Binegar: here's how we do that we take the direction of the we divided by the length of be that creates a vector of unit length it's going in the same direction.
Birne Binegar: And then, when we take this unit vector vector of length one in the direction of fee and we multiply it by the length of the segment we get exactly a vector of length.
Birne Binegar: magnitude of this vector and it's heading in the direction of be alright so here's what we want this sector with computer this length on the preceding slide its length is exactly a dot P divided by the main to be.
Birne Binegar: And now we're going to multiply that length by this direction to get a vector scale or multiple of this unit vector.
Birne Binegar: And all i'm going to do next is.
Birne Binegar: plug in for what we've computed the length was the Doc product of a with D divided by the link to be and the unit vector and the be direction was the divided by its length.
Birne Binegar: And now, when I put this all together.
Birne Binegar: I have two factors of the least squared but the link squared is just adopt product, and so, in the end we get this nice simple formula for this vector a subscript be it's going to be the product of a with be divided by dot product with the with itself times the vector be.
Birne Binegar: All right, alright so.
Birne Binegar: We have effectively I prove the theorem every.
Birne Binegar: Look at an arbitrary vector and lead bb some fixed vector inside our and then we're always going to have a decomposition of a as a vector some have a heart that update is running parallel to be a part of a that's running perpendicular to be the way those two components can be computed.
Birne Binegar: We just saw on the last slide that this guy is certain length times with certain direction when everything is written this simple way as possible turns out that this vector is exactly a dot P divided by P dot B times of direction be.
Birne Binegar: To get the perpendicular component, well, we were given a and we've calculated D, so all we have to do to get the perpendicular part is simply move this term, to the opposite side, and now we have an expression for the perpendicular part.
Birne Binegar: Alright, so.
Birne Binegar: that's how these two formulas tell you exactly how you can compose a vector with respect to a particular direction.
Birne Binegar: here's our vector a.
Birne Binegar: here's our direction be and what we have been trying to do.
Birne Binegar: Is find the part of eight at running parallel to be in the part of a that's running perpendicular to be.
Birne Binegar: And what we see is that a parallel is a dot P divided by D dot B.
Birne Binegar: Times itself and the perpendicular part is simply the original vector a.
Birne Binegar: minus this parallel keys, so that a minus.
Birne Binegar: A de Vida de de de time.
Birne Binegar: Okay, so, like everything is vectors.
Birne Binegar: So.
Birne Binegar: What I want you to note on the right hand side here, I can do this.
Birne Binegar: In a vector space of arbitrary dimension, because I have a formula tells me how to compute dot products when I write vectors as ordered list of numbers so doesn't matter these doctors do not have to be confined to the plane for these formulas to work.
Birne Binegar: Alright, so.
Birne Binegar: that's the simple case.
Birne Binegar: And it's up put it down this the first week of class what's going to be harder it's going to take more work is to generalize this.
Birne Binegar: To what I spoke about before if you have a vector in a three dimensional space and maybe a two dimensional subspace to divide that three dimensional vector up into a part that's running perpendicular to your subspace and apart that athletes running parallel or life inside the subspace.
Birne Binegar: So the geometric picture we're now in a look well you're here i'm just doing this computation.
Birne Binegar: You want to i'm given two vectors A and B that first component is given by this formula, here we just take the dark products.
Birne Binegar: multiply by the vector you see that calculation yield a particular vector inside the three dimensional space.
Birne Binegar: And the perpendicular part is just the original vector a minus the parallel part and then upon computation you get that done there's a sample computation of application of these two formulas to get that the composition, but what i'm.
Birne Binegar: Speaking about next, though, is.
Birne Binegar: When we have a subspace of a vector space of dimension and and we have a particular vector particular n dimensional vector what i'm trying to do now is decompose this vector into a part that lives in the subspace and part that's perpendicular to that subspace.
Birne Binegar: Alright, so.
Birne Binegar: we're going to do that, but the.
Birne Binegar: And in fact what we're going to find, just like in the one dimensional case where a decomposes into two unique vectors set a subscription fee and a perk or add up to a.
Birne Binegar: And this more general situation where instead of.
Birne Binegar: asking how a is parallel to be yes, how a is somehow contained in a subspace w we're going to end up with a similar decomposition an arbitrary vector is going to have a piece that lives inside the subspace and another piece that effectively is perpendicular to it.
Birne Binegar: All right, and so what i'm going to show you next is that there's we can always achieve this decomposition and we'll get nice formulas for both of these.
Birne Binegar: Both these pieces of the but it's going to be a little bit more strenuous because now we're looking at sub spaces.
Birne Binegar: Rather than individual vectors and so the apparatus is linear algebra in particular our ability to solve linear systems using row reduction etc it's all going to come into play.
Birne Binegar: Alright, so, but given us up space and a decomposition of a vector into a part that lives inside the subspace in the part that's perpendicular subspace we're going to call that the orthogonal T composition of the vector be with respect to the subspace.
Birne Binegar: that's.
Birne Binegar: Given.
Birne Binegar: To be.
Birne Binegar: inside.
Birne Binegar: vector space of arbitrary dimension and a subspace.
Birne Binegar: w that sits.
Birne Binegar: Inside the same vector space to.
Birne Binegar: decompose.
Birne Binegar: As he said, lives inside w and plus a piece that's perpendicular to the w.
Birne Binegar: Alright, so that's my goal, but before in order to do that I need to actually introduce a second subspace.
Birne Binegar: So what i'm gonna do is i'm going to create a new set of vectors this is going to be the vectors.
Birne Binegar: inside.
Birne Binegar: The total vector space are in such that let's call these vectors.
Birne Binegar: V, what we want is the to be perpendicular.
Birne Binegar: All vectors.
Birne Binegar: w inside w.
Birne Binegar: so effectively i'm trying to figure out where these guys could live that I want these factors to be perpendicular to every vector it's like w and I want this piece to live inside debbie so this space here it's going to.
Birne Binegar: consist of vectors that are perpendicular to every vector inside w.
Birne Binegar: And what i'm going to show you is how you actually find these doctors.
Birne Binegar: Alright, so.
Birne Binegar: let's suppose.
Birne Binegar: let's suppose our subspace w.
Birne Binegar: Every sub State has a basis, let.
Birne Binegar: be equal i'm gonna write the subscript gabby and the one to be K.
Birne Binegar: The.
Birne Binegar: Patient basis for the sub stage for the given subspace w so that means.
Birne Binegar: Every.
Birne Binegar: w inside w.
Birne Binegar: can be written.
Birne Binegar: At w equals C one C.
Birne Binegar: Plus, plus si K tk and, second, of all the coefficients well.
Birne Binegar: I think you say one of two things either these coefficients are unique or that these vectors are linearly independent so all right all right off to me Downs with.
Birne Binegar: With C one C K unique.
Birne Binegar: And that, of course, it's equivalent to saying that the one.
Birne Binegar: To pk.
Birne Binegar: Are linearly independent.
Birne Binegar: there's more more abstract linear algebra coming into play.
Birne Binegar: And so.
Birne Binegar: My goal here is to figure out what this space is because that's where these guys are all learn these perpendicular components they're all going to have to be perpendicular to every vector inside w So how do I actually write down this space.
Birne Binegar: Well, what we'll do is.
Birne Binegar: If these factors are perpendicular to w and every w can be written in terms of these basis factors, then.
Birne Binegar: If.
Birne Binegar: He is perpendicular to each.
Birne Binegar: Each of the basis factors, then.
Birne Binegar: The is perpendicular.
Birne Binegar: To every factor and w.
Birne Binegar: y why because.
Birne Binegar: The w.
Birne Binegar: let's expand w in terms of the basis that's going to be B dot C one C one.
Birne Binegar: c K.
Birne Binegar: tk for some unique choice of coefficients now my hypothesis is that our vector vs perpendicular to each of the basis factors and so we're going to get see one times D dot the one plus C to D plus pk times d.uk I hypothesis that that.
Birne Binegar: The factors are perpendicular to each other basis factors and so we'll get seen one times zero Plus C two times zero ck time zero.
Birne Binegar: And so.
Birne Binegar: That means that the it's going to be perpendicular to every vector in w.
Birne Binegar: Alright, so that's the condition i'm going to use to.
Birne Binegar: use that to identify this the stakes.
Birne Binegar: Conditions down, so we need.
Birne Binegar: vectors.
Birne Binegar: And the perpendicular subspace.
Birne Binegar: To be perpendicular.
Birne Binegar: To each sector.
Birne Binegar: For gabby so we'll have.
Birne Binegar: started with the hashtag equals zero the to Daddy with the past equals zero last face at factor will be going to have to equals zero.
Birne Binegar: that's condition that i've stated here i'm looking for vectors that are perpendicular to every basis factor, so that those vectors so that any linear combination of them it's perpendicular to.
Birne Binegar: perpendicular to the vector be and so i'm trying to find these factors fee but here's the condition i'm going to put on them.
Birne Binegar: i'm going to demand that these vectors fee are perpendicular total the basis sectors that's going to guarantee that they're perpendicular to every vector inside w.
Birne Binegar: And now, when I do is I view this as a linear system.
Birne Binegar: So here, it is my condition is that my i'm looking for vectors that are perpendicular to wi fi system make these factors perpendicular to all the basis factors.
Birne Binegar: And when I set up this equation here I just listed the equations but now i'm going to put them in inside vectors so on this side I have zero vector in here, I have.
Birne Binegar: Each each component here so Doc product of a basis, Dr with my given vector.
Birne Binegar: This particular vector.
Birne Binegar: Could you can think of it as the matrix that I formed by using.
Birne Binegar: basis vectors be one as rose and taking their dot product with an n dimensional vector.
Birne Binegar: Everything here is living inside our end, but the condition, I have here is i'm taking an n dimensional vector and i'm taking its Doc product with each of the basis factors, but the basis vectors are also elements or in.
Birne Binegar: And so.
Birne Binegar: So this first component here is the first row with this matrix dotted with this first column, and it has to equal zero second row with this matrix is just a second row God, with the one and only call them to the second factor.
Birne Binegar: Why write it this way.
Birne Binegar: What you should see, then, is that, basically, I have a linear system.
Birne Binegar: So.
Birne Binegar: The vectors.
Birne Binegar: D.
Birne Binegar: That are perpendicular to w are going to.
Birne Binegar: Be perpendicular to the basis factors.
Birne Binegar: 50 K.
Birne Binegar: For w and so.
Birne Binegar: That means that.
Birne Binegar: be one written as a row be written as a row UK written as a row multiplying the components that he has to equal.
Birne Binegar: Zero vector.
Birne Binegar: And that I hope you see is up the form of a linear system, a times G is equal to zero, this is your coefficient matrix let me write it a subscription job you.
Birne Binegar: and be because the way I form this coefficient matrix as I take my basis for w and use those as the rows of the matrix.
Birne Binegar: So.
Birne Binegar: But this tells me how i'm going to find these perpendicular vectors they're going to have to satisfy this linear system and I got the coefficient matrix for that linear system, so what we want.
Birne Binegar: A solution.
Birne Binegar: will be vectors.
Birne Binegar: In the north face of that matrix.
Birne Binegar: You have a coefficient matrix times the vector is equal to zero well, the solution set of this linear system thinking of bs and unknown it's exactly what's defined this an old space, I mean this no spaces defined to be exactly the solution set that linear system.
Birne Binegar: And we know how to find these we've been doing this for ages, the find the notes, they have a matrix you reduce to reduce your rational and form and then you convert back to equations expand your solution vectors in terms of free parameters.
Birne Binegar: Then you get asked you not only fine now this no space that you actually get a basis for that no space, at the same time, so now we've got it into.
Birne Binegar: A tractable problem that we can actually solve once we have a basis for subspace we can create this matrix and then we can calculate it's no space, the vectors in this knowledge base are going to be exactly the vectors that are perpendicular to all the factors in w.
Birne Binegar: That space we're now calling.
Birne Binegar: A set of vectors that's perpendicular to every vector inside our subspace w.
Birne Binegar: Okay.
Birne Binegar: So let me point something out about this linear system.
Birne Binegar: here's the coefficient matrix.
Birne Binegar: it's going to have exactly and columns if the vectors V1 to bk or n dimensional vectors i'm just putting their individual components and writing them horizontally it's also got pay rose, because I have a total of K basis vectors from my subspace.
Birne Binegar: So.
Birne Binegar: Great is dimension of w but w itself consists of n dimensional vectors embedded in an n dimensional space it's just that the.
Birne Binegar: Only looking for linear combinations of exactly these factors to create w.
Birne Binegar: So this coefficient matrix it's going to have three rows and columns.
Birne Binegar: Well, and what's going to happen, then, is a vector is going to be perpendicular to all the factors of w if it's perpendicular to each of these faces factors, and that in terms means finding this North face.
Birne Binegar: This face w.
Birne Binegar: Sub super script perpendicular is called the orthogonal compliment to w inside or in it's not the same thing is w in fact it's sort of exactly opposite w what will find if if w has dimension K this space here is going to have dimension and minus K.
Birne Binegar: here's how you see that it's a this matrix that I wrote down is as two rows and columns.
Birne Binegar: So let's since the number of columns is equal to N, what I can do is I can take this matrix in row reduce it to reduce your Echelon form the number of columns that have pivots that's going to be that's kind of correspond to the three parameters in the solution set of this linear system.
Birne Binegar: So that's exactly this dimension of my solution set and then.
Birne Binegar: This corresponds to the number of columns without pivots in the row Echelon form of this matrix and this dimension here is equal to the dimension of the row space or equivalent dimension of the column states it's the number of.
Birne Binegar: columns in a row Echelon form of the coefficient matrix that have pivots so here you're counting.
Birne Binegar: pivots and pounds that have pivots and columns that don't have pivots, although not for the original matrix a.
Birne Binegar: We have here is.
Birne Binegar: The rank.
Birne Binegar: Of a matrix a is equal to the.
Birne Binegar: Number of columns.
Birne Binegar: The reader relational on warm at bay that happens, give it to.
Birne Binegar: The North face.
Birne Binegar: Of the matrix a is the.
Birne Binegar: Number of columns.
Birne Binegar: Reduce roll Echelon warm up a that don't have.
Birne Binegar: Because these columns without pivots correspond to free parameters in the solution.
Birne Binegar: to remind you of that and so that's all i'm saying here is the total number of columns that matrix you can think of that it's being broken up you look at the reduced relational on form of this matrix you.
Birne Binegar: Have a certain number of columns that have pivots and certain number of columns that don't have it, they have to add up to the total number of columns.
Birne Binegar: Alright, so the rank of this matrix we know what that is it's not only the dimension of the row space, but it's also the equal to the dimension of the column space.
Birne Binegar: row space.
Birne Binegar: A it's also equal to the dimensions of the computation a.
Birne Binegar: Because both these dimensions are determined by.
Birne Binegar: The number of basis factors we have and the way we find a basis for the roast basis we reduce a to rational on form and we grabbed the non zero rose.
Birne Binegar: way we find a basis for this basically reduce a to row have fun for you figure out where the pivots live and we go back to the original matrix but both of these numbers just depend on the number of pivots that occur in the row Echelon performance your matrix.
Birne Binegar: So.
Birne Binegar: On the other hand.
Birne Binegar: what's the dimension of the row space.
Birne Binegar: of such a matrix.
Birne Binegar: Well, these are all basis deck your shoulder linearly independent.
Birne Binegar: And that means that effectively the rank of the matrix is a number of dimensions row spaces number of basis vector sheet.
Birne Binegar: These guys are all basis factors are all linearly independent so In fact this rank is exactly equal to the number of K of basis vectors you had for w.
Birne Binegar: And that means the dimension of this perpendicular subspace.
Birne Binegar: we're finding this subspace by solving a linear system but it's exactly the space of vectors we want every vector that sits inside this knowledge base is going to be perpendicular to every vector w.
Birne Binegar: And so subtracting K from N, is going to give me the dimension of the solution, set and what is that what's it going to be and minus K backers.
Birne Binegar: Now, what we have.
Birne Binegar: So.
Birne Binegar: w is gonna have.
Birne Binegar: Yes, w has a basis.
Birne Binegar: Be subscript w equals one up okay.
Birne Binegar: We can find.
Birne Binegar: That perpendicular substation this is going to be vectors.
Birne Binegar: That are perpendicular to.
Birne Binegar: Every w inside our given subspace.
Birne Binegar: We actually can find a basis for this by looking at the north face.
Birne Binegar: Of the matrix reform by using the given list of factors as rose.
Birne Binegar: And what we when I argued here is, I even know the dimension of this solution set.
Birne Binegar: going to have.
Birne Binegar: mentioned.
Birne Binegar: Everything is inside or in So these are all n dimensional vectors and then, when I.
Birne Binegar: removed the rank from the side like, then, I get the dimension of that North base in that dimension is going to be exactly N minus pay meaning.
Birne Binegar: perpendicular subspace.
Birne Binegar: Has a basis.
Birne Binegar: With it exactly.
Birne Binegar: And minus vectors.
Birne Binegar: CW.
Birne Binegar: Be.
Birne Binegar: The basis for w.
Birne Binegar: pk.
Birne Binegar: And then.
Birne Binegar: let's let.
Birne Binegar: The perp be.
Birne Binegar: A basis for.
Birne Binegar: w purpose, which is the north face.
Birne Binegar: Of the matrix reform by using these factors as rose.
Birne Binegar: From this dimension cow, we know that there's exactly N minus K.
Birne Binegar: vectors in this in this solution set.
Birne Binegar: But.
Birne Binegar: And so, let me write those vectors using a notation similar to this i'm going to want N minus K vectors but rather than count from one up to N minus K let's count from K plus one.
Birne Binegar: Up to and.
Birne Binegar: So i've argued that the mention of the perpendicular subspace is N minus K two dimensional.
Birne Binegar: i'm assuming i've calculated this no space by solving the linear system expanding in terms of the free parameters and grabbing those constant factors, they will be a basis.
Birne Binegar: For the no space and now i'm just labeling those guys using indexes that run from K plus one to end, but if you count the number of factors here it's gonna be exactly N minus K vectors.
Birne Binegar: Why did I do that well, the idea is that.
Birne Binegar: From the basis the w the given basis for my subspace and the spaces.
Birne Binegar: Which is now consist of N minus K vectors that start off with the K plus one and up at the end, I can combine these two sets of vectors together.
Birne Binegar: To get a list of vectors that social be one up to be paid.
Birne Binegar: And then continue the K plus one, up to the end and my claim is.
Birne Binegar: Is that this long list of vectors is a basis.
Birne Binegar: For the entire vector space are in.
Birne Binegar: Alright, so let me show you why that's true.
Birne Binegar: If you have.
Birne Binegar: An vectors in an n dimensional space you don't necessarily generate the entire vector space from those and directions, because some of them could be parallel what you need are in independent directions if you're going to generate an n dimensional states so when I claim is that.
Birne Binegar: When I take the given basis for w and I combine it with a basis for this solution set that I end up with a set of vectors that are, I have a total of 10 of them one to end and they're also linearly independent that will make them a basis for our in.
Birne Binegar: The water wildly linearly independent suppose I had a linear combination of them that added to zero, if the only way I can get this time to zero is by taking a coefficient see one to see in all separately equal to zero, then they're linearly independent.
Birne Binegar: So let's see what that would require if I take.
Birne Binegar: This last set of N minus K vectors and I move it to the right hand side I get this identity.
Birne Binegar: And now, let me write the one to be this linear combination that's on the left hand side and let's let be to be the linear combination of factors that appears on the right hand side.
Birne Binegar: Well then, this equation says that the one has to equal minus the two.
Birne Binegar: But from the also from the way I constructed V1 and V2 be one since I formed it by taking a linear combination of the basis vectors for w this vector sits inside w.
Birne Binegar: The right hand side here is constructed by taking linear combinations of basis sector for w PR so this linear combination is going to live inside w PR and so what I.
Birne Binegar: If this linear combination equals 01 thing, it has to happen is that the one half equal minus B to.
Birne Binegar: Let me show you that that can only happen.
Birne Binegar: When see one two ck is equal to zero, and then that's also going to imply that the remaining coefficients are also equal to zero.
Birne Binegar: So if.
Birne Binegar: This the one is equal to minus V2 this guy sitting inside w and this guy sitting inside w perk up well.
Birne Binegar: suppose I take this equation.
Birne Binegar: And I take its dot product with the one on both sides.
Birne Binegar: Right well I don't have the minus sign here, so this equation and play this and what I know about the vectors the tubes and sits inside be perpendicular it's perpendicular to every vector inside w will be one sits inside w so that's equal to zero.
Birne Binegar: On the other hand, this is the link squared.
Birne Binegar: Of the vector V one so cloud can be one of itself, the only vector who slinked is equal to zero is the zero vector so.
Birne Binegar: I take this relation that comes by demanding that I have a linear combination of all my basis vectors that equals zero I put the partner w on the right hand side the partner w PR on the left hand side and then.
Birne Binegar: I get this identity and then, when I take the doctor that this identity with the one.
Birne Binegar: On the right hand side I get zero on the left hand side, I get the length of the one wealth, the length of a vector is equal to zero, that means.
Birne Binegar: The one is equal to zero, but if the one is equal to zero.
Birne Binegar: that's going to imply that the two, which is the negative, of the one is also equal to zero.
Birne Binegar: All right, yeah.
Birne Binegar: But now conclude is that the one has to equal to zero and D to half equals zero if this identity hold.
Birne Binegar: But you only way, these are basis vectors are linearly independent vectors so if this sector equals zero all the coefficient see one two ck happy equals zero if this vector equals zero as we've just shown.
Birne Binegar: All the coefficients of the basis vectors have to separately equals zero, and so, when I found is that see one two ck and si K plus one to see in they all have two equals zero if this identity whole that's how I know that those basis vectors are linearly independent.
Birne Binegar: And now.
Birne Binegar: Now I have a means of actually calculating that.
Birne Binegar: orthogonal D composition.
Birne Binegar: Right so.
Birne Binegar: Here we have be equally.
Birne Binegar: be some bk and then we are joined to that the K plus one would be n.
Birne Binegar: And what we've ended up with is.
Birne Binegar: A basis for the entire vector stage of our w but.
Birne Binegar: We also have the w.
Birne Binegar: pk.
Birne Binegar: And the perp.
Birne Binegar: The tail end of the sequence bk plus one.
Birne Binegar: bk the end that's the basis.
Birne Binegar: For the perpendicular subspace and now, because this is the basis for our end.
Birne Binegar: Every vector V inside our end has.
Birne Binegar: A unique expansion.
Birne Binegar: it's going to be the we expand it in terms of its basis be one see to the to all the way up to ck UK well that's only a basis for the w but adding in these additional basis vectors si K plus one.
Birne Binegar: UK plus one dot dot dot C and D n.
Birne Binegar: These factors in total our basis for the entire vector stage are and, and so I can decompose an arbitrary vector into some of this form.
Birne Binegar: notice that this piece.
Birne Binegar: This piece here is written it's a linear combination is the basic factors are w so this piece lives in w this piece lipson w perpendicular because it's a linear combination of basis factors for this sweepstakes, and so there is your there is exactly.
Birne Binegar: decomposition of the vector D.
Birne Binegar: b equals C one one plus.
Birne Binegar: Okay, plus C plus one.
Birne Binegar: K plus one plus.
Birne Binegar: Plus C and D n.
Birne Binegar: take these factors call that a w.
Birne Binegar: And then call these factors they're all linear combinations of basis vectors for this subspace call that be perpendicular and then that you're orthogonal decomposition.
Birne Binegar: So let's see how much time do I have well i'm at a time okay so i'll stop here today i'll do some explicit computations on Monday, as a way of reviewing it and then also on Monday I make up a little bit about.
Birne Binegar: Finding orthogonal basis and with that we'll finish up the course and we'll start preparing for the final exam after that.
Birne Binegar: Also, be aware that.
Birne Binegar: homework set nine on web assign is the last problem set for the course try to have that done by Monday, so that we can use next week is review and, in particular.
Birne Binegar: If you've done those problems and I can try to help you out with any difficulties you have with finding eigenvectors eigenvalues or diagonal lighting matrices.
Birne Binegar: Okay, so with that have a good weekend and I will see you on Monday.