Birne Binegar: Next week is dead weeks and i'm going to devote Wednesday and Friday of next week, just to prepare you for the final exam and.
Birne Binegar: In between now and next Wednesday, though, we have one last topic to consider and that's notion of orthogonal D compositions so i'm going to introduce this today.
Birne Binegar: And it starts off pretty easy largest speak about how I take a vector and I break it down into a part that's running in one direction, and another part that running perpendicular to it.
Birne Binegar: And then the idea is going to be to extend that geometric picture, where instead of thinking about projection onto another vector we actually project onto a substation.
Birne Binegar: And then.
Birne Binegar: If we have time will will will do some examples of this.
But.
Birne Binegar: started this lecture is going to be a lot easier than what we consider that far, but as soon as like begin to talk about orthogonal projections on the subspace they are the the apparatus we've developed over the courses masters can come into play.
Birne Binegar: So let me start just introducing you to the problem suppose I have two vectors in a plane.
Birne Binegar: let's get that picture falling on screen alright so two factors in a plane, the idea that we're going to examine today is going to be a generalization of this problem.
Birne Binegar: Where, in giving my two vectors, how can I do post now decompose one of them as a sum of two vectors one part that's running parallel to a given direction be and another part that's running perpendicular to that direction.
Birne Binegar: So this will take place in the plane, but so we can use plain geometry to answer this question.
Birne Binegar: But what you see from the diagram and our method of adding vectors geometrically is what we're trying to do is decompose a into a sum of two vectors and.
Birne Binegar: Of course, if you follow the rules for adding vectors geometrically this result and vector here so some of this vector plus this vector so you place the.
Birne Binegar: tail of the second term at the tip of the first term and then you get that vector some.
Birne Binegar: So in this situation.
Birne Binegar: We have some terminology.
Birne Binegar: We say that.
Birne Binegar: This decomposition splitting a up into two parts it's you know domino the composition, we call.
Birne Binegar: A subscribe, be the specter here that's the vector that.
Birne Binegar: The part of a that's running in the direction of be the other component a perpendicular is a component of a that's perpendicular to be and.
Birne Binegar: Another other terminology and get one sometimes apply to the situation it's a call that part of a that's running in the direction of be.
Birne Binegar: We call that the projection of the vector a onto the vector be because if you imagine flashlight held off long distance away shining down directly onto the vector be then effectively.
Birne Binegar: This part of be as a part of aid well it's the shadow with the vector a in the direction of be so we often call these orthogonal projections.
Birne Binegar: in the sense that, when we project directly down perpendicular to be effectively a component that we're sinking here, part of a this parallel to be is exactly the shadow of a on the vector be, hence the nomenclature orthogonal projection.
Birne Binegar: Okay, so let's look at this situation and see what.
Birne Binegar: gear, I have just drawing pictures, but how would we actually compute the things.
Birne Binegar: Using linear algebra How would we have given up order list of numbers, or to order list of numbers, how would, I find the corresponding vectors a subscript be part of a that's running parallel to be in the part that's running perpendicular to be.
Birne Binegar: well.
Birne Binegar: from high school geometry.
Birne Binegar: where you have a right triangle.
Birne Binegar: With an angle feta than the adjacent side you take the link to the high pot news and the link that the adjacent ciders the link to the high pot news times the cosine of the angle in between, and the opposite side is, of course, the link a high pot news time sinus the angle until.
Birne Binegar: Well, I don't have this guy yet as a vector I look at the geometry this situation and see that the length.
Birne Binegar: Of this projection since it's the adjacent side of a triangle angle theta it's just going to be the link to that high pot news, which is the link up the original vector a time the cosine of the angle.
Birne Binegar: And the perpendicular part is well similar except we use the sign because it's the opposite side.
Birne Binegar: All right now.
Birne Binegar: i'm still working sort of geometrically but now will translate into thinking of vectors as ordered list of numbers.
Birne Binegar: One utility of the dark product, even when applied to n dimensional vectors is it allows us to determine links and angles.
Birne Binegar: inside a vector space that and not necessarily three dimensional space or two dimensional space we can find links and angles, in a vector space of arbitrary dimension in.
Birne Binegar: And in fact we have this formula that lakes.
Birne Binegar: That products carried out in our end with this geometric information, the dot product of an n dimensional vector with another in dimensional vector it's going to be.
Birne Binegar: The length of the first director link of the second time project still times to cosine of the angle in between this formula act like you what you prove it, as you.
Birne Binegar: start off working in an n dimensional space, but then you restrict attention to the plane that these two vectors generate now you're working in a plane, and you can use the two dimensional formula for the dot product.
Birne Binegar: Alright, so.
Birne Binegar: what's on the right hand side here i've expressed them geometrically but they're all computer in terms of the dot products, so the length of a vector is just the square root of the dot product with itself.
Birne Binegar: And the angle well.
Birne Binegar: You can take that formula, and you can solve for the angle what you're going to get is cosine of the angle between two vectors it's just going to equal the dot product.
Birne Binegar: divided by the corresponding links.
Birne Binegar: And then surely in terms of.
Birne Binegar: The dot product up to or list of numbers, we can see that this is still keep in the product location, but then this is divided by the square root of a dotted with a and the square root happy with D so.
Birne Binegar: In terms of this dot products that's defined on n dimensional vectors.
Birne Binegar: We can still calculate links and angles, and all we have to do is compute these.
Birne Binegar: These sort of dot products, while we just some appear wife products of each of the components of the vectors and be.
Birne Binegar: Okay, so.
Birne Binegar: So this situation then calculation Lee.
Birne Binegar: reduces this.
Birne Binegar: This link to this vector here is the link to the high pot news times the cosine of the angle in between, but then we can use dot products to compute both those guys and.
Birne Binegar: You can see how that works out i'm substituting for cosine data ad.
Birne Binegar: Using right hand side there.
Birne Binegar: And now i've got an expression for this link that I can compute purely in terms of the dot products have ordered list of numbers.
Birne Binegar: I want more than just the link to the spectrum, though, I want to get its direction as well, but in some sense we already know the direction.
Birne Binegar: The link to the this vector in the direction of B is, of course, heading in the direction of be.
Birne Binegar: So let's represent that direction, not by be itself, but by a unit vector vector of link one that's going in the same direction is be.
Birne Binegar: When you get that if you take your original factor you divide by its length that effectively produces a vector in the same direction, but who's linked is equal to one.
Birne Binegar: Alright, so the way i'm going to get this vector ad I know it heads in this direction and I know it's linked this direction is given by unit vector So in fact.
Birne Binegar: This sector is going to have this part has leaked one, so this vector is going to have the same blank as the link to the unit and it's heading in the right direction as well, so, in fact, this is the vector.
Birne Binegar: This vector eisa be is given by this, and now we'll just fill in our results so let's see.
Birne Binegar: The length of this component, we saw was.
Birne Binegar: Previously previous slide the link to that component.
Birne Binegar: Is.
Birne Binegar: A be its link.
Birne Binegar: was determined to be sorry.
Birne Binegar: A dotted would be divided by.
Birne Binegar: The magnitude or length of beef and then.
Birne Binegar: So i'll substitute in there and then for the unit vector in the direction of Sub to be divided by two to be.
Birne Binegar: Then I can consolidate the nominate a little bit better, but then that link squared is just the duck product deals itself and so now i've got a nicer clothes formula it produces exactly this vector to get that actor.
Birne Binegar: I take the Doc product of a would be I divide by the length of B squared and multiply by the direction of be.
Birne Binegar: Alright, so what this geometry, combined with our.
Birne Binegar: product and our end does is it gives us.
Birne Binegar: A nice clothes formulas for each component of this orthogonal decomposition.
Birne Binegar: When i've just shown you is that the part that's running parallel to be is calculable as this product and add buy to buy that product be with itself times the vector be.
Birne Binegar: To get the perpendicular pump component well all I have to do is I computed ad take a beat to the opposite side and then you get an expression for the perpendicular component as well.
Birne Binegar: So this is how.
Birne Binegar: The orthogonal the competition.
Birne Binegar: Of a vector a let's say this vector a live, not necessarily in the plane or in a three dimensional space an arbitrary vector in an n dimensional space and salty orthogonal typical with respect to another vector.
Birne Binegar: simply call it a direction be inside our end, so now we can do that same with all the composition, we can do it in an n dimensional space and i'll be just will will use the the product, but not the one.
Birne Binegar: Where.
Birne Binegar: A dot P is equal to link to a time link to be time to co sign of data rather what were you using as the dot product, where we.
Birne Binegar: Take each component of a multiply it by the corresponding component of being at things up.
Birne Binegar: Alright, so let's do an example.
Birne Binegar: there's two two vectors there in three dimensional space and what i'm gonna do is i'm going to explicitly COMP compute the decomposition of a into a part that's running parallel to this vector and apart that's perpendicular to that vector.
Birne Binegar: will use the formula we just derived so we compute this vector so you take two products in the numerator and denominator and then you use that as a scale or factor that multiplies the vector be and what you end up with in this example.
Birne Binegar: Is vector one half minus one half zero.
Birne Binegar: And then.
Birne Binegar: They get the other component of the decomposition of pay, with respect to the direction of be all you do is you ship track the the part that running parallel to be you subtract it from a net leaves you with the other perpendicular component alright, so they are explicitly.
Birne Binegar: Is.
Birne Binegar: The computation of.
Birne Binegar: The or the penalty composition of a given vector a with respect to a particular direction.
Birne Binegar: All right now that's all pretty simple and.
Birne Binegar: Everything was was pulled back to simple plain geometry.
Birne Binegar: And are two.
Birne Binegar: different versions of the dot product.
Birne Binegar: The product a.be.
Birne Binegar: can be thought of let's say as a one B one.
Birne Binegar: plus a in the end, so that's thinking about it, if we're vectors are ordered list of numbers.
Birne Binegar: Or we had it's the length of a.
Birne Binegar: Time the length of the.
Birne Binegar: times the cosine of the angle between A and B, and this is when we think of vectors.
Birne Binegar: As directed line segment.
Birne Binegar: The more geometrically.
Birne Binegar: And was by working with these two interpretations of the product that we actually achieve that the orthogonal II composition.
Birne Binegar: Okay, so.
Birne Binegar: The next task.
Birne Binegar: is to generalize these ideas to a different situation now, instead of will look at a vector V, but instead of looking at projection along a particular.
Birne Binegar: vector we're going to look at its projection on a particular subspace so i've drawn this picture in three dimensions, but you do imagine it and being an n dimensional situation where you have a given vector and a subspace.
Birne Binegar: of our end so that's going to be some hyper plane that passes through the origin and what i'm going to want to do is take a given vector.
Birne Binegar: find a part of the vector V that lives inside that subspace and then find another component to be that's running perpendicular to this space and the idea is now to find.
Birne Binegar: A decomposition of the vector V, with respect to the subspace what we're going to want to do with represent D as.
Birne Binegar: A part that lives into subspace plus apart that somehow perpendicular to that subspace.
Birne Binegar: And then that's going to be what we call the orthogonal decomposition of the vector V, with respect to a substate Let me give you a.
Birne Binegar: An application of this this sort of stuff.
Birne Binegar: suppose you have an air foil or something like this and, and you have an airstream hitting that air foil what's relevant.
Birne Binegar: The forces actually exerted on on this air foil it's going to depend not on this direction, per se, rather the force exerted on the surface here.
Birne Binegar: it's going to be a normal force, and so, if you have a force acting on an air foil what you want to do is throw away the part that running parallel the actual force on the surface is that is the normal force, and so you want to achieve up a orthogonal decomposition.
Birne Binegar: Your your i'm just drawing sort of the tangent surface to an air foil you want to you want to calculate the force that this thing actually receives is going to be the perpendicular the component of the force it's actually perpendicular to that surface.
Birne Binegar: But.
Birne Binegar: These notions.
Birne Binegar: are going to have applications beyond this simple sort of.
Birne Binegar: aerial dynamical problem.
Birne Binegar: we're treating it in general, where we simply have a subspace up an n dimensional vector space and what we want to do is D COMP decompose a vector.
Birne Binegar: So that we have a part that lives in the subspace and identify the part that's running perpendicular to that subspace.
Birne Binegar: So what i'm going to show you is that that for any vector once you have a subspace, then you can always break any vector inside our end up into two pieces one device in the subspace in one slide perpendicular to that substation will call in analogy with the.
Birne Binegar: vector case will call this the or thought mental projection of your given vector onto w and D is going to be the perpendicular component.
Birne Binegar: Okay, so let me show you how we do it.
Birne Binegar: What i'm going to do first is.
Birne Binegar: here's the subspace were interested in.
Birne Binegar: My picture up yeah.
Birne Binegar: here's our three dimensional space, maybe we have a hyper plane city in it, and then we have a vector.
Birne Binegar: vector also in your three off.
Birne Binegar: No, no point.
Birne Binegar: Okay, so.
Birne Binegar: What i'm going to do, sorry I was in the middle of a picture I guess.
Birne Binegar: Here we have a subspace w that's this flat sheet embedded in three space we have an additional vector V and what we're trying to do is think of shining a light perpendicular to this this flat surface and we're trying to calculate.
Birne Binegar: The shadow of the vector on this flat surface on this hyper hyper plane.
Birne Binegar: So what i'm going to do.
Birne Binegar: To solve this problem is i'm going to introduce a different subspace.
Birne Binegar: i'll call it w super script not perpendicular and what it is it's going to consist of all the factors inside our end that are perpendicular to every vector inside our subspace.
Birne Binegar: But what i'm going to do is this perpendicular piece is supposed to be perpendicular to w now what that really means is not perpendicular to the individual points and w but rather for any line inside w it should be perpendicular with those so.
Birne Binegar: yeah.
Birne Binegar: And so what what i'll do is simply define a new subspace I demanding that it consistent n dimensional vectors that are perpendicular to every vector that lives inside the subspace.
Birne Binegar: Alright, so.
Birne Binegar: If my subspace has a basis.
Birne Binegar: i'm looking for vectors that are perpendicular to every vector inside w so it's got to be perpendicular to every vector that nw well here's a typical way of writing a vector inside your subspace you just.
Birne Binegar: expand it in terms of its component remember what a basis is it's a set of vectors that generates the entire subspace and also has the property, the vectors are linearly independent so that these coefficients are unique.
Birne Binegar: But if you have a basis, and you can always expand the vector with respect to that basis and what i'm going to do, then, is to in order to find vectors that are perpendicular to every vector inside w it suffices to make sure.
Birne Binegar: The is perpendicular to expansions of the sort, and then we can impose this condition by.
Birne Binegar: Instead of demanding that D is perpendicular to every vector and w it will suffice, just to be perpendicular to these basis vectors.
Birne Binegar: Because if D is perpendicular here's the condition I want these perpendicular to every veteran site w there's an infinite number of factors inside w so that's not going to be possible to check directly.
Birne Binegar: However, if if D is perpendicular to just these case basis vectors then there's only take conditions that we have to check until that's what will impose.
Birne Binegar: will demand in order to find vectors that are perpendicular to every vector and w what we'll do is look for vectors that are perpendicular to each of the sectors to w.
Birne Binegar: And then that leads to a simple linear system here's the equation.
Birne Binegar: Be want to bk are the factors that generate my subspace I want.
Birne Binegar: i'm looking for factors that are perpendicular to every vector and w until they should in particular be perpendicular to each of the basis factors and then, when I set up this equation.
Birne Binegar: basis factor by basis doctrine right it as a.
Birne Binegar: As a vector equation, you see that effectively, we have a linear system you're the zero vector.
Birne Binegar: be stopped products.
Birne Binegar: I could get them the matrix product, I take the first row of this matrix I take a dot product with this first call that gives me the first component with matrix product is exactly this B one B one.
Birne Binegar: And so the condition i'm imposing on D, that is perpendicular to every basis factor is a nice homogeneous linear system, it has the form of the coefficient matrix times variable is equal to the zero vector.
Birne Binegar: And we know how to solve such systems, what you do if you look at the coefficient matrix we reduce it to reduce cholesterol and effectively, what we need to do is calculate the north face of that matrix.
Birne Binegar: But notice we're doing one thing a little bit different than what we normally do normally when I use vectors to form a matrix I normally use them as columns.
Birne Binegar: In this particular instance, though, what we're trying to do is replicate these individual equations but in the context of a linear system, and so it just so happens that norden reproduce this the proper thing to do is take the coefficient matrix using the basis vectors is rose not calls.
Birne Binegar: All right, but then.
Birne Binegar: that's how we do it, and then.
Birne Binegar: what's going to happen is that every vector in a no space this coefficient matrix is going to be a solution.
Birne Binegar: Of this matrix equation this homogeneous linear system and so.
Birne Binegar: that's what we do now, this make this subspace w perpendicular that consist of all vectors in our end that are perpendicular to every vector and five my given subspace.
Birne Binegar: The way I can calculate that is take my basis vectors for w right them so Rosa the matrix and then calculate the north face of that matrix.
Birne Binegar: Alright, so.
Birne Binegar: we'll call this subspace the orthogonal complement the w inside our end.
Birne Binegar: it's not w itself what we do is, we take the vectors that generate w and use them as robe actors and then calculate a basis for the solution, set up a linear system of that form.
Birne Binegar: Alright, so let me call that matrix a subscript gfi so I just use the sub scripts now just to remind you that what i'm creating a set of vectors that is perpendicular to every vector in w but my calculation also can require a choice of basis for that subspace.
Birne Binegar: All right now.
Birne Binegar: i'm not actually going to solve this explicitly, but I am going to use some facts that we know about coefficient matrix sees first of all, each of these basis sectors.
Birne Binegar: Well w it's a subspace are in and and so it's basis factors.
Birne Binegar: It may only the subspace itself may only be K dimensional but it's exhibited as a katie mentioned substation RN so the basis factors they're not katie dimensional vectors.
Birne Binegar: They are.
Birne Binegar: Still n dimensional vectors.
Birne Binegar: It just it by taking linear combinations of these guys I get all these guys.
Birne Binegar: Alright, so we look at this matrix we definitely have K rose, one for each basis vector for w, but we are going to have n columns, and so this coefficient matrix the coefficient matrix of a linear system that we're trying to solve it's going to have exactly in calls.
Birne Binegar: But let's divide the columns of this matrix up into two families.
Birne Binegar: You could take this matrix and you can reduce it to row Echelon form.
Birne Binegar: When once you've done that there'll be columns that happen it's the columns that don't have pits.
Birne Binegar: The cons and habits well they they give you a basis for the column space or equipment Lee they give you a basis for the road space, the number of them it's going to be the dimension of the row stage or did mention up calm space that common dimension is called the rank of day.
Birne Binegar: And then you have the dimension of an old space that so these.
Birne Binegar: You get as many.
Birne Binegar: basis vectors here, as you have columns with pivots in a row Echelon form of this matrix this here is equal to the number of free parameters, you get when you solve the corresponding linear system it's going to be equal to number of columns of outfits and so.
Birne Binegar: The rank of this matrix.
Birne Binegar: Well, we already know what that is.
Birne Binegar: it's the number of.
Birne Binegar: linearly independent rogue actors it's enough number of linearly independent vectors it takes to generate the row space well because these.
Birne Binegar: Are basis vectors they're known to be linearly independent and so these guys are exactly a basis for the row space and so.
Birne Binegar: Since we have a basis we count the number of basis factors that gives us dimension of the row space or the rank of the corresponding matrix that with as many basis vectors, as we have that's going to give us the rank of that matrix and then we have the dimension of no space.
Birne Binegar: added, this is the space that we're interested in that no space here.
Birne Binegar: is exactly.
Birne Binegar: That orthogonal compliment of the original.
Birne Binegar: subspace w this no spaces exactly all the vectors inside our end that are perpendicular to every doctor in the subspace w and what I show you with this little.
Birne Binegar: This little calculation is that the dimension of that no space is going to be exactly N minus K just move this to the opposite side and then you'd see that, as you mentioned there's no space, it has to be.
Birne Binegar: The total number of columns you have here and minus the number of rows you have.
Birne Binegar: Alright, so.
Birne Binegar: We know how to calculate no spaces, we take the matrix We really do should we do show Echelon, or we convert back to the equation separating three parameters from fixed parameters and we expand in terms of.
Birne Binegar: The three variables and you can then we can write down, but we could not only write down the solution we can also find a basis for this space.
Birne Binegar: But what this little calculation tells us, though, is the number of free parameters are the number of basis vectors we're going to have for this North face it's going to be N minus K, so let me suppose that i've.
Birne Binegar: solved the linear system.
Birne Binegar: Let me put full glory in here a job you see so w perpendicular is the solution set.
Birne Binegar: This linear system, and what I argue here is the solution set.
Birne Binegar: And minus K dimensional.
Birne Binegar: And so, because every subspace has a basis.
Birne Binegar: What I know is that once I sold this linear system explicitly there's going to be exactly N minus K basis vectors.
Birne Binegar: Instead of leaving them from one up to N minus K i'm going to start my counting for this basis starting off at K plus one and going up to the end that and.
Birne Binegar: So the total number of vectors in this list here is still going to be N minus K, because I start my counting at K plus one, rather than at one.
Birne Binegar: The reason for doing that it's like you take the basis for w and I could have joined to that the basis for this perpendicular subspace.
Birne Binegar: So here.
Birne Binegar: Sorry, I don't want to write on the wall here.
Birne Binegar: The one to bk.
Birne Binegar: Is a given basis for are given subspace and then the additional vectors the K plus one.
Birne Binegar: To be in.
Birne Binegar: that's how we're denoting.
Birne Binegar: The basis.
Birne Binegar: For the solution set.
Birne Binegar: A w B equals zero and that solution set is exactly.
Birne Binegar: What perpendicular subspace.
Birne Binegar: What i'm going to show you is that, in fact, when I take these factors and I joined them to the original basis for w that back I get a basis for the entire vector space in which everything lives are in.
Birne Binegar: The reason why that's true is.
Birne Binegar: well.
Birne Binegar: it's based on the fact that if you have an linearly independent vectors in an n dimensional space, you have a basis for that space.
Birne Binegar: If you have in linearly independent vectors you can reach every point in your n dimensional space if.
Birne Binegar: you're your reason why the span of the of a given list of factors.
Birne Binegar: If you if you work in an n dimensional space, you have in independent directions to fill up that space and that's that's why.
Birne Binegar: These factors would generate all of our end so long as we knew their linearly independent and they're truly independent directions, as some of those doctors were parallel, then I wouldn't have enough directions to fill up the entire space, so what I want to show.
Birne Binegar: Is that.
Birne Binegar: If I take the original basis and the basis for the solution set for this purpose or the basis for this perpendicular complement.
Birne Binegar: By combining those two bases together, I get a basis for our end now, I got an independent and vectors inside are and, if I can show that they're linearly independent, then I know I actually have a basis.
Birne Binegar: And so, all we have to do.
Birne Binegar: Is check to make sure that the vectors be one to be n that entire list of factors that they're linearly independent.
Birne Binegar: So let's suppose we had a dependence relation, there was a linear combination of these vectors that produces the zero vector.
Birne Binegar: If I can do that without taking all these coefficients separately equal to zero, then they would not be linearly independent.
Birne Binegar: But if the only way I can satisfy this equation is by separately setting see one is equal to zero, see to is equal to zero, all the way up the CNS equal to zero, then i've shown that there are linearly independent and so there'll be a basis for the entire vector space.
Birne Binegar: Alright, so what i'll do is i'll take this condition and i'll split it right between the basis vectors coming from w and the basis vectors coming from the perpendicular to w.
Birne Binegar: i'll just take these last N minus K terms and i'm moving them over to the right hand side of the equation.
Birne Binegar: Alright, so let's take these factors see one do one of two ck bk that's a linear combination of a basis for w so the vector on the right hand side is going to be back to that lives inside the sub state w.
Birne Binegar: On the left hand side, we have a linear combination of patients vectors for w perpendicular, and so the right hand side it's going to be a vector that lives inside w perpendicular.
Birne Binegar: And then, this condition which is equivalent to this condition becomes that this factor inside w has to equal the negative of this vector V to that lives in w perpendicular.
Birne Binegar: Let me keep that condition up so.
Birne Binegar: We said.
Birne Binegar: we're checking.
Birne Binegar: For linear independence.
Birne Binegar: that's why I said see one, the one that that that plus C hey.
Birne Binegar: pk plus C K plus one be K plus one plus dot dot dot plus C and yeah.
Birne Binegar: This is what i'm calling the one, and this is what i'm calling.
Birne Binegar: The to this live is in the perpendicular subspace because.
Birne Binegar: it's linear combination basis vectors that's up space and this guy lives inside w and what we've seen is that the one has to equal to negative that be too.
Birne Binegar: And now, what i'll show you is that, in fact, this condition is going to imply that all the coefficient see one two ck and si K plus one of the end they separately, have the equal zero.
Birne Binegar: Well, you did, that is, I take this equation.
Birne Binegar: And I take it would be one so i'll have the one God it would be one it's going to equal the.
Birne Binegar: Sorry, the one with the to.
Birne Binegar: The negative that because I have a negative sign here.
Birne Binegar: All right now.
Birne Binegar: Remember.
Birne Binegar: That every vector inside the perpendicular subspace is perpendicular to every vector inside w.
Birne Binegar: This guy lives inside w this guy listen to like w this guy lives inside the space of vectors that is perpendicular to every vector and w so.
Birne Binegar: that's got an equal zero because.
Birne Binegar: Every vector.
Birne Binegar: In the perpendicular subspace is perpendicular.
Birne Binegar: To every vector in Devon yeah.
Birne Binegar: But then I have on the right hand side, I have the length of, of the one dot product to be one of itself just going to produce the link squared this link squared is equal to zero, that means that the one has to be the zero vector.
Birne Binegar: Alright, so.
Birne Binegar: So, in fact, if this equation holds then in fact.
Birne Binegar: The vector V, one has to be the zero vector.
Birne Binegar: Well, this implies the one is equal to zero vector and that in turn is going to apply, because these.
Birne Binegar: Are the factors, do you want to decay our basis factors are linearly independent so a linear combination of these factors can only equal to zero vector when all the coefficients or equal to zero, so you get an easy one, is equal to zero t equals zero up to si K after equals zero.
Birne Binegar: But the other consequences that I produced it be one is equal to zero well be two if that if the dependence relation hold it has to equal to negative, of the one, and so the two is also equal to zero it's minus one.
Birne Binegar: minus zero vector, which is the zero vector again.
Birne Binegar: The vector V2 is a linear combination of linearly independent vectors if this linear combination equals zero then all the coefficients have to equal zero.
Birne Binegar: And so the vanishing because the two also has to equal your vector and because it was constructed as a linear combination of linearly independent factors all those coefficients.
Birne Binegar: After separately vanish.
Birne Binegar: All right, and now i've got the statement that I want.
Birne Binegar: You only way a linear combination of this.
Birne Binegar: basis for our in a basis that was constructed by first finding a basis for are given suspects w second basis that was found by.
Birne Binegar: Using the basis vectors of w two form a coefficient matrix and then solving the corresponding linear system that gave us some additional vectors.
Birne Binegar: What the theorem says that when I combine those two sets of vectors I get a set of and linearly independent vectors insight are in, and so a basis school and what i've shown.
Birne Binegar: arguing it's somewhat lengthily is that.
Birne Binegar: This equation implies that these two vectors.
Birne Binegar: live in different substation so being one is perpendicular to be two and then.
Birne Binegar: Taking it taking that product and the ones itself has to equal.
Birne Binegar: Taking that product for both sides will be one we get the link to be one squared has two equals zero so B minus is the zero vector but then the same equation, this is the zero vector implies that this is the zero vector.
Birne Binegar: These two vectors cannot equal zero unless all these coefficients are equal to zero, because the basis vectors we used to expand them are linearly independent.
Birne Binegar: Alright, so now that's the argument.
Birne Binegar: and
Birne Binegar: Now let me return to the original problem.
Birne Binegar: Alright, so what.
Birne Binegar: We have so far Now let me check my time here.
Birne Binegar: Okay six minutes okay so.
Birne Binegar: The goal here.
Birne Binegar: is to.
Birne Binegar: decompose.
Birne Binegar: A vector V, with respect to.
Birne Binegar: A subspace.
Birne Binegar: w.
Birne Binegar: And I haven't done that yet, but we're pretty close to it right now.
Birne Binegar: What I did.
Birne Binegar: Is I just completed a digression.
Birne Binegar: They told me that.
Birne Binegar: I can find.
Birne Binegar: A basis for our and.
Birne Binegar: that's the.
Birne Binegar: Whole basis here by.
Birne Binegar: First of all.
Birne Binegar: Finding a basis.
Birne Binegar: For the subspace debbie.
Birne Binegar: that's going to be these factors be one.
Birne Binegar: To be.
Birne Binegar: The bk.
Birne Binegar: Then, once I find that basis.
Birne Binegar: For for w I use.
Birne Binegar: These vectors V1 do bk.
Birne Binegar: To find.
Birne Binegar: indirectly.
Birne Binegar: A basis for.
Birne Binegar: Not w but the set of vectors that's perpendicular to every gun sector and w.
Birne Binegar: Wait.
Birne Binegar: What this subspace is it it's the no space.
Birne Binegar: Of the matrix I get.
Birne Binegar: Using my given datasets row vectors.
Birne Binegar: Okay.
Birne Binegar: And then, what we found is that the dimension of this no space.
Birne Binegar: it's going to be N minus K dimensional so we write.
Birne Binegar: The K plus one up to bk.
Birne Binegar: As.
Birne Binegar: A basis for that no space or or the perpendicular compliment.
Birne Binegar: The theorem then says that when I combine this set of vectors my basis for w with this set of vectors which.
Birne Binegar: Once we know the vectors V1 these vectors can be calculated by solving a linear system will get a new set of vectors when I join these together, I get a basis for the entire vector space are in.
Birne Binegar: And now I can show you.
Birne Binegar: Precisely how we, we are going to.
Birne Binegar: carry out the decomposition respect to a suffix.
Birne Binegar: What we do yes we've got this basis, do you want.
Birne Binegar: To be K at our basis.
Birne Binegar: For w and then on the joining to that the factors, do you want bk plus one, up to the end.
Birne Binegar: Those are basis.
Birne Binegar: For the perpendicular complemented w.
Birne Binegar: and together we get a basis.
Birne Binegar: For the entire vector space in which both the sub spaces live.
Birne Binegar: Because these guys are a basis for the entire vector space every Dr.
Birne Binegar: d inside our n.
Birne Binegar: can be expressed as be equal some coefficient times first basis vector dot dot dot plus some coefficient times the last.
Birne Binegar: place it factor that lives in w but then we have some additional terms si K plus one times the K plus one now i'm working with the components along nice directions.
Birne Binegar: And last one B, C and.
Birne Binegar: D and.
Birne Binegar: And the point is.
Birne Binegar: i've got these guys.
Birne Binegar: As soon as I group these guys differently i'll get exactly the composition, I want.
Birne Binegar: These first K terms here they're all linear combination of basis vectors for w so let's call that part the subscript w because this vector as a linear combination of basis vectors for w is going to live inside w.
Birne Binegar: Meanwhile.
Birne Binegar: These last N minus K terms they're all multiplying basis vectors for the perpendicular subspace w subscript perk and that will call those hold that vector.
Birne Binegar: The perp and now i've got a decomposition.
Birne Binegar: of an arbitrary vector insight aren't n into a piece that lives inside w plus a piece that's perpendicular to w until that's that's the desire to thumbnail composition, we can now.
Birne Binegar: Take any vector and site RN and split it up into two components heart that lives in or subspace and apart that perpendicular to that subspace.
Birne Binegar: When I run out of time I have, so what i'll do on on Monday so i'll i'll do explicit examples of this, and you'll you'll see the full calculation and then I think i'll talk a little bit about.
Birne Binegar: orthogonal bases and then, then I think we're we're done with the material for the course and we'll just concentrate on reviewing after that.
Birne Binegar: Okay, so have a good weekend homework set nine is going, I guess, going to be the last homework is That said, I assign try to have that done by Monday, so we, we can talk about any typical challenges, you have with finding eigenvectors or dying analyzing matrices.
Birne Binegar: yeah.
Birne Binegar: zero.
Birne Binegar: vector also.
Birne Binegar: also appeared twice.
Birne Binegar: Not necessarily so.
Birne Binegar: So you can happen.
Birne Binegar: You always have at least one eigenvector one linearly independent I get better for every eigenvalue but you could have more than one.
Birne Binegar: So that's why we make a distinction between the algebraic multiplicity and the geometric multiplicity algebraic multiplicity is just tell you how many factors, or how many times that route is repeated in.
Birne Binegar: But the geometric multiplicity.
Birne Binegar: They can't be more than the algebraic multiplicity, but it could be less than that, so you could have when you diagonal if you could have two zero entries, but it doesn't necessarily mean that you have to literally independent eigenvectors.
Birne Binegar: here's a simple example 0100 so when you compute the characteristic polynomial this you're going to get so P, a Lambda is going to be the determinant.
Birne Binegar: minus zero minus Lambda 100 minus Lambda that's going to be Lambda squared.
Birne Binegar: Just Lambda squared and so because the Cross diagonal product zero so you're gonna have only one route and you think of this as Lambda minus zero squared so the the eigenvalues zero has algebraic multiplicity to but as soon as you try to calculate the.
Birne Binegar: The.
Birne Binegar: Zero against space, but you're going to find is that well it's already in relational on foreign how many free parameters, do we have.
Birne Binegar: You only have one call that a pip so you know you see that you can have it most one linearly independent eigenvalue because the system that you have to solve a minus Lambda times I.
Birne Binegar: equals zero to get the corresponding I get space, it only going to have one free parameter so it's only going to be one dimensional so I have two factors here until yeah but I only i'm only getting a one dimensional solution set.
there's.
Birne Binegar: zero.
Birne Binegar: yeah so so the so you have at least two linearly independent eigenvectors what can happen, though, is, you could have to I can vectors that are linearly independent from one another, to the point in different directions, but they have the same eigenvalue that can happen.
yeah.
Birne Binegar: yeah.
Birne Binegar: Let me, let me just stop here for a second here.
Birne Binegar: Stop the slide.