Birne Binegar: Okay um so today we're going to interview, and maybe we'll be finishing up today the topics that are likely to be covered, on the.
Birne Binegar: Second exam we'll see how far we get want to get as far as being able to compose linear transformation, but we may not actually reach this point, I have a fair amount to talk about i'll start by.
Birne Binegar: reminding you what a linear transformation is and how the related to matrices.
Birne Binegar: Then justice we attached a bunch of Sub spaces to him a matrix i'm also going to attach two different substances to a linear transformation that connection between linear transformations and matrices that then can allow me to find bases for these substances remember so.
Birne Binegar: substances are the interesting sets of vectors that occur in linear algebra and bases are essential linear algebra because that's what you need to provide coordinate for points of live.
Jaxon Castillo: Inside subspace.
Birne Binegar: So we'll figure out a way once like attached sub spaces to a linear transformation will use the connection with matrices to figure out a way of actually nailing down.
Birne Binegar: A bases for these two sub faces and that will give us a way of coordinating the points that live in those spaces and simultaneously as soon as you have a basis.
Birne Binegar: And to delete that meant means a set of independent directions that is capable of generating the whole subspace well by counting those independent directions, you get the dimension of the subspace and what we'll find is that.
Birne Binegar: The dimensions of these two sub spaces it's kind of a zero sum game they have the range of a linear transformation it's dimension, plus dimension of this other space, they have to add up to the dimension of the domain.
Birne Binegar: of your linear transformation so that's that's really easy to see once we know make these connections and then finally.
Birne Binegar: What may get delayed until friday's lecture is my talking about.
Birne Binegar: Taking to linear transformation and composing them if you would functions there, so you can go from one space to another to another vector space.
Birne Binegar: Alright, so let me start by just reviewing what we covered last time linear transformation.
Birne Binegar: These are the basic kind of functions that are occurring in linear algebra they take you from one vector space to another vector state.
Birne Binegar: But in addition, they have to behave well for respect to our fundamental operations, so we have a compatibility requirement for scale or multiplication.
Birne Binegar: Whether we do scale multiplication before or after we applied the function T, we should get the same result and, similarly, whether we add vectors before we apply the function or after we apply the function, we should still get the same result, so these.
Birne Binegar: We have a linear transformation we're only looking at functions that have this this property.
Birne Binegar: Right and then what we saw last time, is that we have a function satisfying these two properties, then one way to make it such a fun fitness to take a matrix and what you do is you regard matrix multiplication is giving you this application.
Birne Binegar: Right, so I guess I can write on the black or here.
Birne Binegar: So.
Birne Binegar: If this is going to be an n by n matrix well in order for this matrix multiplication to be defined it's got to be.
Birne Binegar: A column vector so number of columns and the first doctor has to equal number of rows and the second factor, but then what such a matrix multiplication is kind of kick out.
Birne Binegar: So that's the match, you need, and then the dimension of the result, the product is going to be an end by one so effectively this matrix multiplication is taking you from empty mentioned vectors to n dimensional vectors, so it is defining a function between these two vectors spaces.
Birne Binegar: And what happens is that the financing you Curie using matrix multiplication to define it it's naturally satisfies these two properties.
Birne Binegar: Basically, because scale or multiplication commutes with matrix multiplication and matrix multiplication distribute over sons.
Birne Binegar: Alright, so that's one connection between matrices and linear transformation, we can always make a linear transformation from a matrix.
Birne Binegar: But what we saw last time, if we can go the opposite way as well, given any function between rm and our end that satisfies these two properties there's a matrix that exactly replicates the action of that function.
Birne Binegar: Though the statement is that you have a function between two vector spaces that's actually linear transformation, then the way it acts on a vector can be replicated by certain matrix multiplication.
Birne Binegar: And in fact what we saw last time was that how to actually construct this matrix what we do is we look at how he acts on special vectors the standard basis vectors for are in so, for example, one.
Birne Binegar: Is 100 and then the last standard basis vector is the vector 100 all zeros until we reach that last slot So these are your natural coordinate vectors or are in.
Birne Binegar: And what we do to get this matrix is we examine how the function X on our standard coordinate vectors use those results and rewrite them as a columns of the matrix now every time I act on one of these factors, they live inside.
Birne Binegar: Our m, and so this function is going to act on that's going to take them to a new vector and i'm now going to write us a call well that's going to make this matrix a subscription to.
Birne Binegar: The matrix attached to the given linear transformation it's going to be a matrix with n rose because you're writing these results vertically and a total of em calls as many columns that we have basis vectors for the domain.
Birne Binegar: All right, and then it turns out that when we analyze linear transformations the most direct way to get to a calculation of setting is to create this numerical matrix and that's exactly what's done.
Birne Binegar: Alright, so what i'm going to do next, then, is these major changes are going to be essential to analyzing these linear transformation, and so we start today.
Birne Binegar: and looking at substations attached to a linear transformation and then we'll find bases, but in fact our calculations are going to flow through this particular matrix that we can always construct once we were given a linear transformation.
Birne Binegar: Okay, so.
Birne Binegar: For matrices we had three substances naturally associated with them, we could take the road vectors of the matrix and take all possible linear combinations that will.
Birne Binegar: create a subspace of I guess aren't em if we have an invite him matrix each row would have exactly entries, and so the span of the road vectors is going to be some subspace of our em.
Birne Binegar: We also have a column space was exactly sitting inside.
Birne Binegar: Or am the column space well the columns each column it's going to have as many entries, as we have rows of this column bankers, I can be n dimensional vectors and I take all possible in your combinations I get a substation aren't in, and then we had one more.
Birne Binegar: subspace that we can attach to a given matrix rather than looking at the span of sets of vectors we look at the solution set of the corresponding homogeneous system.
Birne Binegar: And for this multiplication to be fine this column vector know die race it, I guess, I did this column vector has to live inside or m so that technically a substation am.
Birne Binegar: Alright, so that's.
Birne Binegar: When I introduced in sub spaces for major season, these are the ones we looked at we're going to do a corresponding.
Birne Binegar: analogous discussion for linear transformations except i'm not going to.
Birne Binegar: introduce the subspace is attached to a linear transformation, by referring to the matrix that correspond to it rather i'm going to find these you find these in a more no more function centric way.
Birne Binegar: So let's start.
Birne Binegar: With the first updates will look at it's called a range.
Birne Binegar: Of the linear transformation, let me first define it it's a.
Birne Binegar: it's oriented or its definition is based on function theory so you're on thinking really abstract sort of high school function theory, you have to set a B and you have a function.
Birne Binegar: That takes elements of a two point F of a that are sitting inside another set.
Birne Binegar: Now, and function theory.
Birne Binegar: there's a notion of of the image of a function, and what that is.
Birne Binegar: it's a subset of all points in your co domain your target space that can be reached by applying this function.
Birne Binegar: So, for example, that's called the image, so the image.
Birne Binegar: F, this is going to be every point.
Birne Binegar: In your co domain that can be reached by applying the function so mathematically that condition is you're looking for points in the co domain, such that B is equal to F of a for some.
Birne Binegar: element, a that lives inside his domain.
Birne Binegar: Now you had F.
Birne Binegar: That sent X to say X squared.
Birne Binegar: The image of this function what's it going to be well, this is, this is a function from on the real line.
Birne Binegar: They taking real numbers to real numbers and if you X gets mapped to.
Birne Binegar: X squared well what in fact happens is you never reach the.
Birne Binegar: Negative numbers, by applying this function right because the square if a real numbers always non negative, it could be zero X itself is zero but it's always other than that it's always positive.
Birne Binegar: Until this map doesn't completely covered the i'm The co domain, it only gets about half of it only reaches the the positive numbers, and so the image set is not generally the entire co domain it's usually a smaller set that sits inside it.
Birne Binegar: But since it's a subset of a larger set that you think there might be an analogous construction which in linear algebra the important subset or sub spaces.
Birne Binegar: And so that's exactly exactly what we do to define this range, we take the functional idea.
Birne Binegar: Finding all points that can be reached by applying a function and we put it in the linear algebra study we're now our function is mapping between two vector spaces.
Birne Binegar: So what will the analog of this construction would be a subset of the of the domain, such that consists of points that can be reached by applying key.
Birne Binegar: And so there's the definition mimics exactly this the range of a linear transformation, or the point in the soap target place the place you land after applying the function that can be reached by applying your function to some point in the domain.
Birne Binegar: Alright, so that sets it up as a story that sets it up as a subset.
Birne Binegar: Of well these guys are all living inside or an.
Birne Binegar: All right now that's a subset it's not necessarily have subspace so what I want to show you next is in fact this subset of our n is closed understand the multiplication close hundred vector edition, and so is the subspace.
Birne Binegar: Sub faces as in most things in linear algebra what we really require is.
Birne Binegar: A nice behavior respect to scale or multiplication and vector edition.
Birne Binegar: And so that's what we use here.
Birne Binegar: So suppose we have a linear transformation between you bet your spaces my claim is that range of tss subspace so what i'm going to have to show.
Birne Binegar: Is that if I start off with any number Lambda and a point that lives inside that subset after I do scale or multiplication i'm still inside that subset.
Birne Binegar: Or if I start off with two vectors living inside the subset and I add them together i'm still inside this subset this is closure under scale or multiplication this is closure under vector edition.
Birne Binegar: Alright, so let's see this is, these are both easy to see.
Birne Binegar: Suppose we going to show you closure under skill or multiplication first so let's choose some any point inside the range and choose any number Lambda what i'm going to show you is that the scale of multiplication of this guy by Lambda is still inside that subset.
Birne Binegar: Alright, so.
Birne Binegar: If y is in the range, well then, it had to come from some point yet well I can just change this picture, a little bit.
Birne Binegar: or m.
Birne Binegar: p.
Birne Binegar: R m so and maybe call this X.
Birne Binegar: T of X and if we start off with some point in this range, it had to come from some X over here well that's what i'm saying here, anything that lives in the range had to come from the domain.
Birne Binegar: ID applications a function key so what i'm gonna do is i'm going to multiply this equation, which.
Birne Binegar: Basically, corresponds to this fact a multiply by Lambda over here this multiply both sides, but the thing is that he is a linear transformation, so the scale or multiplication after I applied the funky cheetah X is equal to.
Birne Binegar: This has the same value which he applied to the scale of multiplication of X by Lambda but now I see that this point Lambda why lives in the range because it's the image of this particular point after after applying T.
Birne Binegar: And transition between these two lines is true because linear transformation is up has a special property that it's in variant under scale it's preserved scale multiplication.
Birne Binegar: And so the fact that Lambda why there's always in the image of a P means that.
Birne Binegar: The wells close to understand the multiplication so you have any number in every point in the range and the scale multiplication at that point in the range by Lambda remains inside that subset, and so the range is closed understanding multiplication.
Birne Binegar: Okay, so it's pretty simple argument is just restating definitions effectively.
Birne Binegar: What about vector addition same idea if we start off with two vectors that live inside the range That means they had to come from some vectors X one X two that lived inside the domain let.
Birne Binegar: Let X one be the got.
Birne Binegar: Some vector inside the domain that produces why one after applying T, this is guaranteed, because, why has to be in the range of P similarly for why to what i'll do is i'll take these two equations i'll add them.
Birne Binegar: I get this and then.
Birne Binegar: Because he has a linear transformation T F X one plus two, you have X two is the same thing as key applied to X one plus x two because linear transformation, you can delay vector addition.
Birne Binegar: Of a summit vectors you can delay that until after you apply the function that addition to laugh, do you play the function.
Birne Binegar: So that shows you that why one plus y to the sum of two vectors that live inside the domain is actually the image of a particular point in the domain, the vector X one plus x two and then that shows you that why one flesh why to also lives inside the range.
Birne Binegar: And so we've got we start off with two factors in the range there vector some still lives inside the range of that function, and so the range of T is closed under vector edition.
Birne Binegar: So to tie things together.
Birne Binegar: We simply have to observe that subspace is a subset that's closed under scaling multiplication and vector addition to slide to go I showed you that was closed understand them application, the preceding slide I showed you the closed under vector edition, and so the range of T is a subspace.
Birne Binegar: Alright, so we've got an abstract substation now, how can we do something concrete with it well.
Birne Binegar: What will bring the subspace into a calculation of seti is by displaying a basis for a way to coordinate, is it.
Birne Binegar: And so that's going to be pretty easy.
Birne Binegar: Suppose we have a linear transformation.
Birne Binegar: Well, we know how to attach a matrix to such a linear transformation, we take the standard basis vectors we apply the function T to them that will give us n dimensional vectors that we can write us call.
Birne Binegar: Putting those columns together in a single matrix we get an n by n matrix attached to that linear transformation, with a property that the action of this function on em dimensional vectors is equivalent to multiplying column vectors of dimension, am I this particular matrix.
Birne Binegar: All right, and then.
Birne Binegar: What i'll show you then is.
Birne Binegar: The range of linear trench mentioned the space of vectors that you can create by applying a linear transformation.
Birne Binegar: To one vector space and ending up in another vector state this range of tea that you think of the shaded sub specialty subset of rm that that's equivalent in fact it identical to the column space of this matrix.
Birne Binegar: and
Birne Binegar: Let me show you why that's true.
Birne Binegar: What i'm gonna do is i'm going to show this.
Birne Binegar: These two sets the range of tea and the columns basically associate matrix the way i'll show you those are exactly the same shut it shall show you that.
Birne Binegar: Every element in the range is also in the column space of the matrix and that every element of the column spaces matrix is he is in the range, so this is going to be sort of like.
Birne Binegar: The logic here.
Birne Binegar: is like.
Birne Binegar: If X is less than or equal to y and y is less than equal to X, then for both of these inequality, to be maintained.
Birne Binegar: You y is equal to X now that's true for for numbers X and y but similarly for sets if you have.
Birne Binegar: If, as a subset of B and b it's a subset of a.
This.
Birne Binegar: These two conditions tell you that the two sets are the same, so this is sort of logic will be applying to show you.
Birne Binegar: That this set is equal to this said i'll show you that every element in the range is a member of the columns base and every element of the column space is an element of the range.
Birne Binegar: And then, since the sets are subsets of one another, then they'll have to equal.
Birne Binegar: Alright, so let's do the first one let's start with.
Birne Binegar: A show you that I start with an element that lives inside the range than necessarily it also lives inside the columns basically associated matrix.
Birne Binegar: So i'll set why i'll choose a point in the range That means it comes from some X in the domain.
Birne Binegar: And now, what i'll do is i'll expand this vector with respect to the standard basis of our and so.
Birne Binegar: When your transformation is going from rm.
Birne Binegar: M and the view associated matrix is going to be.
Birne Binegar: An n by n matrix.
Birne Binegar: So i'm.
Birne Binegar: Here i'm just researching this condition I choosing a point in the range it comes from some X now what i'll do is i'll expand this guy with respect to the basis the standard basis of our end.
Birne Binegar: So i'll just expand it finds the basis vectors now i'll use properties of tea in fact that it's a linear transformation.
Birne Binegar: The specter some can be performed after we apply the linear transformation so that's preservation of.
Birne Binegar: vector addition under linear transformation and then the scale or multiplication inside the linear transformation those can be delayed until after we apply them to your.
Birne Binegar: Information and now we see that any point in the image is a linear combination of these factors to apply to a standard basis Dexter rm.
Birne Binegar: Well, those are exactly the columns of the associated matrix but then now a point in the range is displayed as a linear combination of the columns of the matrix.
Birne Binegar: Well, the set of all possible linear combinations of the column vectors of the matrix is exactly it's called space, so this displays point in the range it's also be an element of the columns based of the associated matrix and we're done.
Birne Binegar: showing this inclusion.
Birne Binegar: alright.
Birne Binegar: So let's.
Birne Binegar: let's say the opposite to region now also choose an element that lives in the column space of my matrix and i'll show you that it corresponds to an element that's also in the range that incidental to it.
Birne Binegar: Though will take us a vector and that's going to live in the span of the columns of the matrix at that we construct from our linear transformation.
Birne Binegar: Well, if it lives in the span, that means it's a linear combination of these factors until we it expands this way and.
Birne Binegar: What we'll do is remember the columns of this matrix way we construct this matrix is as.
Birne Binegar: This first column it's going to be key applied to the first standard basis vector of the domain of the linear transformation T well this column is exactly equal to that last column is exactly equal to to apply to the last standard basis factor and now.
Birne Binegar: That we're just kind of reverse the our men on the preceding slide because he is a linear transformation, the scale or multiplication can be taken inside there the argument of the linear transformation.
Birne Binegar: Of this, this occurs because linear transformations are compatible with scale or multiplication.
Birne Binegar: But they're also compatible with vector additions when I add up all these he applied to numbers time standard basis factors.
Birne Binegar: I can collapse that into a key applied to a single vector, and now it appears that w which was chosen to be an element in the column station a is also in the image of tea because w occurs as T applied to some vector inside the domain.
Birne Binegar: And that puts w inside the range of cheese and so i've effectively done this i've shown you that.
Birne Binegar: The range of tea is a subspace of the columns face.
Birne Binegar: At and i've shown you that a column space.
Birne Binegar: Of at it's a sub set.
Birne Binegar: Of the.
Birne Binegar: range of the corresponding linear transformation and then these.
Birne Binegar: These two inclusions implied that the two sets are equal.
Birne Binegar: range T is actually equal to the column space.
Birne Binegar: At.
Birne Binegar: And that's great because what does that do it actually takes us abstractly defined set.
Birne Binegar: here's a picture of it go from our end to our mo just trying to get a handle on the set of vectors in the co domain that are reachable by applying this linear transformation so it's.
Birne Binegar: defined by this sort of formal picture, but now, what we see is because that range defined abstractly is equal to the column space that is very concrete once we compute this matrix is just a numerical matrix.
Birne Binegar: And we know how to find bases for these four columns space of a matrix so that's going to give us a way of getting.
Birne Binegar: A calculation of handle on the range of tea.
Birne Binegar: This identity gives it gives us the calculation away will handle the range of tea directly, we can work with this matrix a subscript T and it's called space.
Birne Binegar: Alright, so.
Birne Binegar: that's the conclusion.
Birne Binegar: And let's see I may not sure if I do an example yet.
Birne Binegar: Well, because these are exactly the two sets they're also the exactly the two sub spaces, and so what we'll do in order to find a basis for the range of key.
Birne Binegar: off I just need to find a basis for the equivalent subspace, in fact, the same stuff states, and so I find a basis for this that will provide you with a base basis for the range of T.
Birne Binegar: Alright, so I guess I do do an example alright so here's here's the linear transformation that's going to take two dimensional vectors the three dimensional vectors.
Birne Binegar: it's defined by a functional for the appointment coordinates X one X two is getting sent to a point in our three with coordinate X one plus x two X one minus X two X one.
Birne Binegar: What we're trying to do is we'll find a basis for the range of T.
Birne Binegar: So first thing we do is we compute the matrix that corresponds to that linear transformation.
Birne Binegar: Now the domain is only two dimensional so i'll have two columns in this associated matrix the way I compute them as I apply T to the first standard basis vector.
Birne Binegar: Give me first call he applied to the second day basis vectors of domain is going to give me the second call right now notice, even though i've written these things horizontally, when I insert them in this matrix i'm writing them vertically, this has three components when when I.
Birne Binegar: When I act on.
Birne Binegar: First standard basis factor as to apply to one zero well what's that going to be it's going to be one plus zero is one one minus zero is one and simply one.
Birne Binegar: Because X one is that, first, coordinate and similarly TEE up to it's going to be T applied to 01 and that's going to be let's see it's going to be one.
Birne Binegar: minus one and zero.
Birne Binegar: So I get these three these two vectors as the image of my standard basis factors but don't write them as rows convert them the columns to get the correct matrix, and so there it is that's the matrix that will will be working with in order to analyze this.
Birne Binegar: linear transformation between our two and our three.
Birne Binegar: What we need, according to the receiving side to get a basis for the range of tea, I need to construct a basis for the column space of this matrix.
Birne Binegar: That we remind you how that's done.
Birne Binegar: find the basis for the call space.
Birne Binegar: The space of any matrix mother tongue associated to a linear transformation or not, what we do is we reduce.
Birne Binegar: A to.
Birne Binegar: A row or reduce row Echelon porn maybe i'm reading this reading show actual inform or either way it's officially got your reflection on porn, but we cannot go all the way to reduce your actual important to get a unique matrix and then we.
Birne Binegar: identify.
Birne Binegar: Where the pivots live.
Birne Binegar: In this row Echelon for one.
Birne Binegar: And then we grab.
The.
Birne Binegar: Corresponding called of the original matrix.
Regional.
Birne Binegar: matrix a and those columns that we can relate to columns of the row of July and form they have pivot those are going to be our basis factors Doc let's look at this matrix.
Birne Binegar: And let's reduce it if it's pretty easy to reduce this guy very quickly rather reduces to this matrix in reduce row Echelon on what you should see, then, is that both columns in the row Echelon form have pivot so i'll go back to the original matrix.
Birne Binegar: And i'll grab.
It.
Birne Binegar: As basis factors.
Birne Binegar: For the column space when i'll use is well corresponding to this habit in the role I show on form i'll grab this first call 111 and then.
Birne Binegar: corresponding to this second pivot I go back and I grab second call one minus one zero until that's gonna be my basis for the column stage but calling space.
Birne Binegar: Of the matrix a sub T is exactly the same thing it's the subspace range of tea and so these guys are also going to be a basis for the range of tea, however, you were i've written them as column vectors if we go back to the way I define tea.
Birne Binegar: I didn't use make your CC find it rather I just use ordered list of numbers, and so what i'll do is i'll take these columns, this is still an order list of numbers it's just that i've chosen the right that list vertically i'll think of is just a vector not as a matrix and so.
Birne Binegar: This these two column vectors would correspond to.
Birne Binegar: 111 maybe i'll even change my notation slightly using.
Birne Binegar: parentheses distinguish between a list of numbers and and.
Birne Binegar: A rectangular arrangement of numbers, but this column vector corresponds to this order list of numbers and this column vector corresponds to this order list of numbers and these guys.
Birne Binegar: are going to be a basis.
Birne Binegar: For the range of tea.
Birne Binegar: Alright, so.
Birne Binegar: This business here, where I take the column vectors not rewrite them as ordinary vectors I don't do that to confuse you i'm just.
Birne Binegar: I can't really decide whether I always stick with matrices district I vectors or I bounce back and forth yeah.
here.
This.
Birne Binegar: Better or.
This is an image would that be an.
Birne Binegar: This step the yeah so keep this picture in mind, so what the function doing is taking all the points in the domain.
Birne Binegar: and, generally speaking, it doesn't hit every point in the in the co domain, it just reach a smaller subset what we're trying to do is get it get a handle on these points that are that can be reached by applying the function T.
Birne Binegar: So what I found is that this image set here actually corresponds to the column space of the matrix, and so the way I.
Birne Binegar: i'm going to end up cord and enticing these guys is converting to the matrix picture getting a basis for that column space and then using that same basis for for this, because they the the range of.
Birne Binegar: The range of tea is identical to the column space of this matrix.
Birne Binegar: So what we do, we we think of this abstractly from this picture, but here's the calculation of setting by how we actually keep things.
Birne Binegar: Oh, this is typical in this course, and in fact I think it's really the deeper part of the course is our way of.
Birne Binegar: Taking things that aren't you find some other way, besides calculations and but.
Birne Binegar: In order to deal with them, particularly in applications, you need to take your abstract ideas and put them in a place where you can actually.
Birne Binegar: plug things into a calculator and and develop concrete solutions to problems and so that's the big theme of this course, is how do we take abstract things and bring it to a concrete calculation on setting.
Birne Binegar: that's what you're going to need to do in your applications as you leave this course.
Birne Binegar: When a linear algebra setting arises in engineering or in physics, etc, the main thing you want to get from this course as well, how do we calculate that stuff until that's that's what i'm trying to describe.
Birne Binegar: But the other thing that's happening here.
Birne Binegar: Which is just important for your applications is i'm giving you a language but which all this translation from physical settings to the concrete calculable setting all this.
Birne Binegar: All the language is consistent and, in fact, that what you really need is the language, not only to do calculations, but also to understand what's going on.
Birne Binegar: If someone analogous doing calculus.
Birne Binegar: You need to know what a derivative is, but in the end, you really don't have to take a derivative as long as you know how to mount to pass the course you only have to know how to differentiate polynomials maybe and that guarantees you would see in the course well.
Birne Binegar: Underlying yet, though in the reason why.
Birne Binegar: limits or introduced and things like that is that the idea that a fundamental physical law such as F equals ma can only be written down until you know when you're driven it is.
Birne Binegar: that's also happening in this course as well we're developing the language but which physical loss can be expressed or.
Birne Binegar: Physical situations can be expressed and then we're also not being able to sort of mimic physical situation by mathematical situations we're also providing a calculation needs to to exploit that analogy.
Birne Binegar: Okay, so let's see um.
Birne Binegar: So I guess I did this this associated matrix it really reduces to this matrix like go back to the original matrix.
Birne Binegar: And I grabbed the corresponding columns those are going to be basis vectors and so now yeah so this column vector corresponds to this three dimensional vector this this second column vector corresponds to.
Birne Binegar: A second three dimensional vector sitting in the in the domain sorry sitting in the range of tea.
Birne Binegar: Okay, so that's the first step space one, it will attach to a linear transformation.
Birne Binegar: there's another subspace that's very important it's called a kernel of the linear transformation they've given us a function between.
Birne Binegar: Two factor spaces that's compatible with scaling multiplication and vector addition all define a subset of the domain all that range.
Birne Binegar: Is a subset of the young co domain i'm now going to get attached to this thing particular subset of the domain and what it's going to be it's going to be all the points in the domain that gets sent to the zero vector inside the code domain.
Birne Binegar: Alright, so the way it's set up it's clearly.
Birne Binegar: a subset.
Birne Binegar: of our m, because every element in the kernel.
Birne Binegar: starts off inside aren't em and then we put a restriction in it that so only certain vectors inside the domain, are going to satisfy this condition, but it gives me a well defined subset.
Birne Binegar: Of the domain of my linear transformation what i'll now show you that is is that it's not only a subset but it's actually a sub state.
Birne Binegar: yeah.
Birne Binegar: The it's the numbers that that the vectors vectors that where you start off to end up landing on the zero vector inside.
Birne Binegar: yeah after applying the function, so an n dimensional vector under the action of T gets mapped to some n dimensional vector right well, one example of an n dimensional vector.
Birne Binegar: You will let's draw this vector space, a little bit, not as a circle, but more geometrically is somewhere, we got some coordinate grid and.
Birne Binegar: The point of giving set of coordinates it's going to land on what we're focusing on in the kernel or the point.
Birne Binegar: it's going to be a subset here but what's required for the Colonel if they all get mapped not to arbitrary vectors over here, but they all land on the origin.
Birne Binegar: what's this is going to actually represent the little bit later on when we start interpreting linear transformation is that these Colonel athletes represents the information that's lost by applying linear transformation so.
Birne Binegar: What happens here is the all the all the factors here that sitting in are sitting in the kernel you all get mapped to the zero vector which is really the trivial vector.
Birne Binegar: Inside the Co domain, though the instinct vectors inside the Co domain correspond to the range because that's that's where you get non non zero vectors.
Birne Binegar: After applying the linear transformation.
Birne Binegar: know they do every subspace always contains the zero vector but, but the the image or the range it's going to contain many more vectors.
Birne Binegar: On the range we're concentrating on those vectors in the domain that sort of trivialize once we apply the you think they get sent directly to the zero vector.
Birne Binegar: So they don't give you non zero vectors that give you only that trivial vector that doesn't really mean it never does anything like once once you once I get over here and i'm.
Birne Binegar: doing something with the zero vector I scale and multiply get the zero vector add it to another vector like it doesn't do anything so you can think of things in the kernel as ending up here and not making any difference at all.
Birne Binegar: Okay, so so that's the set we're looking at every element in the kernel long to remain but it's also required to have the property that once I apply the function I don't get arbitrary points over here.
Birne Binegar: yeah that's an ar m, but this point here in general it's going to live inside or am I sorry the Co domain, but we're requiring you with the equal to zero vector inside here so that's.
Birne Binegar: that's the definition of the Colonel and as i've set it up i've only specified it as a subset of the domain will now show you that set is closed under scale multiplication and addition, and so the set is actually going to be a subspace.
Birne Binegar: All right, and in fact.
Birne Binegar: What i'm going to show you this in fact this set is identical to the no space of the corresponding matrix and then on what that indirectly showing you is that.
Birne Binegar: The kernel of a linear transformation is equal to this set, but this set is already known to be a subspace when we talked about subspace it's attached to matrix.
Birne Binegar: The solution, set up a subscription fee of X equals zero that's the note space of the matrix and that that we know how to find a basis for, for example by solving the corresponding linear system, so what i'm going to show you is that.
Birne Binegar: Every element in this set can also be interpreted as an element, no space to pay, which this is defined as a set of vectors X inside or m that satisfy a linear linear equation at X.
Birne Binegar: is equal to zero.
Birne Binegar: And now, this when we're talking about matrices and their associated sub spaces, this is one of the three sub spaces, we looked at, and so what i'm showing you now is that.
Birne Binegar: The Colonel subspace define using function theory language is the same as the no space of the matrix which is more using the language of solving linear systems.
Birne Binegar: Okay, so let's see if I can show this really quickly.
Birne Binegar: If we have a vector the star soften the kernel.
Birne Binegar: Then I want to show you that.
Birne Binegar: well.
Birne Binegar: If that's true if he belongs to the Colonel what we're requiring is that T applied to X equal to zero vector, on the other hand.
Birne Binegar: Once we construct the matrix that's attached to this linear transformation well that matrix always has a property that.
Birne Binegar: P of X is equal to matrix multiplication by this particular matrix one we construct this way and so, if P of X is equal to zero.
Birne Binegar: or any point that lives inside the kernel that's going to require Well, this is exactly equal to a TEE of X via this identity.
Birne Binegar: And so, that means that X actually satisfies a linear equation here's a coefficient matrix time to make the column vector variables is equal to zero, so that shows you that.
Birne Binegar: X live in the no space.
Birne Binegar: At which is defined as the solution set.
Birne Binegar: A ti X is equal to zero.
Birne Binegar: And so, in fact, that.
Birne Binegar: The kernel of a linear transformation is in fact identical to the no space at the associated nature, since the no space of a matrix is a.
Birne Binegar: subset of our up an invite and matrix is the subs space of rm so two is going to be this kernel because there's no space is exactly equal to the Colonel.
Birne Binegar: So the two sets coincide and.
Birne Binegar: we'll get them yeah we can just simply identify these two sets.
Birne Binegar: and that in turn it's going to give us a way of computing a basis.
Birne Binegar: For the set of actors that land on the zero vector after applying the linear transformation, all we have to do is calculate the matrix.
Birne Binegar: And figure out what it's no spaces well.
Birne Binegar: That means we have to solve the linear system who's coefficient matrix is exactly this numerical matrix so the next step in order to solve.
Birne Binegar: A subscript ti X is equal to zero, we write the matrix in right write down, and then we reduce this matrix to we do show Echelon form, so that we can actually solve the linear system.
Birne Binegar: So when that's done this matrix reduces the this matrix you see here the pivots they're both equal the one above and below pivot I have nothing but zero that's the reduce row Echelon form of that matrix.
Birne Binegar: So you know when we're solving linear system, once we reach the matrix in reduced relational on form the next task is to convert back to equations so here, I did it.
Birne Binegar: X one plus zero X two minus X three should equals zero second row is going to lead you that equation, we also use this matrix to.
Birne Binegar: break our variables up into fixed variables corresponding the columns with pivot in three variables that correspond the columns about pivots.
Birne Binegar: Remove the free variable coming from these equations to the opposite side, then we write down the solution vector we expand the solution vector in terms of the three.
Birne Binegar: parameters and then we get a basis vector.
Birne Binegar: For well what we constructed as the basis vector for that solution set.
Birne Binegar: But that is no space at at we've already seen to be identical to the kernel of the corresponding linear transformation, so what i've done is i've now found a basis factor.
Birne Binegar: For the kernel of the linear transformation, I found a set of directions inside my domain that will reach every point that that linear transformation T sense to zero.
Birne Binegar: Hope so I guess i'm running out of time, so Actually, this is a good place to stop.
Birne Binegar: What i'm going to do on Friday is i'm going to look again at these two sub spaces and we're going to count the number of basis vectors and that will tell us what the dimensions are and then we'll find.
Birne Binegar: That, in fact, the dimensions never disappear it's sort of a zero sum game that the dimensions of this space that split up.
Birne Binegar: Some of the dimensions are going to be carried over into the image, but they could also lose dimensions as well, and those are getting the dimensions of the going to be dimension up.
Birne Binegar: subspace that gets sent to the zero vector so the kernel that information this last it's all sent to the zero vector but the information that survives is going to be live in the range and when you add the dimension of those two sub spaces, you get exactly the dimension of the.
Birne Binegar: Of the domain OK so i'll stop here and we'll pick up with this discussion on Friday.