Birne Binegar: Okay, just to.
Birne Binegar: person attendees today I don't blame you have a great day out there.
Birne Binegar: So today.
Birne Binegar: I said earlier this week that we're likely to have our first exam immediately after well the week after we come back from spring break I think that's too ambitious.
Birne Binegar: is more likely what's going to happen so i'm going to finish up our discussion of linear transformations today but i'm just barely going to finish that.
Birne Binegar: And then, what i'd like to do is maybe Monday after we come back from spring break to go over homework problems and and try to consolidate things a little bit.
Birne Binegar: Rather than having a reduced session on Wednesday.
Birne Binegar: After we get back from spring break maybe do that on Friday, and then the week after the week we come back from spring break to have the exam probably the Monday the.
Birne Binegar: Second, Monday after we come back from spring break, you have the exam then so i'm not going to schedule today what we'll do is we'll schedule the the second exam on the Monday we get back from spring break.
Birne Binegar: Okay, so.
Birne Binegar: that's kind of good because it means that.
Birne Binegar: You don't have to develop a whole lot of time to do the linear algebra job over the break you do have a homework assignment the homework assignment I guess on linear transformation, I think that's homework assignment seven that's going to be do.
Birne Binegar: The Monday, you get back from spring break, but then you'll still have.
Birne Binegar: At least another week to consolidate your understanding of what we've done since the first exam.
Birne Binegar: Okay, so my goal today is is i'm.
Birne Binegar: kind of limited um.
Birne Binegar: i've already introduced to the topic of linear transformation and the two sub spaces that one naturally attaches to a function between two vectors basis not main purpose of today is to talk about if I have a linear transformation.
Birne Binegar: From one vector stage to another.
Birne Binegar: So R and R and when can I go back and forth between these two So if I have a function that takes me from a given vector space to another one, one can I actually have an inverse map that takes me back so that would be to the minus first this is going to be a really important topic.
Birne Binegar: When a linear transformation has a legitimate inverse function that will take you from the Co domain of the original function back to the domain that's what we call a vector space it more system and those vector space os X or more things are really fundamental in linear algebra.
Birne Binegar: Not so much in the context, so if we're not direct your spaces are our m&r n it's going to turn out that if you have a.
Birne Binegar: linear transformation that has an inverse the only pop maps that are possible or from art and art in itself not to some other vector space.
Birne Binegar: Because effectively what you need in order to have an ice and mormonism or an embeddable linear transformation is you MIT spaces, have to have the same dimension and so effectively you're going to have to have our into our em but we're going to see that today.
Birne Binegar: Just by looking at the underlying assumption theory of linear transformations.
Birne Binegar: Alright, so let me, let me start.
Birne Binegar: get my clicker working there it goes okay so remember linear transformation, these are special functions.
Birne Binegar: Between vector spaces that are compatible with scale and multiplication and vector addition.
Birne Binegar: functions of this type are naturally associated with matrices, this is the topic of my lectures, this week, on the one hand, if you have a.
Birne Binegar: matrix with in rows and columns you can use that matrix to define a function that goes from an m dimensional vector space to an n dimensional vector space that's done.
Birne Binegar: Using matrix multiplication for this multiplication be defined, since I have em rows in this first.
Birne Binegar: In columns in this first factor The second factor is going to have to have it exactly in rows, and so it would be interchangeable as an n dimensional vector.
Birne Binegar: After I do the multiplication though I end up with a new vector that has in components, and that is interpreted as a vector and site or and so as soon as you have a.
Birne Binegar: n by n matrix you can use this matrix multiplication to construct a function that will take n dimensional vectors the N dimensional vectors and it will naturally satisfy these two properties.
Birne Binegar: However, one of the main point of today's or lectures, this week, if you go in the opposite direction as well, if you have a function between two vector spaces to satisfy these two properties, then.
Birne Binegar: You can define and associated matrix that will effectively implement the linear transformation via matrix multiplication so from this, you can define.
Birne Binegar: A matrix.
Birne Binegar: Let me give you the world for what you do if you look at how tx on the standard basis vectors for your domain those standard basis vectors are, of course, first one is 100 it's going to have em components.
Birne Binegar: And the standard basis vectors they just have a one in one of the slots and zeros everywhere else you have a bunch of them.
Birne Binegar: In fact, if you start off in an n dimensional space you'll have exactly em standard basis factors.
Birne Binegar: So what I can do is I can apply my function each standard basic factor that's going to give me an n dimensional vector as a result, and I can write that as a call.
Birne Binegar: I can do that for each of the standard basis factors when I end up with is a matrix with em columns and in rows, and so I get an n by n matrix and what happens, then, is that.
linear transformation.
Birne Binegar: can then be implemented.
Birne Binegar: Even if this was defined by some functional formula once we create this matrix we can replicate the action of that function, the matrix multiplication.
Birne Binegar: So, in fact, we can go from linear transformation matrices and we can go from we can go from he seems to linear transformation that's the first line, and we can go from linear transformations back to matrices.
Birne Binegar: All right now what's nice about that is that sooner, who write down this matrix we just got.
Birne Binegar: A rectangle arrangement of numbers, so if we're in this setting thinking of linear transformation matrices it's we're in this eminently calculable setting for everything you know basic objects are just numbers that we can add multiply etc.
Birne Binegar: Okay, so.
Birne Binegar: Also, if we discussed on Wednesday we're two special substations we can attach to a linear transformation, one is going to be a subspace up the domain and the other is going to be a subspace of the co domain.
Birne Binegar: So the starting point and the ending point at that at that function.
Birne Binegar: The range it's going to be a subset of the co domain what it consists of is all the points in the co domain, that you can reach by applying.
Birne Binegar: The linear transformation so.
Birne Binegar: You may not hit every point here by applying this function but.
Birne Binegar: When you restrict attention to exactly those points that are reachable by applying the linear transformation that's the range and what we saw.
Birne Binegar: This week is in fact that this.
Birne Binegar: The subspace can be identified as the column space of the matrix that's associated with that linear transformation, so that makes this range.
Birne Binegar: easily calculable we take them a linear transformation, we compute it matrix and then we find the column the basis really calm face of that matrix and will give us a basis for the range.
Birne Binegar: there's another subspace that's attached to the.
Birne Binegar: To the to the domain, in this case we focus on the points in the domain that gets sent to the zero vector in the code domain once we apply the function that's called the Colonel.
Birne Binegar: His computer will my referring back to the matrix that's associated to a linear transformation, it turns out it's a no space of the matrix or, in other words it's the solution set up the linear system, you get by.
Birne Binegar: Taking the matrix corresponding to linear transformation multiplying a variable X by it and demanding it that equals zero vector.
Birne Binegar: Though well if these guys we compute in terms of the matrix that's attached to a linear transformation.
Birne Binegar: Alright, so one thing I want to point out is.
Birne Binegar: A dimension and the Colonel here, they have a relationship between your dimensions.
Birne Binegar: I mean i'm.
Birne Binegar: sort of motivate this by recalling a fact we know about matrices.
Birne Binegar: The number of columns of a matrix is always equal to the rank of the matrix, that is, the dimension of a row space or dimension of the columns days, plus the dimension of the north face.
Birne Binegar: Let me show you why OK, so the rank is actually defined as the common dimension of these span of the rows of the matrix or equivalent Lee the number of basis factors, you have in the span of the call of a.
Birne Binegar: Both these dimensions well how do we count them well, we need to get a basis for the row space that's done by reducing a to a row excellent form and grabbing the non zero rose.
Birne Binegar: To get a basis for the column space we reduce a to row Echelon form we figure out where the pivots live and we go back to the original matrix and use those as basis factors.
Birne Binegar: Now basis factors are the number of there's the independent directions that generate the entire subspace.
Birne Binegar: So when we count basis vectors we're basically getting the dimension in fact it's defined as the dimension of the corresponding subspace.
Birne Binegar: So how many basis vectors do I have for the row space well it's good each basic factor corresponds to a non zero row with the relational on form, if you have a non zero which got a pivot in the First Non zero entry net row so.
Birne Binegar: The number of basis factors is going to be equal to exactly the number of pivots you have any row Echelon.
Birne Binegar: or or covertly the number of.
Birne Binegar: yeah either way you count that it's always equal to number if it's in a row Echelon form meanwhile the dimension of the no space of a matrix well that's just dimension up the solution set a of X is equal to zero.
Birne Binegar: And so the dimension of the solution, set is going to correspond to the number of free parameters, you have when you actually write down the solution of this.
Birne Binegar: And that, in turn, is going to equal, the number of columns without pivots well once you try to solve this linear system you write the augmented matrix you reduce reduce show Echelon form and then.
Birne Binegar: In that relational inform you decide which variables are free and which variables are are fixed variables by.
Birne Binegar: Looking at the columns and deciding whether a call contained to pivot That would be a fixed variable or a column, that does not have a pivot that's going to be interpreted.
Birne Binegar: As corresponding to a free variable so the free variables correspond the columns about pivots and the row Echelon form and now, if you look at these numbers, I get the dimension.
Birne Binegar: Upon row face or dimension of the column space that corresponds to the columns of a row excellent form that have it's.
Birne Binegar: The dimension of a no space corresponds to columns without pivots So if I add this number two this number well this week pivots without pivots include both possibilities like just get the total number of calls, and so we had this identity that related.
Birne Binegar: The dimensions of to sub spaces attached to any matrix I want to do something similar for linear transformation, but now I substances are not going to be the rank and novelty of a matrix, but rather the range and kernel of a linear transformation.
Birne Binegar: Okay, so.
Birne Binegar: here's my claim, if you have anytime you have a linear transformation.
Birne Binegar: you're going to have the range, which is going to consist of all the points in the co domain that are reachable by applying the linear transformation.
Birne Binegar: And you have a subspace of the domain, called the kernel which consists of all vectors in the domain that get sent to the zero vector in the code domain my claim is that the dimension of those two sub spaces, is exactly equal to the dimension of the domain.
Birne Binegar: And the way you see, that is.
Birne Binegar: Using the correspondence to every linear transformation there's a matrix So how do we create that matrix.
Birne Binegar: A tea.
Birne Binegar: or even write it down.
Birne Binegar: As we look at the standard basis vectors for rm we evaluate our linear transformation on the standard basis sector and right result as a call second column with the tea applied to eat to written vertical.
Birne Binegar: And then your last standard basis factor is going to correspond to the last column the matrix you see that we have exactly am columns here if we started off with an empty mentioned linear transformation from R amp to our end.
Birne Binegar: so soon, as I have a linear transformation from rm en I get a matrix that has exactly em columns.
Birne Binegar: All right now.
Birne Binegar: We take the columns of this matrix you can divide it up in the columns that have pivots and columns that don't have pivots but.
Birne Binegar: The number of columns that that have pivot so so going to correspond to the.
Birne Binegar: columns that they're being produced basis vectors.
Birne Binegar: For the column space, the recipe is is that you know, reduce this matrix to row Echelon or you look at where the pivots live and you go back to the original matrix and grab the corresponding call.
Birne Binegar: So for each column of a row Echelon form of this matrix that has a pivot you're going to have a basis factor for this column space.
Birne Binegar: and, similarly, for the no space of this matrix as we shown seen on the last slide it corresponds to columns of The roadshow on on corresponding The pre parameters or columns of the row so on on that don't contain pivots.
Birne Binegar: And these two dimensions always add up to the number of columns of the original matrix.
Birne Binegar: But.
Birne Binegar: The relationship between this matrix and this linear transformation is such that dimension of this subspace this in fact this subspace is identical to the range of tea and this subspace is identical to the Colonel T and so.
Birne Binegar: We get this relationship, and now we have this identity that.
Birne Binegar: The dimensions of the domain effectively get broken up into two.
Birne Binegar: two parts one is dimension of the kernel and and.
Birne Binegar: And then, if you add to it to dimension of the range you get exactly the dimension of the domain.
Birne Binegar: The way to think about this.
Birne Binegar: Or at least my way of thinking about this.
Birne Binegar: Is I have a vector station or am.
Birne Binegar: A vector space or and and I have a linear transformation that's taking me between those two vectors spaces.
Birne Binegar: Now.
Birne Binegar: This map doesn't necessarily cover the entire vector space or in most likely.
Birne Binegar: It just filled out.
Birne Binegar: Some subspace of this and that subspace is exactly what we mean by the range of tea, so if I as I move across over all the points in the domain, a statement is I land on some point in the subspace called the range is sort of flat plane that's embedded inside of our em.
Birne Binegar: So that's your range.
Birne Binegar: And then, what the Colonel is the Colonel it's a subspace of the domain.
Birne Binegar: And it consists of all the points oops let me draw this arrow better.
Birne Binegar: You start off on this subspace.
Birne Binegar: And, which is the kernel.
Birne Binegar: of tea, that means you land on the zero vector once you apply that.
Birne Binegar: transformation, so I have a subspace in the co domain subspace in the domain.
Birne Binegar: So there's subspecies of different vectors faces but, just like the column space and row space of a matrix we have substances that live in different vector spaces, but somehow there dimensions are related and so i'm going to show you that.
Birne Binegar: Well, in fact, that's what you described here but.
Birne Binegar: One way to think about this is that what the College tracking is it's the information that's lost from.
Birne Binegar: This is an n dimensional space, so you have.
Birne Binegar: Each point you need a total of em free parameters in order to describe all the points in this domain.
Birne Binegar: But it turns out that some of these guys get sent to the zero vector, which is the trivial vector it doesn't really it's not really a non zero point over here it's exactly one point to all the information that's trapped on the Colonel hyper plane inside domain.
Birne Binegar: lots of different vectors there is all getting set to the zero vector here, so I think of that as last information what I.
Birne Binegar: What I lose a bit in understanding our am if I tried to view it over here.
Birne Binegar: While the range is that sort of represents the information that survives, as you go from R amp are in via this linear transformation.
Birne Binegar: And so the way to understand this formula is the total information in your domain it's n dimensional well, you lose some information.
Birne Binegar: that's those are the factors that gets sent to the zero vector and then you also retain some information, those are the vectors that get mapped over to non zero factors over here.
Birne Binegar: Alright, so that's our dimension formula.
Birne Binegar: and
Birne Binegar: next topic I want to talk about yes.
Birne Binegar: it's going to be about.
Birne Binegar: eventual goal is to figure out when we go from one vector space to another when can we go back and, of course, if we're losing information going over here, then you should not be able to go back.
Birne Binegar: If effectively if think about going from a three dimensional space to a two dimensional space you've lost one dimension of of information in following that that function over over again to explore that next but.
Birne Binegar: Before I can talk about the universal function I first need to talk about what it means to compose functions.
Birne Binegar: And so.
Birne Binegar: let's see did this dimension formula.
Birne Binegar: here's a picture of what i'm talking about when I talk about composing functions.
Birne Binegar: You have a function just between sets napping a set A to B and you have another function that map the set be to set see then you can compose you can go from A to B and then now you're in domain of G, you can apply g to that point and map your way over to.
Birne Binegar: A point in the third set see but then this process of going into jumps to this point here could be done in a single jump that's the come compose function and usually what you do is you compute this by.
Birne Binegar: calculating this and then substituting result into the argument for G, and that would be the compose function.
Birne Binegar: Alright, so what i'm going to show you is how this functional composition, how it behaves.
Birne Binegar: When our functions are actually linear transformation.
Birne Binegar: And so it works, the same way, if I have a linear transformation from an m dimensional vector space to an n dimensional space and another linear transformation that goes from the Co domain of the first one to a new vector stage or P, then I can compose those.
Birne Binegar: But here's the statement that I want.
Birne Binegar: So when I compose to linear transformation I get a new linear transformation and what i'm saying in this theorem is that the matrix that corresponds to the composed linear transformation is actually the matrix product of the major sees corresponding to the individual.
Birne Binegar: interviews individual.
Birne Binegar: linear transformation shelf I have key from.
Birne Binegar: Our end to.
Birne Binegar: End from that.
Birne Binegar: I can get an associate matrix at.
Birne Binegar: II II.
Birne Binegar: II II am.
Birne Binegar: And then, if I also have a linear transformation that's going from one and.
Birne Binegar: RT I also have a a matrix as.
Birne Binegar: And it's going to be s.
Birne Binegar: apply to.
Birne Binegar: All going to use the same location.
Birne Binegar: The first standard basis vector for our end and last one is going to be as applied to the last standard basic vector and write those results vertically, I can attach matrices to each of these guys.
Birne Binegar: And in fact.
Birne Binegar: i'll be able to this is going to be let's see it's going to be n by n matrix this is going to be a matrix with.
Birne Binegar: Key rows and rows and columns i'll be actually be able to multiply these two matrices together, I just have to use.
Birne Binegar: The matrix a subscript s as the less camp factor, because I need the number of columns of the first factor to equal, the number of roses, a second factor.
Birne Binegar: So this product will be well defined in the statement of the theorem is, in fact, the product of these two matrices is going to be exactly the same as the matrix you would get by.
Birne Binegar: First figuring out what linear transformation, this is and then computing it's um.
Birne Binegar: it's corresponding matrix.
Birne Binegar: Let me show you this with an explicit example.
Birne Binegar: Here I have a.
Birne Binegar: function between two factor spaces, that is, in fact, a linear transformation it's taking two dimensional vector producing a three dimensional vector the second function it's going from a three dimensional vector space to a two dimensional vector space and there's the formula for.
Birne Binegar: When I compose.
Birne Binegar: And I found the compose function, the rule is if I first I evaluate T at this point, until i'll just substitute in Wer t of X one X two and then I apply the function F.
Birne Binegar: To that point and that I use the rule for S, so the rule for computing the value of S at a given vector is I add the two first two components so add X one plus x plus x one, and the second component is going to be the third component.
Birne Binegar: Until what's going to happen here is.
Birne Binegar: Some of these first and second component is going to be X two plus x one, and the third component is going to be X one X two.
Birne Binegar: And so there's an explicit computation of the.
Birne Binegar: Composition of this linear transformation it P with linear transformation, yes.
Birne Binegar: let's write down the corresponding matrix that's done by evaluating this function on the standard basis factors.
Birne Binegar: Since I have a two dimensional domain.
Birne Binegar: or two i'll have to standard basis factors, the first one that's going to be one zero so that means taking one for X one and zero for X to evaluate, yet I get zero Plus 110 plus one.
Birne Binegar: One plus zero is one, and then to get the second component I evaluate the same function.
Birne Binegar: On the second standard basic vector where in we take X is equal to one and X one is equal to zero So what do I get the first component, I have one plus zero which one, and the second component, I have zero plus one is one so.
Birne Binegar: that's the matrix corresponding to that composed linear transformation.
Birne Binegar: Alternatively, I can get this matrix by.
Birne Binegar: Looking at the individual linear transformation and calculate their matrices so here we're going from a two dimensional vector space to a three dimensional vector space, so when.
Birne Binegar: Evaluating T on the first standard basis vectors going to be zero in the first slot one in the second slot one plus zero and the third slide 0011.
Birne Binegar: Second, on the second standard basis factor that will be one X two is equal to one X one is equal to zero i'll get 101 negative 101 and, similarly, I can construct the matrix corresponding to F.
Birne Binegar: And all I want to show in this is that if I take this matrix and I multiply the first matrix by it, I get the same matrix here so here's that identity that was on the theorem at the preceding slide the matrix corresponding that composed.
Birne Binegar: linear transformation is just the product of the matrix each of the individual linear transformations.
Birne Binegar: Alright, so that's nice what What that means is that when you have linear transformations and you want to compose them.
Birne Binegar: Calculation Lee you want to get to these numbers as quickly as possible.
Birne Binegar: Rather than dealing with variables all the time, so what we can do is, we can immediately write down the make sees corresponding to these transformations and then we can calculate the matrix not by substituting into formulas, but rather simply by multiplying matrices together.
Birne Binegar: Okay, so.
Birne Binegar: that's how we compose.
Birne Binegar: linear transformation.
Birne Binegar: And reason I introduced that topic.
Birne Binegar: So I can get to the principal result of today, which is.
Birne Binegar: Discovering.
Birne Binegar: exactly when, if I have a linear transformation that takes me for one vector states to another when can I go back.
Birne Binegar: With a well defined function that will take me from my co domain back to my original domain.
Birne Binegar: But, so what happened here.
Birne Binegar: What i'm going to talk about in preparation for figuring out when we can invert a linear transformation it's do.
Birne Binegar: specialize or actually general is the case when we're just looking at functions between two sets and i'll give you a simple criterion that will tell us, given a function from a set a set fee when an inverse function is going to exist.
Birne Binegar: So the set that up.
Birne Binegar: Let me remind you that the image of a function that the third of all point in the co domain, that you can reach I applying the function to a point in the in the domain.
Birne Binegar: We say that a function from one set to another is subjective if the Co domain be is actually the same thing as the image What that means is that every point here is reachable.
Birne Binegar: By applying the function F, there are no point in the co domain that don't come from a point of the original set a.
Birne Binegar: A second property, we can ascribe to a function between to set is this notion of what's called interactivity, this is a different criterion here what the condition is is that a function is injected.
Birne Binegar: Also, sometimes called to be one to one if the value of a function.
Birne Binegar: The two points have the same value if the function is injected, in fact, the point had to be the thing.
Birne Binegar: Let me give you.
Birne Binegar: An example of a simple function that illustrates more clearly exactly what we're talking about here suppose I have a function from the wheel line to the real line.
Birne Binegar: Where if it's going to send X X squared.
Birne Binegar: Alright, so here's the real line.
Birne Binegar: Here, Sir rely now this function is not sure objective.
Birne Binegar: Why is that well subjectivity means you look at the image of the thing, and it should call it if it's subjective then every point in the co domain comes from some point here, but the negative numbers here when you're less than zero here.
Birne Binegar: you're not going to reach any negative number like squaring a positive number right negative one times negative one is equal to one the image of this map it's always a non negative number until these guys here are never reached by applying this information because.
Birne Binegar: Next negative.
Birne Binegar: numbers.
Birne Binegar: In the code.
Birne Binegar: That is this target space.
Birne Binegar: are never.
Birne Binegar: reachable.
Birne Binegar: yeah.
Birne Binegar: So it's not subjective and that's when we say a function is objective that's exactly that's the kind of thing we're trying to be glued we want everything in the domain to be reachable from the Co domain by application of the function.
Birne Binegar: And this function is not injected either.
Birne Binegar: Why because, for example, Fo one.
Birne Binegar: it's going to equal one squared equals one but that's the same thing is F minus one, so I have.
Birne Binegar: Two different points in the domain minus one in one that are getting mapped to the same point.
Birne Binegar: via this function and that's what we try to exclude.
Birne Binegar: When we say a function is injected It means that distinct point have to get mapped to distinct points that can't land on the same phone.
Birne Binegar: So that's what these notions of injected in search at give me the first search activity means you completely cover the The co domain by application of function, and the second one means that distinct point always get mapped to distinct points.
Birne Binegar: All right now.
Birne Binegar: The interest in these two properties.
Birne Binegar: is falling, first of all, if there we have subjectivity and objectivity by activity means both subjective injected.
and
Birne Binegar: get to the point I want to get to.
Birne Binegar: We say that a function is in verbal if you cannot only go from A to B, but you can also go backwards from.
Birne Binegar: The co domain back to the original set a lie, the application of another function a label F minus first what you want, for this new function is that if you.
Birne Binegar: apply it to a point in be and then you follow it with the original function you basically have the identity function takes B2B.
Birne Binegar: Or if you compose in the opposite order, starting with appointed a going to be and then using this function to go back to a you get back to where you started with.
Birne Binegar: that's what it means to have an invisible function that you can pass back and forth between the two, and when you do so, you just get the identity.
Birne Binegar: Alright, so here's the.
Birne Binegar: here's the the theorem that holds for anytime you have a function that by active.
Birne Binegar: If it's my objective then it's going to be inverted.
Birne Binegar: And the reason for that is i'll show you how you actually.
Birne Binegar: abstractly defined this inverse function in the case when a function is in verbal.
Birne Binegar: We have a set a.
Birne Binegar: Set the function from a.
Birne Binegar: To B and going this way, what I want.
Birne Binegar: is to figure out.
Birne Binegar: A function F terminus first that will go in the opposite direction.
Birne Binegar: Alright, so here's how it's done i'm going to assume that this function F is both subjective and objective because.
Birne Binegar: and start with a point B in the domain of this.
Birne Binegar: function that's going to end up being the inverse of egg we start here well because it's subjective this point had to come from some point here so.
Birne Binegar: If be as a point in be then it has to equal F of a for some some point of bed, but because it's subjective, we only have one choice here if we had.
Birne Binegar: We had two different points in a that landed on the same point here, then you would not be injected so we're actually eliminating by stating that the function is not only subjective but injected.
Birne Binegar: we're eliminating the possibility that this particular point be here could have come from two different points so it has to come from one point because it's rejected, and it can only come from one point because it's also injected.
Birne Binegar: And so that guy does.
Birne Binegar: That auxiliary point doesn't occur, and so, in fact.
Birne Binegar: Although I don't have a formula for the inverse of the I know that they exist, simply because I the function is subjective and injected.
Birne Binegar: Okay, so.
Birne Binegar: The way you want you to understand this, is that.
Birne Binegar: Just to know that a function has an inverse it suffices to check that you can reach every point of be by applying the function to some point of the first set a.
Birne Binegar: And you need to verify that two points of the original set neverland on the same point in the in the co domain if those two conditions hold, then the function it's going to have an inverse.
Birne Binegar: OK now let's apply.
Birne Binegar: This function theory to the case at hand what we're looking at functions between.
Birne Binegar: Between vector spaces linear transformations.
Birne Binegar: Well, so first statement is that linear transformation is going to be subjective only if the range of the linear transformation is the entire code domain.
Birne Binegar: that's that's not really much of a lemma just chasing definitions.
Birne Binegar: Thinking of this as a function is going to be subjective if every point here gets map to the image set or in well the image set are in is exactly the range and so.
Birne Binegar: This is just this is just terminology if the entire co domain is the image of the linear transformation, then.
Birne Binegar: The linear transformation is subjective.
Birne Binegar: All right, what about injecting.
Birne Binegar: Next lemme tells us that, if we have a linear transformation between two backer spaces, then it's going to be injected only if the kernel is equal to zero.
Birne Binegar: I can make I can show this real quickly.
Birne Binegar: first thing i'm going to point out is that anytime you have a linear transformation, the zero vector always gets mapped to the zero vector so.
Birne Binegar: No matter what this function it when no matter what this linear transformation is, I know its value at the point X equals zero inside the domain.
Birne Binegar: The reason for that is it applied to zero, I can rewrite that as to apply to X minus X any vector X is difference between X, in itself, is always going to produce the zero vector but because it's the linear transformation.
Birne Binegar: He have X minus X exactly he have X.
Birne Binegar: plus x minus X, you can think of it this way here.
Birne Binegar: Plus minus one times X.
Birne Binegar: And then.
Birne Binegar: Let me write it that way minus one X, I can do this because, let me your transformations are compatible with vector edition, and so I can take that vector some and I can delay it until after I played the function but it's also compatible with scale multiplication.
Birne Binegar: So you have to X minus.
Birne Binegar: X and that shows you.
Birne Binegar: P zero.
Birne Binegar: Point evaluation.
Birne Binegar: Always equals zero vector because in the end you're going to be P of X minus t of X.
Birne Binegar: Alright, so this shows you.
Birne Binegar: That one thing that always lives inside the range.
Birne Binegar: Of linear transformation is the zero vector.
Birne Binegar: Alright, so now let's let me show you that occurred and give you a criteria for injected it suppose the kernel of a linear transform mom supposed to use injected.
Birne Binegar: I went up calculate the Colonel and so let's suppose.
Birne Binegar: The kernel consists of all the vectors to get sent to zero so i've already shown you that zero.
Birne Binegar: Its value is always going to be equal to the zero vector in the in the code domain suppose I have another vector that lands on the same spot in in activity what we're trying to make sure is that two vectors don't land on the same spot.
Birne Binegar: So I know zero lands on zero would have excellent on zero but X is not equal to zero well, that would contradict.
Birne Binegar: Keeping objective if.
Birne Binegar: i'm.
Birne Binegar: Here I have.
Birne Binegar: Two points in the domain landing on the having the same value and if they're not the same point, then I contradicted the fact that it's injected.
Birne Binegar: So.
Birne Binegar: So, in other words the kernel of tea it only consists of a single point the zero vector.
Birne Binegar: And so, if T is injected the necessarily the kernel of tea the set of points Atlanta on zero, it can only be the zero vector.
Birne Binegar: in the opposite direction now i'm going to show you the Colonel is equal to zero, then necessarily he is injected.
Birne Binegar: So let's.
Birne Binegar: Look at the condition for objectivity, if you land on the same spot so T evaluated X one is equal to X two if that happens, then necessarily.
Birne Binegar: Just move this term, to the opposite side by subtracting it from both sides, then you get zero is equal to one minus 32 because you have a linear transformation, you can delay that.
Birne Binegar: scale and multiplication and vector addition and help you could equally well do it before you apply the linear transformation sorry that's not a plus sign just the mouse is on top of it, I can do that vector subtraction before I applied, but then.
Birne Binegar: Now I have zero.
Birne Binegar: is equal to T applied to X one minus two, so the value of T applied to this particular vector equals zero, but my hypothesis is that the only way I can reach zero on the.
Birne Binegar: In the Co domain is by applying T to the zero vector and so now.
Birne Binegar: Its value of this value of tea applied to this vector is equal to zero in the kernel of tea consist of only the zero vector did X one minus X to have two equals zero, and so we get this condition which tells you that X, one has to equal to X two so if.
Birne Binegar: The Colonel is equal to zero.
Birne Binegar: Then, and give X one is equal to X two that's going to imply that X, one has to equal X two.
Birne Binegar: And so.
Birne Binegar: And that means if whenever you have the same value, you have to come from the same point, that means that he is injected.
Birne Binegar: Okay whoops i'm just about out of time, but let me just give you the punch line here.
Birne Binegar: A linear transformation between two vectors spaces, is going to have an inverse well from function theory.
Birne Binegar: We know that.
Birne Binegar: Just general function theory if a function between two sets is injected in subjective, then you have an inverse.
Birne Binegar: What we just develop is.
Birne Binegar: criteria for a linear transformation to be subjective and objective.
Birne Binegar: If the range of T is equal to our end that's going to imply it subjective if the kernel of T is equal to the zero vector that's going to imply to injected but if a function is both subjective and injected, then it has an inverse.
Birne Binegar: And so there are theorem a linear transformation between two vector spaces, is going to have an inverse so long as the the function T completely covers the Co domain, are in a subjectivity and simultaneously, the only vector.
Birne Binegar: That gets sent to the zero vector is the zero vector itself of the domain if those two conditions hold then.
Birne Binegar: The linear transformation is going to be in vertical.
Birne Binegar: Well, I guess i'll finish today by just.
Birne Binegar: Introducing a definition we say that a linear transformation is an ice amorphous exactly when these two conditions hold exactly when there's an inverse linear transformation.
Birne Binegar: Okay, so i've got the series presented today, but actually a lot more discussion that that should take place so that you actually get some more intuition, as to what we're requiring here it's going to turn out that vector space is amorphous and.
Birne Binegar: Really, the key structure that makes linear algebra so useful it's not so useful in the context that we've been working with thus far, but we're looking at vector spaces that are just ordered list of numbers, what you're generally speaking when you're dealing with them.
Birne Binegar: affect your space in an application, it might be a solution, set up a set of linear differential equations it might be solution set up some linear system.
Birne Binegar: What we like to do is have.
Birne Binegar: vector space ice amorphous that can take an abstract that your space and map it into a concrete calculable that vector space that's where vector space ice and more systems are really crucial.
Birne Binegar: And so what else i'll do once you get back from spring break so we'll talk a little bit more about how vector space ice and more systems are used to now bring.
Birne Binegar: Take our real of what constitutes a vector space and makes it much broader until that it can actually handle all possible applications of linear algebra as well.
Birne Binegar: Okay, so i'll stop here and we'll pick up with this discussion on Monday, when you come back from spring break.