Birne Binegar: yeah.
Birne Binegar: Years really that long it's just.
Birne Binegar: it's just.
That.
something's.
Birne Binegar: Free chapters.
Birne Binegar: yeah.
Birne Binegar: So much I.
Birne Binegar: Like.
Birne Binegar: To pass.
Birne Binegar: yeah.
Birne Binegar: Operations research.
Birne Binegar: process.
Birne Binegar: lecture before spring break Oh, maybe hope you guys can get home safely have today like your.
Birne Binegar: So what i'm going to do today is I have i'm not going to be that ambitious, I just want to talk a little bit about more how we interpret.
Birne Binegar: linear transformation using basic function theory.
Birne Binegar: The main goal here would be to figure out when a linear transformation option between two vectors spaces, is going to be inverted.
Birne Binegar: So you can not only go from one vector space to another, but you can go back again, and this turns out to be now really important topic and, once you leave this course, one of the main things you study or vector space ice and morph isms.
Birne Binegar: Not so much from our end to aren't in but rather from more abstract vector spaces into our end where you can do calculations.
Birne Binegar: So let me start i'll start today by reminding you what the range and kernel of linear transformation is we'll.
Birne Binegar: figure we're going to figure out the dimensions of these guys that's something I did not cover on Wednesday.
Birne Binegar: And then, a pretty simple criteria to figure out when a linear transformation a function between two vectors spaces actually isn't vertical so you can go back and forth between the two vector spaces.
Birne Binegar: Okay, so so first of all linear transformation is a function between two factor spaces that are compatible with scale or multiplication and with vector edition.
Birne Binegar: That up a little bit and then.
Birne Binegar: What we have been exploiting this week has been the connection between linear transformation and matrices.
Birne Binegar: On the one hand, you have an n by n matrix you can always create a function between two vector spaces, that has these properties.
Birne Binegar: All you have to do is regard the function is being defined by matrix multiplication so for this a's and N by n matrix in order for this multiplication to be defined, you have to take X to be.
Birne Binegar: A column vector with the same number of.
Birne Binegar: entries, as you have columns of the first first factor, which would be in so m would have to be an m by one column vector, and so it would naturally fit inside rm but after you do the matrix multiplication you end up with a column vector that n by one, and so you end up over here so.
Birne Binegar: You simply have a matrix and rows and columns you can always create a linear transformation that takes you from R amp D and D have this rule.
Birne Binegar: Going the other direction, if you have a linear transformation between two vector spaces, you can use it to create a corresponding matrix that will have a similar property is this.
Birne Binegar: What we do is we apply the linear transformation to each of the standard basis vectors have a domain that will give us a column vector.
Birne Binegar: will write the result as a call and then what we end up with is an invite him matrix via this construction and I guess the point of that construction the explicit is.
Birne Binegar: This new function that you're we define sorry this new matrix that we define this way it's going to have the property that T fx is exactly equal to at effect.
Birne Binegar: So that will give us an alternative way of implementing this function, we can do it directly by whatever formula prescribes the end components in terms of the image point in terms of the end components of the.
Birne Binegar: original vector well.
Birne Binegar: We could also implement that same function just for using a matrix.
Birne Binegar: And do matrix multiplication.
Birne Binegar: OK, so the upshot of this is that linear transformation are basically one to one with matrices and which one you use in any.
Birne Binegar: particular circumstances just going to depend on that circumstance which is which presentation of linear transformation is more efficient What we do know about.
Birne Binegar: And we represent, when your transformations in terms of matrices that these guys are just ordered list of numbers so they're eminently calculable.
Birne Binegar: Right now.
Birne Binegar: In addition to having matrices associated with the linear transformation, we also have some sub spaces associated with linear transformation.
Birne Binegar: The first one so called range.
Birne Binegar: Let me draw a picture up here will be using throughout today's lecture.
Birne Binegar: distinguish between two cases function theory.
Birne Binegar: which you think is more general than.
Birne Binegar: linear transformation.
Birne Binegar: So this is going to be a special case of function theory, but we shall have the General said, you have a function.
Birne Binegar: F, it takes you from one set to another set called a set the.
Birne Binegar: call this set a.
Birne Binegar: We have some subsets of the domain and co domain just from this picture, first of all, you have.
Birne Binegar: liberty rider over here.
Birne Binegar: Just because you have a function between two sets doesn't mean that every point in the co domain is reachable by that function.
Birne Binegar: Inside your co domain is going to be.
Birne Binegar: what's called the image of your function and that's the set of points that you can think of sort of as the shadow of a inside be what I mean by shadow, is it.
Birne Binegar: Every point inside this shaded area your to regardless has been F of some a so here's the Omega hey gets mapped over here in Atlanta ends up inside this image all right now that's just from general function theory.
Birne Binegar: In linear transformation, rather than these.
Birne Binegar: sort of blobs these lens diagrams of set rack like dealing with vector spaces that are getting mapped to other vector spaces.
Birne Binegar: This might be on em and going on in and now we'll talk about linear transformation to special functions mapping between two vector spaces and.
Birne Binegar: What corresponds to this notion of an image of a function it's actually going to be.
Birne Binegar: A subspace of your co domain are in, but in the setting of linear algebra you don't have just some blob over here rather your image is going to be at subspace so represented a sort of a hyper plane.
Birne Binegar: And in fact we have a use a different name, we don't call it the image of the linear transformation, we call that the range.
Birne Binegar: Of theater all the points in the co domain your target space that you can reach by applying your linear transformation to a point in your domain.
Birne Binegar: Right so that's called the range.
Birne Binegar: and
Birne Binegar: What what we saw last time, is that we can actually calculate that thing directly if we can one the matrix that corresponds to our linear transformation and that subspace range of P, is going to be identical to the column space at the associated matrix.
Birne Binegar: Now there's another.
Birne Binegar: there's another special subspace.
Birne Binegar: attached to a linear transformation that doesn't really I mean I can develop a function theory analog of it, but I just assumed describe it directly, we have a subspace.
Birne Binegar: Of the Co domain.
Birne Binegar: That consists of all the points in the domain.
Birne Binegar: To get map to the zero vector in the co domain so starting here in the co domain you play the function to you brought over to the Co domain what we're focusing on here it's a subset of points in the domain that always land on the zero vector there that's called.
Birne Binegar: That should be a vector there that's called the Colonel the linear transformation and it's also relatable to the.
Birne Binegar: matrix that corresponds to the linear transformation it's In fact the know space of that matrix meeting to find this kernel you have this you can find it by solving this linear system.
Birne Binegar: OK, so those are the two sub spaces, I talked about on Wednesday, what I wanted to do today, initially just tell you what their dimensions are.
Birne Binegar: So.
Birne Binegar: Remember, if you have n by n matrix we had a theorem that said that the number of columns of this matrix is equal to the rank of the matrix plus the novelty of the matrix.
Birne Binegar: What are those precisely well the rank is either the dimension of the row space or the dimension will calm space is going to be the same number.
Birne Binegar: And the reason why the same number is because dimension is the number of faces factors number of independent directions, you need to reach every point in your subspace.
Birne Binegar: Well, the way we get basis factor for the row space is really grabbed the non to rows of a rule Echelon for every non zero is going to have a pip.
Birne Binegar: That you can counter dimension of this guy by simply counting the number of pivots in a row Echelon form of the of the matrix.
Birne Binegar: But similarly for the columns based the way we get a basis for that is we figure out where the pivots live in the row Echelon born, we go back to the original matrix and we grab the corresponding calls, but the number of columns that's gonna.
Birne Binegar: Number of columns were using it basis for the columns face of a is equal to the number of pivots and this role Echelon form so.
Birne Binegar: basis vectors for the row space and basis vectors for the column space.
Birne Binegar: are constructed two different ways, but they're starting point is the same you look at the pivots and the regression line form.
Birne Binegar: But you'll have the same number of basis, doctors, whether you're looking at the call and space to the row space, so the dimension that the same we call that common dimension, the rank of a matrix.
Birne Binegar: And then another space of a matrix well that's the solution set of a times X is equal to zero and.
Birne Binegar: The dimension of that space well can be equal to the number of free parameters, you have in this linear system after you solve it and.
Birne Binegar: Number three parameters exactly equal to the number of columns that do not contain it in the real life any row excellent point A.
Birne Binegar: And so what you should notice here, as I count the number of faces factors in each of the sub spaces they're all are related to the number of pivots we have in the row Echelon form.
Birne Binegar: To get the rank we can we count the number of columns with pivots to get the novelty we get the number of columns without pivots.
Birne Binegar: If I add these two numbers columns with pivots comes without pivots I get the total number of calls, so that when I have here and i'm just saying that like.
Birne Binegar: add the number of columns about pivots the number of columns with pivots I get it, I just some these two, so we have this relation between.
Birne Binegar: The number of columns up through matrix and the dimensions of these two sub spaces.
Birne Binegar: Well, similarly.
Birne Binegar: We have a linear transformation we're going to have a relationship between the dimensions of the range of tea.
Birne Binegar: upset subspace over here in the co domain and the dimension of the Colonel team that's a subspace of your of your domain and it turns out that the dimension of the Colonel plus dimension of the range is exactly equal to the dimension of your domain, which will be in.
Birne Binegar: And it's pretty easy to see that all we have to do to get this formula is express the range of tea in the kernel key.
Birne Binegar: In terms of the matrix that's attached to the linear transformation, what will have is m well what's em when I form like make the matrix corresponding this lemon only your transformation I get exactly in columns one for each standard basis director for me.
Birne Binegar: And then.
Birne Binegar: I can split those columns up into columns that contain pivots when I take the roadshow on format that matrix and columns that don't contain pivots when I take the relational and form of this matrix.
Birne Binegar: But this number number of columns of any row epsilon form about pivots that dimension of the column space, the rank or dimension, the columns based and number of columns will.
Birne Binegar: Without pivots is dimension of the solution, set at of X is equal to zero, with a dimension of no space, but the call calm space of this matrix is exactly the range of tea and the null space of that matrix is exactly the kernel of it, and so we do get that.
Birne Binegar: This identity that the range of these two particular sub spaces always has to add up to the dimension of the domain.
Birne Binegar: later on.
Birne Binegar: When we start interpreting linear transformation, this is sort of important and the way I like to think about it is.
Birne Binegar: The domain is an n dimensional space, but this function does not carry all the information about this implement n dimensional space over to this new vector space are, and it could lose some information well the subspace.
Birne Binegar: Here that get mapped to the zero vector that gets trivialized all these vectors here get mapped to this zero vector that really has no effect on on anything over here so.
Birne Binegar: I regard the the kernel of linear transformation that's the information that's lost when I moved from one vector space to another and then.
Birne Binegar: What remains is a range, and so the what you should think of the range of tsp it corresponds to the information or the dimensions that are carried over into the new vector space.
Birne Binegar: So this is sort of information loss and the information that passes through.
Birne Binegar: yeah.
Birne Binegar: Right right so that means that if you were simply wanted to know dimensions, you might have to calculate the dimension of the range, but then, once you calculate that you would just subtract that number from him, and then you would get the dimension of the kernel.
Birne Binegar: or vice versa, you can find the dimension of the kernel and then then subtract that number from em and you get the dimension of the range so that's really convenient it gives us some you don't necessarily have to find bases for both sub spaces to get their dimension.
Birne Binegar: Alright, so that's that's one thing I wanted.
Birne Binegar: to present to you today a second thing I wanted to do is to show you how you compose linear transformations so let me remind you first what it means to compose two functions.
Birne Binegar: Suppose i'm a function now we're talking just about general set of a function from a set a set be and another function from the set be to.
Birne Binegar: Set see I can combine those together well actually I can even be looser than this, all I really require is that the the image of F sits inside the domain of this function.
Birne Binegar: But.
Birne Binegar: Once I have that situation, I can define a new function that can take me from my original domain to the Co domain of the second function.
Birne Binegar: So i'll go directly from A to Z that's called composition is how you do it, you take a point in your domain you calculate.
Birne Binegar: Its value of effort that point that puts you in be but be isn't a this guy here it's going to end up being into the domain of G, so I can apply g to that point and I get a new value so here's here's the picture.
Birne Binegar: Three set a B and C function between a be a function between B and C and the new the compose function is where you go directly from A to C D E, F G F of a.
Birne Binegar: Okay, so we can this this works for.
Birne Binegar: In general function theory, it also works for linear transformations what I want to show you is.
Birne Binegar: What happens to the associated matrix under these composition rule and what we'll see is that the.
Birne Binegar: If I have to linear transformation linear transformation ghost say from our am to rp and another linear transformation that's going from our end sorry from our end to our end is T and our end to our P is.
Birne Binegar: linear transformation, so I can compose the tools, on your transformation and end up about linear transformation that goes from rm rp.
Birne Binegar: just doing again functional composition, the statement of the theorem here is that the associated matrix it's just going to you can attain that.
Birne Binegar: I calculating the corresponding matrices for these two guys and then just multiplying matrices together rather than figuring out exactly what this function is and then constructing its matrix.
Birne Binegar: So let me give you an example, so she more explicitly when i'm talking about.
Birne Binegar: Oh let's see up here, I have a function that they're taking two dimensional vectors the three dimensional vectors.
Birne Binegar: The second line, I have a function that are linear transformation that's taking different three dimensional vectors to two dimensional vectors when I compose them well that's done by substitution, as you would do in and say a calculus course.
Birne Binegar: bring us down a little bit.
Birne Binegar: To compute this count compose function, the first thing I do is I.
Birne Binegar: apply teed to a point with coordinates X one X two well that proof is this result and now what i'll do is i'll apply, so this thing.
Birne Binegar: And, but i'll use this formula for S, so the result of this is i'm gonna have to add the first two coordinates together, so the first coordinates going to be X two plus x one yeah it is, and then the second coordinate it's just the third variable sorry.
Birne Binegar: Where we end up.
Birne Binegar: yeah it's just the third variable X one plus x two so that's the composition and now that we have an explicit function taking a two dimensional vector to two dimensional vector we can calculate the corresponding matrix how men, let me remind you how that's done what you do.
Birne Binegar: Is a matrix corresponding to a given linear transformation you form that by taking the standard basis vectors of your domain, in this case domain is two dimensional so we have two standard basis vectors.
Birne Binegar: He and.
Birne Binegar: 01.
Birne Binegar: And what we do, then.
Birne Binegar: Is we figure out where this sector goes and we write it edge components vertically to get a call this matrix similar and we find out where this goes and we write it.
Birne Binegar: columns vertically so let's see one X one is one and X two is zero what happens over here we get zero plus one one plus zero or we get one, one that that explains this call, and the same thing is going to happen for the second.
Birne Binegar: One the second standard basis vector which has components zero and one it's going to get mapped to zero, plus one and.
Birne Binegar: Zero plus one or it's going to again equal one one okay so that's.
Birne Binegar: that's.
Birne Binegar: composing these functions by hand by substituting the value of tea into the argument of S and then working out the details, and then I have a linear transformation that I can compute it matrix.
Birne Binegar: What i'll show you next is that I could also calculate the matrices for the individual.
Birne Binegar: components of this compose function so let's see for the matrix for T, I take the first standard basis factor one zero I evaluated, that would be 011 until 011 and then for second column would be.
Birne Binegar: I guess 101 and yeah I got that right and let's see for s for standard basis vectors going to go to 101 plus 00 second standard basis vectors going to go to one and zero and the third standard basis vector that's what only why three is non zero is going to go to zero and one.
Birne Binegar: And what you'll.
Birne Binegar: Think page here, maybe.
Birne Binegar: No still there.
Birne Binegar: What i'll show you now.
Birne Binegar: This compose function is matrix to be determined from.
Birne Binegar: The matrix each corresponding of the functions from which is composed with keywords was formed and, in fact, if you take this matrix and the matrix for us.
Birne Binegar: And then you multiply by the matrix for key, not only is the matrix multiplication going to be defined it produces the correct matrix so what I get by multiplying these two.
Birne Binegar: matrix each corresponding to the linear transformation pfs together what I get is the matrix for the compose representation or compose function, so it all works out this um this business of going.
Birne Binegar: Creating linear transformation that matrices or matrix Jesus linear transformation that's all compatible one another.
Birne Binegar: The basic statement here is this composition of functions is now going to be at the level of matrices it's going to be matrix multiplication and that can be a lot easier than to compute.
Birne Binegar: To this matrix by multiply meet matrix together than it would be to now let's imagine a.
Birne Binegar: linear transformation, with a large number of variables so maybe from our tend to our 25 if I was going to compose.
Birne Binegar: function between two large matrices then i'd have to offer a lot of substitutions to do i'd have to subtract the value into the new function and then.
Birne Binegar: You know you have to substitute and see and see where everything goes and that's sort of an error prone calculation because it's more algebra rather than just manipulating numbers, but once we convert over to.
Birne Binegar: This matrix formulation.
Birne Binegar: it's easy to calculate the individual matrices you can basically just read it off the definition of the function and then they get the composition.
Birne Binegar: it's really easy just multiply those matrices together.
Birne Binegar: Okay, so.
Birne Binegar: It may be that next week i'll show you a practical example this.
Birne Binegar: One one thing I like to do that I don't think i'm going to do today is.
Birne Binegar: If you have a.
Birne Binegar: linear transformation.
Birne Binegar: I guess it looks like this sine theta cosine data if you have a linear transformation, whose matrix look like this.
Birne Binegar: What this matrix represents is up it's a matrix that.
Birne Binegar: implements.
Birne Binegar: A rotation.
Birne Binegar: High angle theta in the plane.
Birne Binegar: Hopefully, whether your physics or engineering classes you've seen rotations implemented the matrix multiplication you take any vector here X and y and you apply this matrix to that calling back to what you get is a new vector that would be this vector rotated by an angle data.
Birne Binegar: What you can do if you can take these two matrices and you can now let's call that a data.
Birne Binegar: And a.
Birne Binegar: fee would be a similar matrix.
Birne Binegar: data.
Birne Binegar: Sorry cosine fee sign free minus sign up.
Birne Binegar: see what you could do, if you could the combined action of a rotation by data, followed by a rotation by the angle be would be the composition of the to.
Birne Binegar: Non linear transformations or in this view would be adjusting multiply two matrices together and, in fact.
Birne Binegar: What should happen if you multiply by an angle theta and then you follow it by rotation by an angle fee, the total rotation is going to be theta plus fee right you go an angle theta and then you go further by an angle fee.
Birne Binegar: And so the total rotation would be fee plus D, if you multiply these two matrices together what's going to get kicked out by that are the trig identities that tell you.
Birne Binegar: What the cosine of fee plus data is actually going to be, but i'll demonstrate that you know, maybe next week it's a it's a nice way to drive a trig identity.
Birne Binegar: Just all you have to do is write down the matrices and multiply two matrices together and you get both the trig identities for sign of data, plus fees product does cosine and sine functions of D amp D and you also get the trig identities relating.
Birne Binegar: The cosine of a sum of angles from this simply multiplying these two matrices together if I do that calculation i'm gonna not finish today's lecture so i'll do that on Monday, maybe.
Birne Binegar: What I wanted to do.
Birne Binegar: today.
Birne Binegar: Is instead talk about invert ability of functions.
Birne Binegar: So.
Birne Binegar: If you have functioned between two set.
Birne Binegar: Remember that the image.
Birne Binegar: Of the function it's not necessarily the entire co domain images usually some subset.
Birne Binegar: it's all the points in the co domain that are reachable by applying your function to.
Birne Binegar: All the points in a you map them over in general, you get some subset of the co domain, we call that subset the image it's a in the case of linear transformation, we call this range.
Birne Binegar: But i'm going to introduce some new nomenclature here anytime we have a function.
Birne Binegar: Between to set, and in fact we can reach every point in the co domain by applying the function to some point in F, then we say function is subjective or onto What that means is that every point in the co domain is reachable by applying the function to some point in a.
Birne Binegar: This this notion it's going to be related to.
Birne Binegar: Whether or not this function from A to B actually has an inverse function.
Birne Binegar: So it's going to depend on two criteria, the first one is this notion of subjectivity.
Birne Binegar: Just basically says that.
Birne Binegar: Every point in the co domain is reachable by applying the function.
Birne Binegar: And then there's a second ocean.
Birne Binegar: and basic function theory called inject diversity.
Birne Binegar: that's a different condition.
Birne Binegar: We say the function between two such as objective if anytime the value of the function at two points is the same that in fact they had to come from the same point.
Birne Binegar: So what we're eliminating here, we talked about a function that's not objective.
Birne Binegar: is to have a point A and appoint a double prime function between the two set.
Birne Binegar: What we're eliminating here is that.
Birne Binegar: two distinct points in the domain getting mapped to the same point in the co domain by the application of the function so that's what we mean by not objective.
Birne Binegar: My objective, you only way the values can stay the same is is if they have to if they come from the same point.
Birne Binegar: Alright, so those are sort of.
Birne Binegar: abstract definitions about functions between two sets that are explained largely using pictures but what's so what's really surprising about it is that.
Birne Binegar: These notions of injected in church activity are sufficient.
Birne Binegar: Well, actually, let me give you some example these things here's my favorite example function X goes to X squared.
Birne Binegar: This function is neither objective or subjective it's not.
Birne Binegar: it's not.
Birne Binegar: subjective because.
Birne Binegar: The domain is consistent all real numbers, but when you square real number, you only get non negative numbers, so the negative numbers here, cannot be reached by applying this function to real numbers until the the this function is not subjective not.
Birne Binegar: The negative numbers in the co domain cannot be reached by applying this function.
Birne Binegar: norris that injected because.
Birne Binegar: The numbers one in minus one they land on the same value in the real numbers so because these two.
Birne Binegar: Values X have the.
Birne Binegar: have the same value as a as a function of F this thing is not injected, but these two properties subjectivity and objectivity, if you don't have them.
Birne Binegar: You can often.
Birne Binegar: impose them by simply restricting your domain, so the reason why.
Birne Binegar: The reason why this was not subjective is because this set was too big yeah it contains the negative numbers and the negative numbers for never reachable So if I restrict them the Co domain to the only non negative numbers, then this problem.
Birne Binegar: Having numbers in the code domain, that are not reachable by this function i've eliminated that i've eliminated all the numbers that are not reachable by redefining my co domain.
Birne Binegar: And Similarly, I can eliminate this duplicity the fact that one and minus one land at the same point one I can eliminate that by restricting my domain to just throw out all those negative numbers and so.
Birne Binegar: If you start off with a function that neither subjective or injected I read the funding your domains, you can typically make it that way, so.
Birne Binegar: And so.
Birne Binegar: But they're important properties because.
Birne Binegar: If a function is both injected in search active nutcase we call it by get did it has the property that it always will have an inverse function.
Birne Binegar: That there'll be a function F minus first that when I it's going to go from your original co domain, who that back to the original domain it's going to.
Birne Binegar: When you compose it with F you get back when you started with or you do the opposite combinations you go to F to be and then use the invert the F minus first you go back to a you get back to where you started with.
Birne Binegar: So.
Birne Binegar: here's the point of introducing all that abstraction it turns out that you're guaranteed and an inverse of a function whenever that function is both.
Birne Binegar: subjective and objective.
Birne Binegar: show you how that works, and in fact it's really it's it's an existence all argument I doesn't I don't actually calculate in this argument what this inverse is i'm just going to argue abstractly that it does exist yeah question, yes, so whenever there's a subjective value.
Birne Binegar: The value, does that mean you're values.
Birne Binegar: that's what you're eliminating.
Birne Binegar: If you have different values different values come from different points, naturally, but what you don't have two different points landing on the same point so you you yeah So if you start from two different this.
Birne Binegar: distinct point and the function is objective, that means when you map both point over they land on different points they don't court and they don't care less they sort of collapse into a single point.
Birne Binegar: So, so this is the situation that you will avoid when it's injected two different point landing on the same point.
Birne Binegar: and
Birne Binegar: right because, so if F of X equals X squared then F of minus one is equal to one, but at the same thing is equal to F of one right, so I have two different points in the domain they're getting mapped to the same point, so this is not injected.
Birne Binegar: So the reason i'm introducing this concept is when i'm really interested in this one, can I have a function from A to B when I can when can I go from the back to a.
Birne Binegar: And it turns out that I can answer that question simply by observing whether the function is over injected interjected here's the rationale.
Birne Binegar: suppose I so suppose I tried to define this function F terminus first that's going to actually be the inverse of this function, so that.
Birne Binegar: I can go pass back and forth between these two set well if it's injected if it's subjective anytime I choose a point be here, it had to come from something over here so that would call this i'll go back here, and maybe this is.
Birne Binegar: F a pay for some a.
Birne Binegar: And then.
Birne Binegar: I had another point here be prime that would be equal to F, say, have a prime.
Birne Binegar: For some.
Birne Binegar: A prime now what I want to do from this picture is define a function that will take me from being a well so here, I have two different point to be I don't want them.
Birne Binegar: landing on this point over here, because then, then I wouldn't sorry, you know if I start off with two different points I should end up to two points here and so these guys.
Birne Binegar: Can each be pulled back, but they can't land on the same point because, because the subjectivity if I start off with distinct points here they had to come from distinct point here.
Birne Binegar: And, but now I can go back and forth, this goes to here, and then it goes back to there, and this goes back to here, but then it goes back to there, and so, not simply from this drawing really you can prove the serum that the.
Birne Binegar: For every point here, there does exist, a unique point over here that that lands on that point and only one point over here can land on that point, because it's injected.
Birne Binegar: So you're guaranteed that it comes from some point and that's shared activity and then interactivity guaranteed guaranteed there's no you couldn't have.
Birne Binegar: A choice of point over here from which to call the inverse of a.
Birne Binegar: yeah so sure so far function has to have a unique value so if I had if I had two points landing on the same point here I wouldn't know whether to choose this as F, the inverse of this point or, this is the inverse of this point, but because it's a subjective.
Birne Binegar: I don't have to make that choice there's only one choice.
Birne Binegar: So you get this this theorem.
Birne Binegar: really just by looking at this diagram and arguing logically.
Birne Binegar: But dumb.
Birne Binegar: what's Nice is that.
Birne Binegar: will have a nice linear algebraic interpretations that situation.
Birne Binegar: that's going to be related to the kernel and range of a matrix.
Birne Binegar: So.
Birne Binegar: Remember linear transformation is the function so it's still inside the setting of this general function theory and all we're requiring us to be it's compatible with scale and multiplication in that tradition.
Birne Binegar: So the first thing was point out this now regarding a linear transformation as a function it's going to be subjective if, and only if the range is equal to the entire co domain.
Birne Binegar: that's that just definitions, the range of tea is exactly the same thing is the image of tea the tea is a function, rather than as a linear transformation and so.
Birne Binegar: function is subjective if it's co domain is equal to the image, but the image is just equal to the range the range equals the image is equal to the Co domain, then.
Birne Binegar: Then.
Birne Binegar: You have subjectivity.
Birne Binegar: i'm.
Birne Binegar: It doesn't have to equal em so, for example, you could have.
Birne Binegar: A two dimensional plane mapping on to the line it and maybe maybe you take it think of this as X and y space and the function is going to just map to the X coordinate well here you have a two dimensional space mapping to a one dimensional space, but it completely covers that space.
Birne Binegar: So, but you're you're you're you're pretty close to to an important point if if we actually have a linear transformation that's both subjective and objective it's going to have an inverse and then the then these two spaces are going to have to have the same dimension.
Birne Binegar: But.
Birne Binegar: Look till we have an inverse you don't know that yet.
Birne Binegar: Alright, so.
Birne Binegar: So this is this is just easy, this is just.
Birne Binegar: realizing that the range of T is exactly the image we think of it as a function between to set and that the definition of subjective simply saying is that the image is the same as the co domain.
Birne Binegar: Right so.
Birne Binegar: We have that and then.
Birne Binegar: linear transformation is subjective well that's the requirement that we don't want.
Birne Binegar: To different points in the co domain or two different points in the domain landing on the same point in in the co domain.
Birne Binegar: A claim here is that linear transformation is going to be injected if, and only if the kernel of that linear transformation consists of just the zero vector.
Birne Binegar: Get that up there.
Birne Binegar: Alright, so i'll prove this to you.
Birne Binegar: The goal here is to show.
Birne Binegar: That.
Birne Binegar: X equals X one equals.
Birne Binegar: two.
Birne Binegar: Only if.
Birne Binegar: X.
Birne Binegar: is equal to X two.
Birne Binegar: So this is.
Birne Binegar: linear transformation is subjective It means, if you have the same value, you had to come from the same same point in the domain and what i'm going to show you is that this condition is implied by the fact, if you if the kernel is actually equal to zero.
Birne Binegar: So the first thing i'm going to show you is that anytime you have a linear transformation you evaluated at the zero vector in your domain, you have to end up at the zero vector and the co domain why.
Birne Binegar: So here's T applied to zero, I can rewrite zero as X minus X, but because it's holding your transformation, I can delay that vector addition or.
Birne Binegar: subtraction until after I play the linear transformation and get that and then you're subtracting the same vector from itself, I get zero.
Birne Binegar: And so anytime you have a linear transformation, the zero vector always gets mapped to the zero vector, but this is the zero factor in like the domain, this is the zero vector inside the code domain.
Birne Binegar: I feel that's one observation.
Birne Binegar: And now, what i'll show you let's see here's the Colonel at the set of points in your domain that get mapped to the zero vector inside your co domain.
Birne Binegar: let's suppose we had a vector besides zero such that he of X is equal to zero.
Birne Binegar: Well, if P of X is equal to zero, we just seen in the previous slide that t of zero always equals zero and my hypothesis here is, I had.
Birne Binegar: Another vector besides here, or they get to map to the zero vector but if its objective.
Birne Binegar: That can happen i'm landing, I have two different vectors landing on the same spot, so the key is objective.
Birne Binegar: Two factors that land on the head the same value after the application of tea that cannot happen when you're injected so.
Birne Binegar: If X is not equal zero, then we would get a contradiction, so that means that the only vector that map to zero is actually the zero vector itself.
Birne Binegar: And that's just a statement that the kernel of T, the only vector that map to the zero vector has to be exactly the zero vector don't what i've shown on this slide is that P is injected then necessarily the kernel of tea consist of a single point the zero vector.
Birne Binegar: All right, let's go the opposite way supposed to Colonel P consists of a single vector zero vector.
Birne Binegar: And let's i'm going to now show you the function T is injected so suppose we, I have two points in the domain that land at the same spot, so the values of the linear transformation is the thing.
Birne Binegar: Well, I can move this term, to the opposite side.
Birne Binegar: end up with this term, to the opposite side I subtract.
Birne Binegar: T of X two from both sides of P of X one from both sides of that equation, you get this relation.
Birne Binegar: subtract one from the other you get zero is equal to the difference and then because it's linear transformation, I can delay that vector subtraction I can actually carry that out before I applied the linear transformation.
Birne Binegar: And then, when I have is that zero has to equal.
Birne Binegar: He applied to X one minus X to my hypothesis here is that the kernel of tea consists of a single vector zero so that tells me that.
Birne Binegar: If the kernel of T is equal to zero and T applied to X one minus X two is equal to zero, that means that Well, this is the only guy whose value evaluates to zero, so that means X one minus X two half that equals zero well that's true, that means that X, one has to equal X two.
Birne Binegar: So what's The upshot of this.
Birne Binegar: If T is.
Birne Binegar: subjective so here's T from our am sorry, in general, linear transformation if P s.
Birne Binegar: Is subjective.
Birne Binegar: Well, that corresponds to a range of T.
Birne Binegar: equaling the entire co domain.
Birne Binegar: saying the set is entirely covered by applying T and tea is injected.
Birne Binegar: With this slide shows you that happens exactly when.
Birne Binegar: The kernel of tea.
Birne Binegar: consist of only the zero vector.
Birne Binegar: And so.
Birne Binegar: If both these conditions hold or equivalent Lee these two conditions whole than T is going to be both subjective and injected, and so it will have an inverse so there.
Birne Binegar: If he from our m to our end yes.
Birne Binegar: one.
Birne Binegar: You have the range of T.
Birne Binegar: equals.
Birne Binegar: The entire co domain and, second, of all.
Birne Binegar: That subspace Colonel a P is equal to just the zero vector.
Birne Binegar: Then you have an inverse lip transformation.
Birne Binegar: All right now, what good is this to us if we have a function between two vectors faces, I know how to calculate this the range of tea just turned out to be the column space of the associated matrix.
Birne Binegar: The kernel of tea that turns out to be the null space of the associate matrix so once I have linear transformation, I can easily determine these two sub spaces and then I can define.
Birne Binegar: I can see whether or not the linear transformation actually has subspace I just have to calculate those two things and verify, I get the right result, a sense yeah so both candice both these conditions.
Birne Binegar: Are calculable.
Birne Binegar: From the associated matrix.
Birne Binegar: Until I don't actually have to explicitly construct this this is inverse matrix and know that it exists, rather I could just look at the associated matrix verify that.
Birne Binegar: The range isn't the all of the co domain and that the kernel of T consistent only the zero vector, then I know that that exists.
Birne Binegar: Okay, so oops I ran three minutes over so let me stop here.
Birne Binegar: and wish you guys a.
Birne Binegar: restful spring break and we'll pick up again.
Birne Binegar: Next Tuesday Oh, by the way, I said that the second exam was likely to be.
Birne Binegar: The week, right after spring break i'm gonna relax that I think it's actually more likely that we do it, maybe Monday.
Birne Binegar: Monday the week after the week we come back from spring break because I want to go over this material better on on the Monday afterwards and.
Birne Binegar: I didn't just assume give you another class meeting to sort of consolidate things before I test you on it so what's likely to happen after spring break is we'll come back and i'll.
Birne Binegar: rehash some of this material, maybe go over homework problems in the like on Monday, maybe do new material on Wednesday, and if we do have the exam on the following week we can do a review session on Friday, I think that that would work out best for everyone okay.
Birne Binegar: All right, so have a good break and i'll See you in a couple of being a week or so.