Birne Binegar: Okay, I guess we'll get started, so today what i'm going to do is first of all i'm going to review real quickly the topics we covered on Wednesday.
Birne Binegar: About sub spaces attached to a matrix in there dimensions of those substances and then i'll go do a long well not too long, but next any the example.
Birne Binegar: That can be very much like the problems on homework set six which by the way, is due on Monday do in the sense that you want to keep up with the course, you should turn do the problems by Monday, but of course i've on web assign itself, I put the due date all the way at the end of.
Cole Aldridge: The semester.
Birne Binegar: So that students can turn them there.
Birne Binegar: The homework when they're able to but i'm just take concurrent where the course you try to get those problems done by Monday and i'll show you today how simple, they are.
Birne Binegar: And then, well then we'll move on to a new topic.
Birne Binegar: functions be between vector spaces and back will specialize very quickly to special family functions called linear transformations which.
Birne Binegar: You should come to understand is just this sub spaces are very special subsets of a vector space they behave well respect to the vector space operations.
Birne Binegar: So too will be looking for functions to behave well respect the scale of multiplication and vector addition and those functions will be linear transformation so i'll introduce that topic today, we will spend most of next week, talking about linear transformations.
Birne Binegar: And then, if you actually have been.
Birne Binegar: Looking at the sequence of lectures are looking at the syllabus you'll note that after we cover linear transformation that's about when the second midterm is to occur.
Birne Binegar: So what i'm thinking is is it probably will spend next week on linear transformations and there'll be a homework assignment associated with that the following week it's going to be spring break so i'll have an exam then.
Birne Binegar: Now what the exam will occur, the week after spring break most likely i've already had one request to not have it on the Friday after spring break I think that's the 25th of March, so it could, but on the other hand, since I don't want to have it on the Monday after spring break that's.
Birne Binegar: We do a review beforehand that means there's a week before the review and the actual exam so it could be that one today, which I think is march 23 that could be a possible date for the second exam, but it could also I could push it.
Birne Binegar: To the following weekend as well, but if I do that then i'm just going to start covering new material material that won't be explicitly on the second exam but what would be covered up on on the second exam will the material up to linear transformation, I think.
Birne Binegar: Okay, so any questions on that where we're going.
Birne Binegar: Okay, so um and once I talked about.
Birne Binegar: ascribing sub spaces to a matrix there's three different ways we can do that through standard ways anyway, given a a matrix with and rows and columns.
Birne Binegar: i'm going to not only tell you what the sub spaces are, but how to find the basis the sort of Nice directions inside the subspace that are capable reaching every point in the subspace.
Birne Binegar: In a unique way will seek a linearly independent set of factors, such that when you look at all possible linear combinations they actually generate the subspace in question.
Birne Binegar: So what i'll do i'm going to do in the next series of slides is talk about the sub spaces, we can attach to a matrix and then how to find their bases.
Birne Binegar: What I showed you on Wednesday is that the find the bases for any of the sub spaces, but a central part of the calculation is to compute a reduce row Echelon format that matrix.
Birne Binegar: So the first one is the row space of a matrix it's the span of all the road vectors so you know.
Birne Binegar: Each of these ordered list of numbers, I listed here are the entries of a row of a matrix.
Birne Binegar: Then what you could do once you have these factors, you can take all possible linear combinations that's the span and then it creates a subspace.
Birne Binegar: If you notice the Rose all have em components that means each row vector is going to be an n dimensional vector and still the space that's generated by taking linear combinations these guys is going to be live in the empty or m set of all ordered list of em numbers.
Birne Binegar: Okay, so that's the subspace and the way we find a basis for it, if we take the matrix and rewrote reduce it to.
Birne Binegar: Its devices to go to a row Echelon form but for some of these other sub spaces, we have to go all the way to reduce rational on form, so what i'm.
Birne Binegar: calling a double prime is the fully reduced version of the original matrix a reduced row Echelon form, and so the way we get a basis for the road space of as we reduce a to.
Birne Binegar: relax Echelon form or reduce rational and porn and grab the non zero rose.
Birne Binegar: When you reduce you could end up with some zeros at the bottom, those are the Rose you ignore the non zero rows are the ones you'll keep when you try to find a basis for the road stays.
Birne Binegar: Right now, we have the columns space constructed in a similar way we write all the column vectors and then we take every possible linear combination of them that creates a new subspace.
Birne Binegar: But since the columns are going to have as many entries, as we have rows the column space of an n by n matrix is going to be a subspace on our end.
Birne Binegar: Now, when we rub reduce.
Birne Binegar: The same matrix a to reduce your national and form we can use that calculation not only define a basis for the road space, but also basis for the column space.
Birne Binegar: But it's not going to be the non zero columns of The reduce reliance on on rather the prescription for finding a basis for the calm space, and you, you go from a.
To.
Birne Binegar: Reduce relational on board.
Birne Binegar: And then.
Birne Binegar: here's where you'll have zero road you'll have columns with pivots columns about pivots we have we've analyzed matrices in Russia and porn for a long time now, what we do here is we find.
Birne Binegar: A columns.
Birne Binegar: Of the reduce your actual on or that have pivots.
Birne Binegar: Now fully we're solving a system of linear equations the columns of The reduce your show on phone they have pivots they correspond The pre variables.
Birne Binegar: And so, essentially when as soon as you find these columns you're also a flank actually finding the fixed variables.
Birne Binegar: If you pivot contains the column contains a pivot then it's a fixed variable columns with outfitters are free variables so effectively we find the free variables here and then.
Birne Binegar: What we do is we go back to the original matrix a and we grabbed the corresponding columns.
Birne Binegar: And these columns of a will be sufficient to generate the entire column space of bay and more over there automatically be linearly dependent so they'll provide a basis.
Birne Binegar: Alright, so that real reduction calculation taking a day the already determined real quickly over basis for the roast base of a in a basis for the column space of a and, of course, the reduce rotational on porn with a is also essential to finding.
Birne Binegar: A basis for the no space of a, that being the solution, set up a set of linear equations that have a as a coefficient matrix.
Birne Binegar: The way you find the basis for this subspace is well you solve the equations.
Birne Binegar: And the only new thing you have to do is interpret those constant factors that are being multiplied by the three parameters of the solution.
Birne Binegar: interpret those constant vectors as the basis vectors for this subspace so as soon as you write down the solution of the linear system as a hyper plane then you'll have.
Birne Binegar: A variable X free variable say X three times a certain content vector maybe another free variable X four times a different content vector it's those constant vectors that are going to provide you with the basis for the North space.
Birne Binegar: Alright, so that's how we find bases me um let me supplement this by.
Birne Binegar: First of all, reviewing what the dimension of them subspace is.
Birne Binegar: Well, these basis vectors there, the number of independent directions, you have in your space so it's natural to think of them as a representative dimension, indeed that's the definition of the dimension of the subspace so number of vectors you have any basis what's.
Birne Binegar: hidden in this definition, are two theorems the first theorem is is that the number of vectors you have well, first of all, every subspace has a basis.
Birne Binegar: And the second the theorem is that every basis for a given subspace has the same number of vectors so, because every sub base has a basis and every.
Birne Binegar: Every basis has the same number of vectors we can ascribe a number of basis vectors as the dimension of any subspace.
Birne Binegar: All right now for the subspace is attached to a matrix i've already shown you how to count how to find the basis factors, all we have to do now let's count them.
Birne Binegar: And so, for the row space we found a basis for the road space, by getting a row excellent form of a grabbing the non zero rose.
Birne Binegar: Well, a non to row is going to have a non zero first entry and so it's going to have the pivot and so, in fact, the dimension of the row space is just the number of pivot in any real Echelon form.
Birne Binegar: But actually that same number is it the dimension of the column space of the matrix why well the column or.
Birne Binegar: a recipe for finding a basis for the column space is to figure out where the pivots live in the real life long form and go back to the original nature, you can grab those columns.
Birne Binegar: So for every pivot you have here you're going to have a basis vector for the column space, so these two numbers are actually the dimension, the Rose basis always equal to the dimension of the column space, because both are equal to the number of pivots in a row Echelon born.
Birne Binegar: Right and so.
Birne Binegar: Then, lastly, we have no space.
Birne Binegar: well.
Birne Binegar: Think of what it means to solve a linear system, we take the matrix we reduce it to reduce volitional on form, and then we expand.
Birne Binegar: Our solution in terms of the three parameters, how many free parameters, do we have well as many as we have columns of the row Echelon form that don't have pivots.
Birne Binegar: So, in fact, the dimension up no space it's going to equal, the number of free parameters in the solution.
Birne Binegar: Which is going to equal, the number of columns that don't have pivots so here i'm counting the columns that do have pivots in here the columns that don't have pivots so from these numbers.
Birne Binegar: What we find first of all is the row space in the column space they even though they live in different vector spaces.
Birne Binegar: If you have a is equal to anyone one a one m a n 1am.
Birne Binegar: Just putting in the corner entries, this is an n by n matrix the row space.
Birne Binegar: Well, it consists of n dimensional vectors.
Birne Binegar: So that's going to live inside the vector space or m the column space each column is going to have as many entries that we have rows so the columns space.
Birne Binegar: is going to be a subset of our in these two subspace is live in different vector spaces, if our row and call them dimensions or differently, yet.
Birne Binegar: The hyper plane that they generate that that the actual subspace sitting inside.
Birne Binegar: Whatever vector space contains it, they have the same dimension, so I have.
Birne Binegar: mentioned these two sub spaces, even though they live in different vector spaces, is always the same, and the dimension of this guy well, you can actually get it as soon as you know, the dimension of either of these.
Birne Binegar: So.
Birne Binegar: No it's not because their unit directors of the reason is is because we have a fixed number of columns and it's a zero sum game.
Birne Binegar: Some of those columns are going to contain pivots summer not going to contain pivots and so when you add both of those numbers up you get the total number of columns.
Birne Binegar: But actually you know it says very simple identity that just comes from this description of how we find the basis factors and.
Birne Binegar: Once we got that recipe simple the count how many vectors we're going to get and then we see this identity which is.
Birne Binegar: This again become important later on next week, when we talk talking about linear transformations and but suffice to say right now it's just that we have these three sub spaces attached to a given matrix ease in their relationships, amongst other dimensions.
Birne Binegar: The first two are exactly equal and the third one, even if I didn't calculate this I would know it's dimension, simply by knowing the dimension of, say, the aerospace i'll have to do is subtract from the number of columns of a number of.
Birne Binegar: basis factors of the road states and then i'm left with a naughty.
Birne Binegar: Okay, so.
Birne Binegar: That much is was meant to be or reduce the things we talked about on Wednesday, or I want to do, primarily started lecture today is to work through an example.
Birne Binegar: The side kind of questions that are posed and web assign problems that six where you're typically given a matrix and you're asked to find a basis for the row space basis for the columns space let's.
Birne Binegar: list the things a basis for the no space and then we'll also find the dimension of the row station, you mentioned the call today so so called rank of a and the dimension of this subspace and all space that's called a note here.
Birne Binegar: So the rank is.
Birne Binegar: The dimension of either one of these spaces and the novelty if the dimension of the no space.
Birne Binegar: But the let's proceed when i'm giving you here in this problem it the meat of the calculation.
Birne Binegar: The row reduction, where I take this matrix and I play elementary real operations to like reach it reduce relational on form that i'm not displaying this example, but.
Birne Binegar: After the first exam you're all supposed to know how to do the road reduction.
Birne Binegar: Actually, prior to the first exam you should know that but um.
Birne Binegar: So what i'm going to do, simply assume is that this three by three matrix each row reduces to this matrix it that's the brunt of the calculation.
Birne Binegar: What i'm going to do on this next series applied is use this result to explicitly write down and basis for the growth based based on the columns basic cetera.
Birne Binegar: Okay, so this top line it's nothing new i'm just repeating my my assumption that a row reduces to a given matrix.
Birne Binegar: Now, how do we get a basis for the row space of this matrix I look at a row Echelon one or the reduce row excellent warm and I see two non zero rose.
Birne Binegar: Those non zero roads are going to be a basis for the road space until all I have to do is identify them write them down and put them inside a.
Birne Binegar: curly braces to indicate a set of vectors so there's our basis for the road space.
Birne Binegar: By the way, we can see that the aerospace is two dimensional right because we have to basis actors.
Birne Binegar: All right, what about the column space well the same calculation, but just the end part of the calculation changes once we've reduced a to.
Birne Binegar: A matrix general Echelon warm the recipe for finding a basis factor for the column space if the look at where the pivots live oh here's a pivot there's a pivot in the third column here, there is no pivot.
Birne Binegar: You might want to try and put them in there, but this is a zero entry pivots are always the first non zero entry don't In fact this matrix rule excellent form only has.
Birne Binegar: To pivot but go back to the original matrix first pivot and the first column i'll grab that first vector the second pivot and the second column means I grabbed the second column vector of a and I use these first two columns as a basis for the column space.
Birne Binegar: So, and again it's two dimensional why well because we had two columns from which we were going to.
Birne Binegar: Well, we have to pivot that are going to lead us to two different columns of the original vector a, but what we should note is, we have the same number of.
Birne Binegar: basis factors, as we have non zero vectors to and so that's why the dimension of the road space is always the same as the dimension of the columns face.
Birne Binegar: Right and then the null space okay So here we have to do a little bit of extra work but it's we're familiar with this business, if you have them.
Birne Binegar: You want to solve a linear system.
Birne Binegar: You form the augmented matrix and you reduce it to reduce real Echelon born.
Birne Binegar: Well, what I gave you on the preceding slides with the reduce rational and foremost the coefficient matrix not the augmented matrix but there's very little difference between these two.
Birne Binegar: If you know what the real Echelon form of a is then the role isilon form of a joins with the zero vector is just going to be watchful unfortunate a a join with zero vector.
Birne Binegar: Because the operations that take a to this guy or are not gonna have any effect at all, and this last column.
Birne Binegar: Alright, so what did we end up with you end up with this augmented may free to reduce cholesterol on form which.
Birne Binegar: Is the augmented matrix corresponding to the solution of the original into your system.
Birne Binegar: So we stare at this guy and and see that column one in column two again the columns will pivot these are tied to the fixed variables in the solution.
Birne Binegar: The third column has no pivot so it's going to be associated with a free parameter X three and write down the equations.
Birne Binegar: corresponding to each row with this augmented matrix I move the free variables over to the right hand side now, I have the fixed variables expressed in terms of free variables.
Birne Binegar: I use these equations to write down a solution vector substituting where I can for the fixed variable and then, finally, what I do is I expand my solution vector in terms of the pre parameters, so this factor is simply X three times this constant vector.
Birne Binegar: Constant vectors what we're after here we're looking for a basis for the no space, and here it is it's a single vector in this case, if we had two free parameters, then we would have had another vector to deal with.
Birne Binegar: All right, so what happened here, well, actually, we can also see that the dimension of the no space is one because we only have one basis factor well.
Birne Binegar: And then, here we have a column for pivot call with a pivot that common sense, we have to pivots the rank is equal to two, we have one column without a pivot That means that the novelty, the dimension of space it's just going to equal one in there and there's its basis factor.
Birne Binegar: Alright, so um.
Birne Binegar: So yeah this kind of fell off the bottom, but the sector that kind of fell off the bottom of the page is exactly this minus B minus five three and one.
Birne Binegar: Right and then yeah now the rank of a which is he does it dimension of the row space or the dimension of the column face because I had to pivots and the raw salon form it's the rank is to and the novelty.
Birne Binegar: it's number of columns about pivots we had a single call that a pivot yet rotational on form so which one.
Birne Binegar: All right.
Birne Binegar: All the problems on homework sets it.
Birne Binegar: Up you basically doing this analysis, over and over again, and hopefully.
Birne Binegar: Once you've done homework that's it you don't have to worry about this so much it's pretty easy to do, once you know what to look for.
Birne Binegar: Okay, so.
Birne Binegar: Any questions so far.
Birne Binegar: Okay, so what i'm going to do.
Birne Binegar: In the remaining time today is introduce a new topic.
Birne Binegar: The whole purpose of them talking about southern spaces was well up to the first exam we dealt some calculations tools, but after the first exam what i've been concentrating on i'm.
Birne Binegar: sort of filling out the mathematical framework of linear algebra I began by talking about special sets now the so called sub spaces and we in fact we never really consider sets that are any more complicated than sub spaces and linear algebra.
Birne Binegar: Well, the next thing if you have sets next thing you might want to deal with mathematically you talk about functions between sets, how do you relate sets and turns out justice sub spaces are very special kinds of sets they're pretty simple kinds of sets as well.
Birne Binegar: will be justice restricted when we consider functions in linear algebra will only get to look its functions that are really well behaved with respect to the two fundamental vector space operations that scale multiplication index tradition.
Birne Binegar: So, first let me just establish some nomenclature about functions, if you have a function from one set to another, we usually call.
Birne Binegar: The place where you start a set which you start we usually call that the domain and the set that you end up with after applying the function, we call that the Co domain.
Birne Binegar: Oh, I introduced from the start, just so that I can point to the sets and call them Pacific specific names, but here's the picture function is just a rule that are given X inside your domain it's a rule that identifies a unique element of your co domain, we indicated by apply to X.
Birne Binegar: Alright, so.
Birne Binegar: um what am I going to talk about next is a very special family functions when X is a vector space, why is it, why is the vector space.
Birne Binegar: So a linear transformation it's going to be a function be here are our vector spaces, what are list of em numbers in order list and numbers.
Birne Binegar: And we'll see any function that takes an ordered list of em numbers and produces an ordered list of n numbers it's going to be a linear transformation if it has a property that I can do scale multiplication before or after I apply the function and I get the same result.
Birne Binegar: Or, and I also use this is making sure that the function is well behaved with respect to scale a multiplication.
Birne Binegar: Similarly we're going to demand that the function is really well behaved with respect to vector addition meaning, I can add vectors and apply the function, but if I do, that I get the same result as if I applied the function to the individual vectors and then added them.
Birne Binegar: So one thing you want to notice here is that biggest most obvious in.
Birne Binegar: The second case this is vector addition but it's occurring you're adding n dimensional vectors to one another over here you're already landed in this target space The co domain.
Birne Binegar: And now you're adding in dimensional vector so it's it's actually also in saying that somehow the two fundamental operations between these two vectors faces are linked to one another.
Birne Binegar: And so i'll frequently refer to these two properties of saying linear transformation preserve scale or multiplication meaning this identity hold and they preserve vector addition meaning this identity holds.
Birne Binegar: All right, so.
Birne Binegar: One thing that's nice about linear transformations is.
Birne Binegar: To be a linear transformation, it has to behave well respected back tradition and scale multiplication.
Birne Binegar: When I take a linear combination of factors I do generally do both scale multiplication and vector addition.
Birne Binegar: And what's going to happen here is that if you have a linear transformation, not only is it going to behave well respect to just doing vector offer vector edition, or just doing to get scale and multiplication sorry I should have a multiplication.
Birne Binegar: it's also going to be behave nicely for respect the linear combinations of vectors.
Birne Binegar: So let me show you what i'm saying here suppose we have the and.
Birne Binegar: Maybe i'm expanding it in terms of them some other vectors let's call it w and a call to BC one plus C to the tube plus ck Ek so here's a particular vector that's represented as a linear combination of K other vectors if T.
Birne Binegar: From say or in.
Birne Binegar: rm tomorrow and is a linear transformation.
Birne Binegar: Then what's going to happen is that he applied to w.
Birne Binegar: it's going to be well he applied to see one be one dot dot dot plus ck.
Birne Binegar: UK but now, because not linear transfer nations behave well respect to that tradition, I can delay the vector.
Birne Binegar: sons, I can delay them until after I applied the transformation, then I haven't seen One the one plus T see to the two plus similar terms until I reach the last guy less seeking.
Birne Binegar: Clarity plus.
Birne Binegar: ti si K times DK.
Birne Binegar: linear transformation also behave nicely respective scale or multiplication so I can fact pull out these constants as well T.
Birne Binegar: The one plus C to D of the two dot dot dot plus C K T V K so in some sense linear transformation kind of distribute themselves over linear combinations.
Birne Binegar: This is basically when you have a linear combination of factors and you act on it like a linear transformation, you get another linear combination of vectors.
Birne Binegar: which you could have obtained by applying the transformation individually, the vectors V1 V2 v3 and then doing them the scale of multiplication after you figured out where be one goes in the two goes.
Birne Binegar: So.
Birne Binegar: So that's that's displayed on this line I see it as a limb, you have linear combination of vectors and you apply linear transformation you end up with a linear combination.
Birne Binegar: That works really the same linear combination, you have exactly the same coefficient it's just that be the vectors be I have been replaced replaced by their values rather than the vectors itself.
Birne Binegar: OK.
Birne Binegar: Though.
Birne Binegar: kind of function to we're going to be dealing with.
Birne Binegar: I haven't yet given you an example of a linear transformation well, let me show you some a couple things that are not linear transformation.
Birne Binegar: First example.
Birne Binegar: nonlinear transformation.
Birne Binegar: let's take a favorite function from calculus.
Birne Binegar: Say F will think of it as a one dimensional vector space going to one dimensional vector space, but the way I phrase, it is that X is getting map to X squared.
Birne Binegar: So this function very basic function, you see very early on in calculus one, and it is what example a nonlinear function right because we're squaring the variable rather than having it appear linearly.
Birne Binegar: This is not going to be a linear transformation so, even if I interpret business vectors.
Birne Binegar: like this by putting it inside an ordered list, even though it has only one element, even this it's not going to be a linear transformation because, for example.
Birne Binegar: F of Lambda X.
Birne Binegar: Well, what it should be yes Lambda X squared which is going to be Lambda squared X squared, on the other hand, Lambda times the value of F of X is going to equal Lambda times X squared and because this is squared, and this is not squared they're not equal and so.
Birne Binegar: even have the most non trivial First Non trivial function.
Birne Binegar: First Non trivial polynomial it's not a linear transformation and, of course, you got head cubes or ahead assign function if I had an exponential function, I could not use it to construct a linear transformation even between these really simple vector spaces, or one.
Birne Binegar: However.
Birne Binegar: If I had like two X here, I had linear function.
Birne Binegar: Then, in fact, it would work my head.
Birne Binegar: If F was actually a linear function.
Birne Binegar: Then it would work.
Birne Binegar: F of Lambda X, in this case, would be to Lambda X.
Birne Binegar: Two times Lambda X, on the other hand, Lambda times F of X, if this is the rule would just be equal to Lambda times to wax and these are the same.
Birne Binegar: And so.
Birne Binegar: One basic idea that one in order to generate examples of linear transformations might be tried to generate this example generalize this example to higher dimensional spaces.
Birne Binegar: But I actually chose a pretty special linear function here if I change that slightly suppose my function F sent X two two X plus one well it's going to turn out that this constant term here it's going to destroy it, if I try to interpret this as linear transformation because F.
Birne Binegar: Lambda times X, we end up being two times Lambda X plus one wow, which is to Lambda X plus one well Lambda times F of X is going to be Lambda times.
Birne Binegar: Two X plus one it's going to equal to Lambda X plus Lambda times, one is one, so these things are not equal so it's almost right That way we.
Birne Binegar: can create a linear transformation between two vector spaces, is to use linear functions that Matthew from one to another, but in fact we have to be even more restrictive we can't allow any constant terms, either.
Birne Binegar: So then, how in general we're going to.
Birne Binegar: set up a linear transformation.
Birne Binegar: Well, first of all, what we need is a way of writing down a function between two spaces, so he is going from.
Birne Binegar: Our in.
Birne Binegar: Our m let's think of points in this space.
Birne Binegar: As having coordinates next one through xm and they're getting sent well they're gonna have to end up in an n dimensional space sorry and N dimensional space and so on, when i'm going to need are a total of n components in the results here let's say the first component is call it.
Birne Binegar: yeah, let me just call it F1 but there's kind of depend on the variables in the domain it's going to depend on X one.
Birne Binegar: X m but that's only the first component of the result and vector I need a second component also depending on the original coordinates X M and will have similar components until we finally.
Birne Binegar: reached the total dimension of it co domain, then we'd have it function F and depending on X one X m.
Birne Binegar: My statement here is is that if these functions F one F m are all linear functions of their variables and they don't contain any constant term.
Birne Binegar: Then, this will be a linear transformation, but if I had anything more complicated for these component functions F one F n anything more complicated it's been ball defined function of involve.
Birne Binegar: cosine functions exponential function, the square root yeah I would not get linear transformation at that these are really special by linear function.
Birne Binegar: So F is linear.
Birne Binegar: Yes, when you.
Birne Binegar: write it in terms of its variables.
Birne Binegar: It looks like some number of times the first variable some number times a second variable in fact what you what you what i'm writing down here it's a linear combination of variables.
Birne Binegar: And m m now for linear function.
Birne Binegar: I could also add in a constant tyranny which still be a linear function, but these things, these constant terms are not allowed.
Birne Binegar: The constant term.
Birne Binegar: not allowed for a linear transformation.
Birne Binegar: But that's how you make them you take linear functions of a given a list of variables and then you arrange them as as a vector and then.
Birne Binegar: However, many components, you have how many, many functions, you have going to give you, you got one for each component and that's going to give you a linear transformation from your original set of variables to however many components, you want to include here.
Birne Binegar: alright.
Birne Binegar: So even this it's not very explicit example.
Birne Binegar: So i'll give you an explicit example I had.
Birne Binegar: T of X one X two.
Birne Binegar: equals.
Birne Binegar: X one minus X two X one plus x two X two, this would be a rule that.
Birne Binegar: sent a two dimensional vector to a three dimensional vector.
Birne Binegar: This component functions are all linear functions, they don't have constant terms, this will be a linear transformation.
Birne Binegar: Okay.
Birne Binegar: But in fact.
Birne Binegar: Even though this is a simple way to write down examples of linear transformation, this is usually not how they occur in linear algebra.
Birne Binegar: i'm going to give you another construction of linear transformation.
Birne Binegar: let's see if I.
Birne Binegar: OK, so I sort of.
Birne Binegar: As I was working on the blackboard I worked my way a little bit ahead of the slide here at this stage i'm just talking about.
Birne Binegar: I have a function between two vectors spaces like take two variables and i'm get it gets mapped to a three dimensional variable.
Birne Binegar: I have nothing but linear functions here, this is a linear function of tea of essence a linear function of tea.
Birne Binegar: And this is a linear function key of F, but because of that constant term this particular function is not going to be a linear transformation and I just show you the calculation that we do, and you do the scale multiplication.
Birne Binegar: Of this function, you did scale multiplication before you get one result.
Birne Binegar: The constant in the last point is one if you delay the scale of multiplication till the very end that constant term and the last component turned out to be the scale or multiple not one, so this is not a linear transformation.
Birne Binegar: But.
Birne Binegar: here's the example I want to get to this is the most common way that we're going to see linear transformation inside linear algebra is that we can attach a matrix to them.
Birne Binegar: Actually, the attachment the the creation of a matrix corresponding with linear transformation is something i'm going to cover on Monday today we're going to do i'm just going to show you, you know n by n matrix how do I construct a linear transformation.
Birne Binegar: So let's start with it.
Birne Binegar: A matrix which.
Birne Binegar: columns and rows.
Birne Binegar: And what i'm going to do is, I have a.
Birne Binegar: vector.
Birne Binegar: With em components, then a X is define.
Birne Binegar: him components here because.
Birne Binegar: They do for this product to be fine i'm multiplying in n by n matrix time to column vector and by one i've got the match and I need.
Birne Binegar: And what's going to get kicked out by this is a column vector.
Birne Binegar: And components and also enrolled in a single call alright so.
Birne Binegar: What I want you to see here is that the matrix multiplication I can take a list of em numbers and I can produce a list of n numbers.
Birne Binegar: And in fact that's exactly the way we create a linear transformation okay here's the matrix multiplication.
Birne Binegar: What I do it's all defined a linear transformation, corresponding to this matrix by seeing it in action on a given n dimensional vector is going to be the vector that's produced when I do matrix multiplication.
Birne Binegar: So what's going to happen, I have an n dimensional column vector be multiplied from the left by an n by n matrix it's going to produce an end by one column vector or an n dimensional vector.
Birne Binegar: And so I get a map that takes me from em dimensional vectors to n dimensional vectors that's a function.
Birne Binegar: And it's easily seem to be a linear transformation because scale multiplication commute with matrix multiplication so here i'm applying the definition of this function.
Birne Binegar: And the next in the middle i'm using the fact that matrix and scale multiplication commute with matrix multiplication and then i'm identifying a of X is just being here subscript a of X and I get the identity, I want.
Birne Binegar: scale multiplication before applying the function you'll see the same result is when I delay the scale of multiplication you till after I played the function.
Birne Binegar: And for.
Birne Binegar: Preservation of now, or the.
Birne Binegar: Conservation of vector edition that's also clear from this for this function, because when I.
Birne Binegar: apply the function to a summer vectors that's the same thing, according to this definite as multiplying the some of these two vectors from the left by a matrix egg but that matrix multiplication is going to distribute over the some and then I see that I get.
Birne Binegar: The result of applying this function to some of factors is equal to some of the of the values of the two vectors.
Birne Binegar: Okay, so.
Birne Binegar: I think what i'll do is stop with the theory here.
Birne Binegar: will delay that until next week, but what I want to do.
Birne Binegar: sort of.
Birne Binegar: leave you with before we get back to the stuff stuff on Monday is showing you why we're interested in linear transformation.
Birne Binegar: further down the road, because it turns out, these linear transformation that's one of the reasons why you're you're in linear algebra.
Birne Binegar: So.
Birne Binegar: So here i'm just looking ahead.
Birne Binegar: don't we've been talking.
Birne Binegar: About order list of numbers is be a basic vector space i've always concentrated on this, because this is your calculation or arena, even when you consider more general objects that you want to regard as vector spaces, so this is our basic calculations.
Birne Binegar: vector space.
Birne Binegar: But in the first week of class I told you that other things that behave like doctors and so.
Birne Binegar: An example for more general.
Birne Binegar: vector space.
Birne Binegar: Is the following.
Birne Binegar: on making.
Birne Binegar: A rigorously defined set but that's the rigor in here, just so I can take derivatives.
Birne Binegar: So if I have a set of functions on the real line that but which I can take as many derivatives and, as I want that also is a it has the structure of a vector space.
Birne Binegar: Well, how so it doesn't look anything like an ordered list of numbers but what's important here is that if you have a function inside this set.
Birne Binegar: Then there's a notion of scale or multiplication.
Birne Binegar: Lambda times to F is going to be a new function.
Birne Binegar: And it's going to be defined as being a function, whose value at X is just Lambda.
Birne Binegar: times the value of the input function, so this is creating a new function from an old function like do we scale or multiplication and similarly.
Birne Binegar: If you have a summer functions, if you have two functions, you can create a new function that's called the sum of F and G and it's just it's values are just going to be the some of the values of F and G, so my claim, and this will become clear at the end of the course is that this set.
Birne Binegar: Is a vector space in an abstraction just as much as this it's a set where I have both scale or multiplication injector addition well defined.
Birne Binegar: What I want to show you is the linear transformation on the set.
Birne Binegar: I claim.
Birne Binegar: The derivative operator.
Birne Binegar: takes you from differentiable functions, the differentiable functions.
Birne Binegar: Is a linear transformation.
Birne Binegar: When I think of these guys is vector spaces, why.
Birne Binegar: Because when you take the derivative.
Birne Binegar: Of a constant times a function you just get the constant times the derivative of that function.
Birne Binegar: So it does behave well respect the scale or multiplication and when you take the derivative.
Birne Binegar: Of a summer functions, while you can just differentiate turn by turn, can you can def dx plus.
Birne Binegar: The GD X, so, in fact.
Birne Binegar: This notion of linear transformation is going to allow once we expand our horizons to consider more general vector spaces they're going to occur naturally in sort of a surprising way now we're talking only talking about.
Birne Binegar: objects in calculus is being governed by linear algebra break rules, and in fact that's the whole major point of this course is that, in fact, everything you learned in.
Birne Binegar: math 20 to 33 elementary linear algebra are typically when you get to linear differential equations if really most athlete handled using linear algebra.
Birne Binegar: And that that's that's coming up, but what I want to leave with you today it's just the fact that these linear transformations functions between vector space it's.
Birne Binegar: really crucial they're really crucial to the utility of linear algebra, in fact, what will end up i'll show you later on when you try to solve differential equations basically what you're trying to do is figure out how to.
Birne Binegar: How to deal with it, that linear transformation in its and as long as you're dealing with a finite dimensions, you can handle a differential equation, as if it were a matrix and you work, you know, working with a matrix.
Birne Binegar: Okay, so that's all coming down the road.
Birne Binegar: i'll stop here wish you a happy weekend and i'll see you again on Monday.