Birne Binegar: jack.
Birne Binegar: Okay i'm just waiting for the clock to pick up to 930 Okay, so what I want to do today is not even as ambitious I put on the agenda.
Birne Binegar: What I want to talk about mostly today is substations attached to a basic I talked about that on Wednesday, but I didn't really give you a full example, so the main purpose.
Birne Binegar: which may take up 1520 minutes of today's lecture is to remind you what's up spaces, we can attach to a matrix and how we find their bases.
Birne Binegar: One thing is saved before even start is.
Birne Binegar: Remember macy's or just arrangements of numbers so attaching a subspace to a major.
Birne Binegar: Underlying that That problem is the problem of how do you compute things in general and typically what we've done in this course, whether we're solving linear equations or whether we're trying to figure out if a set of vectors are linearly independent or not.
Birne Binegar: We translate problems into a matrix setting where we can actually do calculations so that's really the point here as well.
Birne Binegar: Until the sub spaces, the row space to call and say no sacred matrix they're not going to make a whole lot of sense to you, yet, because I haven't give you the couch.
Birne Binegar: The situations where you need to find a substation and then, once you figure out what you need typically introducing a matrix is your first step towards calculating it.
Birne Binegar: Okay, so let me, let me start by.
Birne Binegar: Just i'm.
Birne Binegar: Here we go.
Birne Binegar: we're doing this up spaces that one can attach to a matrix that pretty simple to construct So if you have a matrix and wrote an n calm war and N by n matrix.
Birne Binegar: Suppose we're given one, and let me indicate by double prime i've taken this matrix and I reduced it all the way to reduce relational on.
Birne Binegar: Such as, I might if I were trying to solve a linear system involving a but it turned out that this a row Echelon form of a and even better at the reduced relational and form of a.
Birne Binegar: will give us an explicit construction of the bases for the subspace attached to this matrix.
Birne Binegar: So, for example, you can take the rows of the matrix.
Birne Binegar: Take those numbers in each row when I think of them as ordered list of numbers so each since we have in columns will have.
Birne Binegar: em numbers nitro until each row vectors going to be an empty mentioned vector when I take the set of all possible linear combination of these row vectors I generated a subspace.
Birne Binegar: Of the space in which those vectors live that's called the road space of a matrix and what we found last time, is that we can find a basis for this subspace by looking at the.
Birne Binegar: A row Echelon porn or the reduce request form and just grabbing the non zero vectors of that matrix and relational on on, then you have a basis for the road space of your matrix.
Birne Binegar: Certainly, we can calculate a basis for the column space of the matrix the span of all the column vectors of the matrix.
Birne Binegar: We find a basis for that by again looking at the original on form of the original matrix figuring out where the pits lives and then going back to the original matrix and grabbing those calls so.
Birne Binegar: The way you find a basis the column space if you're still row reduced your matrix to rationalize form, you see where the pivots are living in that row actual important you go back to the original matrix and grab the corresponding columns.
Birne Binegar: And then, finally, the third subspace one attaches to a matrix well once you have a matrix you have enough effective Leah coefficient matrix from which you can make a.
Birne Binegar: System of homogeneous linear equations the solution set up the set of linear equations that correspond to your matrix that's called the no space of the matrix and.
Birne Binegar: Finding a basis for this subspace of our LM each vector here's a veteran or m, so the subspace is a subset of our.
Birne Binegar: The way we find a basis for this is, we simply sold the linear system express our solution so hyper plane and then grab the constant vectors.
Birne Binegar: That are being multiplied by your free variables alright so that's a verbal description of what we're going to do let's do an explicit example where I demonstrate on a particular matrix exactly how we find bases for these three.
Birne Binegar: Sub sectors, but actually even before that, let me give you a little bit more nomenclature.
Birne Binegar: Every subspace has a basis and every basis for subspace has the same number of vectors we use the number of vectors in any basis that's what we'll learn to call the dimension of space and it makes sense of up your all your usual notions of bases.
Birne Binegar: All your usual notions of dimension is the number of independent directions that are actually going to live in the state, just like we know that.
Birne Binegar: Physical spaces three dimensional because we have three independent directions that will from that we can reach any point well those independent directions are effectively the directions prescribed by basis vectors.
Birne Binegar: Okay, so.
Birne Binegar: In particular, for those row spaces, or for those sub spaces attached to a matrix we can get the dimension of the those guys can be the number of basis factors.
Birne Binegar: Now for the row space what you do is you wrote reduce a to relational on form and you grab the non zero rose well every non zero row is going to have a pivot the first non zero entry in a row, and so to count the number of basis vectors you just.
Birne Binegar: calculate a row Echelon on your matrix and you count the number of pivots in that row excellent for each pivot is going to correspond to a non zero, which is going to correspond to a basis sector for the subspace.
Birne Binegar: Similarly.
Birne Binegar: The column space its basis vectors are determined, also by the rotational on form of a but indirectly what we do here is we reduce a to a row Echelon warm.
Birne Binegar: figure out where the pivots lives go back to the original matrix and grab those corresponding columns but the number of columns you actually grabbed them here it's going to be determined by the number of pivots you have in your row Echelon form.
Birne Binegar: And so the dimension of the colon space is again going to equal, the number of pivotal role Echelon born, which, when I want you to note here is.
Birne Binegar: Proceeding slide if I started off of it n by n matrix then each row would have in vector so that would be a subspace or m meanwhile sorry the row space is is.
Birne Binegar: We have the row space.
Birne Binegar: of you.
Birne Binegar: If a is as in rows and columns, then the row space of a is going to be a subspace on in the call space.
Birne Binegar: A is going to be a subspace.
Birne Binegar: Or, and because, in the columns you have an entry for each row and then.
Birne Binegar: Michael put up here as well, and all space of a well in order for this multiplication a time for this equation, to be defined.
Birne Binegar: This vector of variables has to be well multiplying an n by n matrix is going to be a column vector is going to have to be an m by one column vector for that matrix multiplication to be defined.
Birne Binegar: And then, what gets out what's going to be on the right hand side isn't in by one zero vector but don't.
Birne Binegar: These.
Birne Binegar: Were these variables X live they live in an n dimensional space so that's why the no space is subspace of our am so I noticed that, while these two guys.
Birne Binegar: A basic subspecies of the same vector space, the call and space is different, and in fact the fact that these guys live in.
Birne Binegar: Different vector stations that's what makes it a little bit surprising that the dimension of the road space of a matrix is always equal to the dimension of the column space.
Birne Binegar: Though, even though these rogue actors in column vectors live in different spaces spaces generate exactly the same dimension.
Birne Binegar: All right, and so.
Birne Binegar: For more nomenclature on this slide.
Birne Binegar: Just names for the dimension of the road space to call him space.
Birne Binegar: And the knowledge base of matrix.
Birne Binegar: The common dimension of the road space in the common space we call that rank up a matrix and then the dimension of the no space, the dimension of the solution, set a X is equal to zero that's called a novelty of a matrix.
Birne Binegar: And what i'm once simple identity that comes out basically just from these definitions are actually more precisely from.
Birne Binegar: The way we compute them is that the number of columns of the matrix is always equal to the rank in a novelty, because what the rank is is well.
Birne Binegar: figure out what it is you just count the number of pivots of any row Echelon form same for the call so if the relational on form of my matrix is going to determine well i'm just count puppets but.
Birne Binegar: The novelty of a matrix dimension of the solution, set remember as we solve linear systems.
Birne Binegar: The idea to take this guy and row reduce to reduce row Echelon form, and then we.
Birne Binegar: Aside from actually furnishing us with the equations that of the solution we also get from the rational and form.
Birne Binegar: partitioning of the variables into free variables in fixed variables, the number of free variables is equal to the number of columns in the.
Birne Binegar: rotational on on.
Birne Binegar: that do not have pivot so effectively this rank.
Birne Binegar: Its novelty it's going to equal to the number of columns.
Birne Binegar: row Echelon form of a that don't have pivots.
Birne Binegar: because those are going to correspond directly to the free parameters of the solution if you know how many parameters, you need to describe the solution set effectively, you know the dimension at least you know.
Birne Binegar: In a formal way or informal way, but the rank of a matrix rule that's determined by either the number of non zero what number of pivots and so that'd be the same thing as the number of columns with pivots.
Birne Binegar: And so I basically take my columns a and I can divide it up into two sets those that contain pivots those that don't don't contain pivots but then respective numbers have to add up through the number of calls, and so this turned out to be a fundamental identity.
That.
Birne Binegar: If you know the rank of a you also almost immediately know that now the dimension of the no space as well, because all you have to do is subtract the rank from the number of columns to get the dimension of that no space.
Birne Binegar: Okay, so.
Birne Binegar: Well, that it was just nomenclature that we covered on Wednesday when principal thing I wanted to do today is do an explicit example.
Birne Binegar: I want to get this.
Birne Binegar: Situated a little better.
Birne Binegar: Alright, so i'm going to start with a three by three matrix and i'm gonna.
Birne Binegar: assume that i've already done, the brute force calculation, where I take a matrix and I perform elementary row operations until I.
Birne Binegar: get a matrix it's in reduced row Echelon for all right, so I did that by computers and that's the result now from this single row reduction calculation.
Birne Binegar: i'm going to figure out what a basis for the row states of this matrix is what a basis for calling state is a basis for us no space that are solution set of eight X is equal to zero and i'm going to get the rank up the matrix and it's noti.
Birne Binegar: Alright, so each of these is going to be really easy to do with do it basically by inspection of this matrix.
Birne Binegar: So, for example, this top line is exactly what I had before, this is my assumption is that this matrix reduces to this matrix.
Birne Binegar: Alright, so here's a row Echelon form my original matrix it's non zero rose provide a basis for the row space of that matrix so basis for the road space of a simply look at the real Echelon or find the non zero rose and listen and that will be my basis.
Birne Binegar: For the columns space here you you again work with the relational on form here ritual matrix, but let me lower a little bit, so I can.
Birne Binegar: draw on it right.
Birne Binegar: Here again, you look for the pivot and your idea for finding a basis for the columns space was well you don't use these calls.
Birne Binegar: That won't that won't work what you do do is you figure out where the pivots live in its raw form and you go back to the original matrix and you grab these two columns these two columns are going to be a basis for the calm state of the original matrix.
Birne Binegar: No, in fact.
Birne Binegar: This row Echelon form it's called space is actually going to be in general, different than the column face at the original matrix because elementary row operations perfectly preserved the row space, but they're not preserve the columns face.
Birne Binegar: Nevertheless, when I saw last time, is that we can use the reduction of a matrix to each row Echelon form we can still use that calculation to find a basis for the columns days.
Birne Binegar: All we all we have to do is figure out where the pivots live in the row Echelon form and then go grab the corresponding columns and the original matrix so.
Birne Binegar: We Center this.
Birne Binegar: Then.
Birne Binegar: For this particular matrix a a basis for the columns states of that well it's going to be in the row Echelon form the pits live in the first two columns I grabbed the first two columns of a one to one in 132 and then i'm done i've got the basis for the columns face.
Birne Binegar: But notice, of course.
Birne Binegar: Each each column, it has a pivot also lives in or non to a row so that's why we always a number of basis vectors for the columns based always equals this number of patients sectors for the roast beef, even though they're completely different vectors and sub spaces.
Birne Binegar: All right, and then, finally, we have the no space and here's what we may have to do a little bit of calculation it's not that we have to redo the matrix reduction of our of our matrix a in fact.
Birne Binegar: That same role reaction, reducing a to it's relational on form reduce your watch will inform presented here that same calculation could be used to.
Birne Binegar: reduce the augmented matrix corresponding to the linear system to each row Echelon one it just all you have to do in this coefficient part just insert the reduce relational and form of a because, under your role reaction.
Birne Binegar: algorithm nothing ever happens in this last column, so all the.
Birne Binegar: Elementary real operations applied to this matrix are going to effectively be the same thing is the real operation apply to a alone augmented by that zero metrics.
Birne Binegar: All right, but So what we need to do to find a basis for the no spaces, we need to figure out a basis for the solution set at that system So the first thing we do is, we take the augmented matrix and reduce electronic form, and we convert back to equations.
Birne Binegar: And simultaneously what we do is we look at the form of the reduce your national on form and the divide our variables up into free variables and fixed variables.
Birne Binegar: Six variables correspond to columns with pivots X one X, to the third column here doesn't contain a pivot so it's gonna it's corresponding variable it's going to be free, so I take the equations.
Birne Binegar: Coming from the augmented matrix and reduce your national on I moved the free variables over to the right hand side, giving me expression for the fixed variable and then I substitute in for the fixed variables a general vector that's.
Birne Binegar: Excellent is equal to minus five X three and X is equal to three x one and these equations are solved, and this guy is going to be a solution of these equations which is equivalent to the original linear system.
Birne Binegar: Then.
Birne Binegar: seeing how the free variables occur I extract the three variables, or more precisely, I expand my vector solution in terms of the free variable.
Birne Binegar: And the constant vector that be multiplied by the three variables, those are going to be your basis factors in this case we have only one variable so we have only one constant vector we have only one basis sector.
Birne Binegar: Alright So although it's kind of slipped off the bottom of the page a basis for the no space of a solution set of.
Birne Binegar: homogeneous linear equation with coefficient matrix a.
Birne Binegar: Its basis it's going to be given by this factor this constant vector minus five three and one.
Birne Binegar: Okay, so that's yeah.
Birne Binegar: So you.
Birne Binegar: You basically solved the system, so it just point you this is what you did on the first exam is you.
Birne Binegar: Had a matrix and reduce relational on point me you wrote down the solution there's a hyper play the basis vectors going to be these constant factors that are being multiplied by the three variables, so you just identify.
Birne Binegar: More or less by inspection, do you know how to interpret this guy then you've got the basis factor.
Birne Binegar: So.
Birne Binegar: So the point of this examples is for the me to show you explicitly how that role reaction to Rochelle and form find two bases for all three sub spaces and once you.
Birne Binegar: The hard part is that of course doing the row reduction doing that carefully so you don't make mistakes, but once you reach the reduce relational on form, then all the information just POPs out from that answer.
Birne Binegar: Okay So what about the the dimensions of the sub spaces, the rank, this is the common dimension up on the row space or the home space.
Birne Binegar: But another way it's a number of basis factors, you have, for instance those sub spaces, but since we had.
Birne Binegar: Every pivot in a row so on form of a is going to give me a non to a row I count the number of pivots I get the dimension of the road stage.
Birne Binegar: But also every pivot of the relational on form can also give me a basis factor, it just the rule for getting it is different, you figure out where the pivot lives and you go back to the original matrix but for each pivot in the row Echelon form, I have a corresponding basis factor and.
Birne Binegar: So going back as soon as I looked at this.
Birne Binegar: augmented matrix accounting to two columns with pivot I know that the the rank is equal to two years, I to basis vectors for.
Birne Binegar: The row space and the two basis director for the column space would be figuring out where those two columns came from, and the original matrix a.
Birne Binegar: And then the dimension of the no spaces equal to the number of free variables and that's going to equal, the number of columns without pivots I only have one column without pivot in this role Echelon form of a so my no space is going to be one dimensional and.
Birne Binegar: Confirming.
Birne Binegar: That remark that I made it.
Birne Binegar: towards the beginning of the lecture, then the rank of a matrix plus the novelty of a matrix two plus one is equal to three is equal to the total number of columns of the original matrix.
Birne Binegar: Okay, so with with that example I think you can really easily carry out all the problems in homework set six, which is due on Monday.
Birne Binegar: And while i'm thinking of logistics, a little bit.
Birne Binegar: The second exam is going to come up and in the near future, I don't not before spring break.
Birne Binegar: I have one more topic that I want to cover that will be linear transformations and we'll we'll actually start a little bit on that today.
Birne Binegar: But i'll probably be boat all of next week to linear transformations and there'll be a homework assignment associated with that as well.
Birne Binegar: Then we have spring break the following week, I think, is the week of March 21 that week is a likely spot for the second exam.
Birne Binegar: And i'm thinking a logical place to put it would be on Wednesday of that week that way we can come back from spring break have a review session on Monday and then do the exam on Wednesday.
Birne Binegar: i've already had one person request that no please do not have an exam on Friday so i'll, at least initially i'll try to avoid that slot and if it turns out that.
Birne Binegar: Wednesday works out to be a bad day, then it may be that I actually push forward the exam until Wednesday of the following week.
Birne Binegar: But if I if I do that then i'll just start talking about new material.
Birne Binegar: When we get back from spring break and i'll still have a review session the class meeting directly before the exam.
Birne Binegar: Okay, so.
Birne Binegar: Good so Now let me proceed to will be the last topic we cover before the second exam and that's going to be about linear transformation.
Birne Binegar: So what we've been doing here, since the first exam is i'm been setting up.
Birne Binegar: We we had the.
Birne Binegar: vector spaces and first thing we did after the exam we talked about sub spaces and subsets of vector spaces, that would be relevant to linear algebra we're thinking be just being mathematically relevant.
Birne Binegar: Once you have a bunch of sets the next natural thing to consider next would be functions between the two sets, and indeed this is what we're going to turn to our attention to next functions that take us between vector spaces.
Birne Binegar: But, just like we were very, very specific about what kind of sets would be interesting to linear algebra so to when we consider functions between vector spaces we're going to be very, very choosy about what kind of functions are we want to use.
Birne Binegar: And you might guess.
Birne Binegar: Since i've been harping about do basic vector operation since the first week of class that what's going to distinguish between our special functions and more general functions between two vectors spaces is their behavior of respect to scale or multiplication and vector addition.
Birne Binegar: To so let's first just talk really briefly about functions.
Birne Binegar: So a function.
Birne Binegar: is just a rule.
Birne Binegar: for signing.
Birne Binegar: The each member of a given set at a certain member of another set why don't we.
Birne Binegar: Do we generalize this notation the function from X to y means that for every element of X F of X is going to be some element of why we call.
Birne Binegar: The first set the domain of the function and the place that you end up after applying a function, we call that the code domain now here's your picture that you've been seeing ever since high school, I hope, and certainly in calculus functions take you from one set to another.
Birne Binegar: Alright, so.
Birne Binegar: What we're gonna be interested in linear algebra is a very special family functions they're gonna they're gonna be well behaved respect the scale or multiplication and vector addition.
Birne Binegar: Specifically, what we're going to require is well we'll have we'll be looking at functions that take us from one stage to another, not necessarily the same vector space.
Birne Binegar: And what we're going to require of them is that well they behave well respect the scale of multiplication in the sense that whether I do a scale of my multiplication before I apply the function or after I apply the function, I get the same result.
Birne Binegar: and similarly.
Birne Binegar: With respect to vector addition, if I add two vectors before I apply the function so add two vectors I get a new vector in the domain, then I apply the function, or if it's a linear transformation statement is that whether I add vectors before or after I apply the function.
Birne Binegar: Between the two vectors states is, I get the same result.
Birne Binegar: And so.
Birne Binegar: often say that a linear transformation is a function between vector spaces that preserves scale or multiplication index tradition so i'm not exactly sure what this notion of preserving is, but it is clear that somehow.
Birne Binegar: The function function T here is sort of an oculus.
Birne Binegar: rift respect the scale of multiplication I can.
Birne Binegar: Do the scale of medication before or after I play the function without consequence or I can add two vectors before after I apply the function without consequence, I get the same result.
Birne Binegar: Okay.
Birne Binegar: So.
Birne Binegar: The first thing to note about.
Birne Binegar: linear transformation these special functions, is that, if they preserve individual acts of vector addition and scale multiplication they're also going to.
Birne Binegar: preserve the combined action scale and multiplication and vector addition, so, in fact, if you have a linear combination of factors and you apply a linear transformation to it well because.
Birne Binegar: These linear transformations are compatible with vector addition I can delay these summing up these vectors until after I apply the linear transformation.
Birne Binegar: And well actually I should do this in two steps.
Birne Binegar: The first thing I should point out is that this is so, forget about this line for the time being, this is the same thing is key times are One the one plus.
Birne Binegar: Key times.
Birne Binegar: are to be two plus dot dot dot plus T times I guess last one is rk bk on using here is the fact that I can delay these vector sums until after I apply the linear transformation.
Birne Binegar: But then I can.
Birne Binegar: Now I have linear transformation apply to a scale or multiples of vector because it's linear transformation, I could delay that Scott Gala multiplication until after I apply the function, I can pull that constant factor through.
Birne Binegar: And I get p.
Birne Binegar: One was similarly I can do that for the second term are two times teeth something to.
Birne Binegar: I guess the last term it's going to be.
Birne Binegar: rk T.
Birne Binegar: V K, which is what I originally wrote here.
Birne Binegar: So in fact.
Birne Binegar: If you have a linear transformation and it, acting on a linear combination of vectors well, the result is going to be a linear combination, in fact, the same linear combination, as you started off with it just the vectors these have been replaced by T acting on on them.
Birne Binegar: Okay, so.
Birne Binegar: So in some sense if you have a linear transformation, it also preserves linear combinations in this sense.
Birne Binegar: Okay.
Birne Binegar: So.
Birne Binegar: I told you what it is, I haven't given you any examples of what linear transformation might be.
Birne Binegar: let's see on this slide.
Birne Binegar: Actually, let me.
Birne Binegar: Let me give you an abstract example the sort of candle if you why the things are so important.
Birne Binegar: So.
Birne Binegar: So this is.
Birne Binegar: An upcoming interpretation.
Birne Binegar: A linear transformation.
Birne Binegar: So let's let's.
Birne Binegar: Think of our concrete vector spaces or alicia that numbers think of replacing this by a more general vector space.
Birne Binegar: What I want you to think of our functions on the real line that you can differentiate.
Birne Binegar: Now order list of numbers at first you think don't think have anything to do with functions on the real line, but the actual point is, is that these functions on the real line they have notions of scale and multiplication in vector addition.
Birne Binegar: If you want to.
Birne Binegar: If you have a function on the real line, you can create a new function by simply taking the old function and multiplying its values, my common number Lambda and if you have.
Birne Binegar: Two functions on the real line, you can create a new function that's some of them like demanding that its value is going to be some of the values of the individual summon functions.
Birne Binegar: Right, so we do the set does have a notion of scale multiplication and vector edition, and what we'll see.
Birne Binegar: Towards the end of the course, in fact, this is a back your space, just as much as this is a vector space, because it is a set with these two fundamental vector space operations, what I want to point out to you is that, on this set, there is a.
Birne Binegar: an easy thing to say, easy to see is a linear transformation.
Birne Binegar: The derivative operator.
Birne Binegar: If you take them.
Birne Binegar: If you act on a function.
Birne Binegar: or even a function this be multiplied by a number that's the same thing as Lambda times the derivative of X because differentiation ignores Constance and similarly.
Birne Binegar: um I guess i'm writing below the black over here below the desk here if I take the derivative of the summer functions what I get is the some of the derivatives.
Birne Binegar: So in the applications that are coming down the road.
Birne Binegar: The analog of a linear transformation on this space could be something like a.
Birne Binegar: differential operator, because the derivatives are going to effectively the linear transformations of the space of functions.
Birne Binegar: So i'll come back to this example later, but I just want to show you that there is actually a lot of substance hidden these definitions it's just we haven't really gotten, to the point where we can see.
Birne Binegar: The applications that bring this out, but in fact my purpose in introducing linear transformations is just so that we can then begin to.
Birne Binegar: See how the theory of differential equations how that is influenced by linear algebra.
Birne Binegar: Okay, so let's go back to.
Birne Binegar: A topic at hand.
Birne Binegar: How would you how would I actually write down a linear transformation between two vector spaces well one thing you might suspect and it's easy to confirm is that you're not going to have a linear transformation.
Birne Binegar: Unless the functions well I guess i'm going to explain this more carefully if i'm.
Birne Binegar: If I have a linear transformation that taking me dimensional vectors.
Birne Binegar: To n dimensional vectors.
Birne Binegar: Then up a point in this space.
Birne Binegar: it's getting map to.
Birne Binegar: A point in an n dimensional space, so I use a notation there yeah i'll call it first component it's going to depend on where it came from so X one.
Birne Binegar: X n second component still going to depend on where it came from depend on the coordinates X one X M and because you're ending up in an n dimensional space you'll have a total in coordinates in this image X one X m.
Birne Binegar: So.
Birne Binegar: When I get down to the nuts and bolts of actually writing down a particular function from an n dimensional vector space to an n dimensional vector space what that involves is me giving you a list of for each component here, I need a function of the variables in here.
Birne Binegar: All right, and so.
Birne Binegar: What i'm trying to say on this slide is these.
Birne Binegar: This will be find a linear transformation, so long as these functions why one through y n functions.
Birne Binegar: For the individual components of the vector that you end up on those functions have to be linear functions in the sense of calculus you can't have if you had something like.
Birne Binegar: let's just do a one dimensional example we go from our one.
Birne Binegar: On one if this worse X getting map to X squared it would not be a linear transformation because, for example, he.
Birne Binegar: called us T.
Birne Binegar: So P of Lambda times X would be Lambda X squared.
Birne Binegar: equals Lambda squared X squared but that's not the same thing as Lambda times to X, because this would just be Lambda times X squared those things are not equal, so one thing is.
Birne Binegar: Certainly, clear, just when you're looking from a one dimensional vector stage to one dimensional vector station you're not going to be for these component functions you're not going to.
Birne Binegar: be allowed any nonlinear function, because this property is going to break down scale multiplication kind of break down and also vector edition is going to break down so.
Birne Binegar: One thing it's going to be necessary for these component functions in order for key to be a linear transformation is that these guys have to be linear functions of the variables otherwise.
Birne Binegar: This problem that occurs just mapping our one to one it's going to be.
Birne Binegar: multiplied when we just have more components to consider.
Birne Binegar: All right, but in addition they not only have to be linear functions.
Birne Binegar: So functions linear linear.
Birne Binegar: function is linear.
Birne Binegar: If it has a form.
Birne Binegar: let's say a function X one X m is linear if it has the form.
Birne Binegar: F of X equals some constant times the first variable some constant times a second variable.
Birne Binegar: Some constant Times last variable and you could also add in another constant.
Birne Binegar: that the reason why they're called linear functions take the one one dimensional case.
Birne Binegar: F of X is linear.
Birne Binegar: If F of X equals some constant times X plus B.
Birne Binegar: And then the reason why it's called linear because in calculus if you graph such a function it'd be a straight line so, hence the word linear a straight line with slope a in why intercept be.
Birne Binegar: And all this form does is to generalize that construction of linear function for one variable to a construction of a linear function of multiple variables.
Birne Binegar: What i'm saying here, though, is that if i'm going to use a bunch of linear functions to create a linear transformation between two vector spaces, what I need are these component functions to be linear.
Birne Binegar: And I need to get rid of those constant terms as well i'll show you an example, how about the appearance of a constant term.
Birne Binegar: In one of these component functions which screw up the linear the mile the ability to interpret this as a linear transformation.
Birne Binegar: Right so here's what I mean by a linear function without constant term for each variable it's The worst thing that happens to me to see multiply by some number and we add it to similar contributions from other variables, but we're not going to include any constant terms because.
Birne Binegar: it's a simple example i'll try to position, a little bit better.
Birne Binegar: If I have.
Birne Binegar: A function from a two dimensional vector stage two or three dimensional vector space, so he is a linear function of s&t so it's a linear function of s&t.
Birne Binegar: This is also a linear function of the variables, but now we have a constant term and i'll show you how this constant term screw things up.
Birne Binegar: let's pick an arbitrary point in the domain.
Birne Binegar: When we apply.
Birne Binegar: function T to this point, this is the result, just given by that same formulas, I had above.
Birne Binegar: But now, when I do a scale of multiplication of a vector and then apply the function well now i'm applying it to to it to new coordinates RS and RT, but I still have the same room so for to substitute RT SS up to.
Birne Binegar: or s and then here, we have one plus R s T i'm just substituting are s s and our T for T into this formula, on the other hand.
Birne Binegar: If I take this formula and I multiply both sides from the left by the number of our I get this and now scale multiplication of this vector by the number of our I move that are through the each component, and here we get an are not a one, so what happened here is that.
Birne Binegar: The scale or multiple of the.
Birne Binegar: TV my art is not the same thing as to apply to our times D, because one it's not going to equal are, in general, this would have to be true for all on not just ours equal one.
Birne Binegar: Okay, so appointed, this is that that constant term.
Birne Binegar: prevent.
Birne Binegar: From be.
Birne Binegar: A linear transformation.
Birne Binegar: If that had not occurred there, then in fact I would have gotten the same result.
Birne Binegar: Alright, so.
Birne Binegar: So on this slide my point was just to show you that yes.
Birne Binegar: You can make linear transformation out of linear functions, but you have to be careful, you can't include any constant term, because that will destroy.
Birne Binegar: A property that the function that you're creating would be well behaved with respect to scale or addition, if I.
Birne Binegar: If the value of my function after I do scale a multiplication it's not the same as the value of applying directly your function directly to a vector and then doing this yield more if these two are not the same same doctor, then he is not a linear transformation.
Birne Binegar: Okay Now let me give you an example, and in fact this is really the basic prototype for linear transformation, if I actually want to create a linear transformation.
Birne Binegar: Without talking about linear functions are all all you really need is a matrix, by which I can construct the linear transformation so here's how this is going to work.
Birne Binegar: If you have an n by n matrix, then you can always multiply it from the lap by a call back to that has the same number of entries, as we have sorry.
Birne Binegar: It has the same number of rows, as we have columns the first factor now, this is just matrix multiplication of the column vector by n by n matrix.
Birne Binegar: But what you can in the way you can interpret this calculation is it's taking an n dimensional vector and it producing a vector with end components and so you are going from an n dimensional space.
Birne Binegar: Coming n dimensional space and your results are you start off with a vector and you get a vector in a different vector space.
Birne Binegar: So let's use this matrix in to define a function between these two vector spaces well.
Birne Binegar: Such a function is automatically going to be a linear transformation because.
Birne Binegar: DNA applied to Lambda so I put the subscript here just to emphasize that.
Birne Binegar: My particular linear transformation is defined in terms of the input matrix a.
Birne Binegar: But when I do scale multiplication of a vector and then apply that linear transformation that's the same thing as multiplying my vector like the matrix.
Birne Binegar: But scaling multiplication commutes with matrix multiplication so I can pull this factor through the day and then I have Lambda times X or Lambda times to a webex and i've got the identity, I want.
Birne Binegar: This this function here preserve scale and multiplication.
Birne Binegar: And it also particular naturally vector edition, because when I.
Birne Binegar: play the function, which means multiplying by the column vector corresponding the variable by the matrix a when I applied the function, I can distribute that matrix multiplication over the sum and then I get well, which is effectively P of X one plus P of X two and.
Birne Binegar: The extreme size of these two equations verify the essential properties of the linear transformation that they preserve scale multiplication and vector addition.
Birne Binegar: Okay, so I guess yeah.
Birne Binegar: Well, you will actually use that that so one thing is sort of hard to do you see this a little bit in calculus to where you start taking compositions of functions you go from one space to another space to another space, and you have tried to.
Birne Binegar: write to compose something down by substituting results into result well what we're going to find is that when we interpret linear transformation as major sees that instead of having to compose functions, by doing all that substitution you just multiply the corresponding matrices together.
Birne Binegar: And so yeah filled in so yeah it's related to that.
Birne Binegar: Okay, so I think what i'll do is um I just managed to introduce linear transformation next week we'll spend the whole week on on on dealing with them.
Birne Binegar: Okay, so i'll let you guys go do what I do try to do problem set fix on web time before Monday and i'll be happy to answer questions on that on Monday, as well.
Birne Binegar: Well, I mean it's do anytime before the end of the Semester really because i'm leaving web assign you will still accept the assignment, but if you want to keep up with the course that should be done on Monday.