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Current time:0:00Total duration:25:51

Video transcript

let's say I have a linear transformation T that's a mapping between our n and our M we know that we can represent this linear transformation as a matrix product so we can say that T of X so the transformation T let me write a little higher to save space so we can write that the transformation T applied to some vector X is equal to some matrix equal to some matrix times X and this matrix since it's a since it's a mapping from RN to RM this is going to be a M by n matrix because each of these entries are going to have n components they're going to be members of RN so this guy has to have n columns in order to be able to have this matrix vector product well-defined so we can go back to what we've been talking about now in the last couple of videos we've been talking about invertibility of functions and we can just as easily apply it to this transformation because transformations are just functions and we just use the word transformations when we start talking about maps between vector spaces or between sets of vectors but there is essentially the same thing everything we've done the last few videos have been very general I never told you what the set of our domain and the set of our Co domains were made up of now we're dealing with vectors so we can apply the same ideas so T is a mapping right here from RN to R M RM so if you take some vector here X T we'll map it to some other vector in RM call that ax if we take this matrix vector product right there and this is the mapping T right there so let's ask the same questions about T that we've been asking in general about functions is T invertible is is t invertible so invertible invertible and we learned in the last video that there's two conditions for invertibility t has to be has to be on two on two or the other way the other word was surjective surjective that's one condition for invertibility and then T also has to be one-to-one has to be one-to-one and the fancy word for that was injective in injective right there so this video I'm going to just focus on this first one so I'm not going to prove to you whether t is invertible but will at least be able to try to figure out whether T is Sahn to or whether it's surjective so just as a reminder what does on to or surjective mean it means that you take any element in RM you take any element in your codomain so you give me any element in your codomain let's call that element b it's going to be a vector the fact if t is surjective or if t is on two that means that any be that you pick in our codomain there is always going to be some vector at least one vector in our domain that if you apply the transformation to it you're going to get B or another way to think about it is the image the image of our transformation is all of our M all of these guys can be reached so let's think about what that means so we know that the transformation is ax it's some matrix a so the transformation of X I'll just rewrite it is equal to some matrix a that's an M by n matrix times the vector X times the vector X now we're if we want if we if T has to be on - if T has to be on to that means that ax this matrix vector product has to be equal to any member any member of our codomain can be reached by multiplying our matrix a by some member of our domain so what's another way of thinking about it another way of thinking about it is for any B so on two implies that for any vector B that is a member of RM so any be here there exists there exists at least one solution to a times X is equal to B where of course X is where X is where X is the vector X is a member of RN it's just another way of saying exactly what I've been saying the first part of this video you pick you give me any B in this set and then there has to be if we assume that T is on 2 or for T to be on 2 there has to be at least one solution to ax is equal to B there has to be at least one X out here that if I multiply it by a I get to my B and this has to be true for every maybe I should write for every set of any but there's the same idea for every B in RM we have to be able to find at least one X that makes this true so what does that mean that means that that means that a times X has to be equal to has to be able to you can construct and any member of our M you can construct any member of RM by taking a product of a and X where X is a member of our and where X is a member of here now what is this if X is an arbitrary member of RN let me write it like this we know that matrix a it look like this it'll be a bunch of column vectors a 1 a 2 and it has n columns so it looks like that that's what our matrix a looks like so we're saying if you take this product you have to be able to reach any guy any member of RM but what does this product look like if we're taking this product instead of writing an X there I could write X like this x1 x2 all the way to xn so this product is going to be x1 times the first column vector of a plus sorry x1 of you write plus x2 times the second column vector of a all the way to plus n times the nth column vector of a that's what this product is and in order for T to be on to this combination has to be equal to any vector in RM well what does this mean this is these are just linear combinations of the column vectors of a these are just linear combinations of the column vectors of a so another way to say that is for T to be on 2 so for T to be surjective or on to on 2t the column vectors of a have to span have to span RM have to span our codomain they have to span this right here you have to be able to get any vector here with a linear combination of these guys right and the linear combination is set up because the weights are just arbitrary members of the real numbers right this vector is just a bunch of arbitrary real numbers so for T to be on 240 to be on to the span the span of a1 a2 all the way to a n has to be equal to RM has to be equal to your co-domain has to be equal to codomain that just means that you can achieve any vector in your codomain with the linear combinations of the column vectors of this well what's the span of a matrices column vectors by definition that is the matrices column space so we could say so that means that the span of these guys have to be RN or in other ways that the column space of a let me switch colors the column space of a has to be equal to has to be equal to R M so how do we know if a vectors column space is equal to RM and so here here maybe it might be instructive to draw it to think about when can we not find solutions to the equation a X is equal to B so whenever you're faced with this type of a an equation what do we do we can set up an Augmented matrix that looks like this where you put a on this side and then you put the vector B on the right-hand side and then you essentially put you perform a bunch of row operations and you have to perform on the entire rows in both cases on both sides and we've done this multiple times and your goal is to get the left-hand side into reduced row echelon form so what you want to do is eventually get your Augmented matrix to look like this where the left handed side is let me define our big capital R to be the reduced row echelon form of a and we've done many videos on that that's just you know you have a matrix where you have your pivots and the pivot will be the only nonzero entry in its column but not every column has to be have a pivot in it maybe you have you know maybe you have an on this is will be a free column or non pivot column and then they can have a bunch of zeros and maybe this has a pivot this would have to be 0 if there's a pivot there these would have to be 0 and so on and so forth and maybe the next pivot is right there these would have to be 0 and you get the idea you could have some columns that don't have pivots but whenever you do have a pivot they have to be the only nonzero entry in their column this is reduced row echelon form reduced row echelon form so what we do with any matrix is we try we keep performing those row operations so that we eventually get it into a reduced row echelon form and as we do that we're performing those same operations on the right hand side we're performing on the entire row of these augmented matrices so this guy this vector B right here I guess I could write it as a vector it's eventually going to be some other vector C right here you know if this is 1 2 3 maybe I'll perform a bunch of operations this will be 3 2 1 or something of that nature now when does this when does this when does this not not have a solution have a solution and we went on this we we reviewed this early on the only time where you don't have a solution remember there's three cases there's three cases there's hat where you have many solutions many solutions and that's the situation where you have three variables we've talked about that before you have the case where you have only one unique solution one solution that's the other case and then you have the final case where you have no solutions no solutions and when do you have no solutions what has to happen to have no solutions to have no solutions when you perform these row operations you have to eventually get to a point where your matrix looks something like this you know I don't know what all of this stuff looks like maybe there's one here a bunch of stuff there's one there's a one here and a zero but if you have a whole row at least one whole row of zeros so you just have a bunch of zeros like that and over here you have something that is nonzero you have something here that is nonzero this is the only time that you have no solution so let's remember what we were not even talking about this stuff for we're saying that our transformation is on - if it's column vectors or if it's column space is RM if it's column vectors span RM and what I'm trying to figure out is how do I know that it spans RM and so essentially you can give me for it to span RM you can give me any be here any B that's a member of RM and I should be able to get a solution so we asked ourselves the question when do we not get a solution so we don't get a solution in to situate well we we definitely don't get a solution if we have a bunch of zeros in a row and then we have something non-zero here that's definitely not going to be a solution now there's the other case there's the other case where we have a bunch of zeros so there's the other case where we do have some solutions where it's only valid for certain B's so that's the case let me let me draw it like this let me start it like this so let's say I have my matrix a and I have my B 1 B 2 let me write it all the way and then you have B M remember this is a member of RM and we produce it we do our reduced row echelon form with this augmented matrix and a goes to its reduced row-echelon form and let's say it's reduced row-echelon form has a row of zeros at the end of it so it just has a row of zeros right there everything else looks like your standard stuff ones and zeros but the last row let's say it's a bunch of zeros and when we perform the row operations here on this generalized member of RM this last row has some function maybe just looks like you know b2 b1 + 3 B - B - B I'm just writing a particular case it won't always be this - b3 and it'll be a bunch of it'll essentially be some function of all of the B's so let me write it this way this would be I'm writing a particular case in here maybe I shouldn't written a particular case this will be some function of b1 b2 all the way to BM now clearly if this is nonzero we don't have a solution and so if we don't have a solution for some cases of B then we are not definitely not spanning RM so let me write that down if we don't have don't have a solution solution for some cases of B some cases of B then we don't we don't span our M let me I don't know if I'm if I'm overstating something that is maybe obvious to you but I really want to make sure you understand this anytime you just want to solve the equation ax is equal to B and remember we want to make sure that this can be true for any B we chose what we could do is we just set up this Augmented matrix like this we perform a bunch of row operations until we get a we get this matrix a into reduced row echelon form as we do this the right hand side is going to be a bunch of functions of B so maybe the first row is b1 minus b2 plus B for something in the next row will be something like that we've seen examples of this in the past and if you end up by doing the reduced row echelon form with a row of zeros here the only way that you're going to a solution is if your vector B if its entries satisfy this function here on the right so that this thing equals zero so it's only going to be true for certain for certain B's and if this is if this only has a solution for the certain B's that make this equal to zero then we definitely are not spanning all of our M let me do it visually so if that is our M and if we put if this is only zero for a couple of bees for let's say for some handful of bees then these are the only guys that we can reach by multiplying a times some vector in RN and we definitely won't be spanning all of our M in order to span all of our M when we put this in a reduced row echelon form we have to always find a solution and the only way we're always going to be finding the solution is if we don't run into this condition where we have a row of zeros because when you have a row of zeros and you have to put the constraint that this right hand side has to be zero so what's the only reduced row echelon form where you don't have a row of zeros at the end well any row and reduced it row echelon form either has to have Z all zeros or it has to have a pivot entry in every row so the only way that you span so T is on 2 T is on 2 if and only if if and only if it's the column space of its transformation vector is equal to RM its column vectors span all of RM and the only way that that's happening is if the reduced row echelon form the reduced row echelon form of a has has a has a pivot entry in every in every row in every row and how many rows does it have this is an M by n matrix has m rows and n columns and n columns so it has a pivot entry in every row that means that it has to have M pivot entries it entries right there now what's another way of thinking about that remember we earlier on several videos ago we thought about how do you figure out and this might confuse things a little bit how do you how do you figure out the basis for your column space so how do you figure out the basis for your column space so the basis for for the column space of a matrix and this is a bit of review what we did is we say look you take your matrix and you put it in reduced row echelon form so you put in a reduced row echelon form of your matrix and then essentially let me draw it a little bit different here well you put in reduced row echelon form so let's say that's its reduced row echelon form and you look at which columns have pivot entries you look at which columns have pivot entries and the corresponding columns in your original matrices in your major Uhl matrix forms the basis for your column space so let me draw that out so do a particular instance so let's say that you know it has this column vectors a1 let's say a 2 all the way to a n that's what a looks like and when you put it in reduced row echelon form let's say that this column right here has a pivot entry that column has a pivot entry let's say that this one doesn't let's say it's like a 2 there I'm just picking particular numbers let's say that the let's say that let's say that none of the other ones have any let's say there's a 3 let's say all of these are non pivot entries and then our last entry n is a pivot entry so it just has a bunch of zeros and then a 1 like that how do you determine what are the basis vectors for the column space well obviously the column space is everything that's spanned by all of these guys but what's the minimum set you need to have that same span while you look at which one has a corresponding pivot entries or pivot columns in say I have a pivot column here and I have a pivot column there so the basis from my column space must be this column in my original matrix and that column in my original matrix and then we extended we said well how do you define the the the dimension it's columnspace and you just essentially count the number of vectors you need for your basis and we call that the rank of a so the rank of a this is all review the rank of a was equal to the dimension the dimension of the column space of a which is equal to number of basis vectors number of basis vectors for the column space and this is how you determine it you essentially figure out how many pivot columns you have the number of pivot columns you have is the number of basis vectors you have and so that's going to be the rank of a now the whole reason why I talked about this is we just said that our transformation T is on two if and only if if and only if it's column if and only it is column spaces are M which is the case if it has a pivot entry in every row and it's reduced row echelon form or since it has m rows it has to have M pivot entries M pivot entries so every for every row you have a pivot entry but every pivot entry corresponds to a pivot column every pivot entry corresponds to a pivot column so if you have M pivot entries you also have M pivot columns pivot columns which means that if you were to do this exercise right here you would have M basis vectors for your column space or that you would have a rank of M rank of M so this whole video was just a big long way of saying that T is on to that T is on to and another way of saying that is that if you have your you have your domain here which was RN and you have your codomain here that is RM that every every member of RM can be reached by t by some member of RN that you apply you any guy here there's always at least one guy here that if you apply T to it you get right there there might be more than one now we're not even talking about one to one yet so we say that T is on two if and if if and only if if and only if the rank the rank of the rank of its transformation matrix a is equal to M so that was the big takeaway of this video it's actually let's just actually do an example because sometimes when you do things really abstract it seems a little bit confusing when you see something particular let's see let me define some transformation s let's say the transformation s is a mapping from r2 to r3 and let's say that s applied to some vector X is equal to the matrix one two three four five six times the vector X so this is clearly a three by two matrix let's see if s is s on two is s on two well based on what we just did we just have to go and put this guy in reduced row echelon form so let's do that so if you put this guy into reduced row echelon form so let's keep see one two three four five six now let's keep the first row the same it's one two let's replace the second row with the second row minus three times the first row actually let's replace it with three times the first row minus the second row so three times one minus 3 is 0 3 times 2 minus 4 that 6 minus 4 is 2 now let's replace the third row with five times the first row minus the third row so five times one minus 5 is 0 five times two is 10 minus 6 is 4 now let's let's see if we can get a 1 right here so I'm going to keep my middle row the same actually let's just well let's just divide the middle row by 2 or multiply it by 1/2 so you get 0 1 and we have a 0 4 1 2 and now let's get try to make these 0 again reduced row echelon form so I'm going to keep my middle row the same 0 1 and let's replace the top row with the top row minus 2 times the second row so 1 minus 2 times 0 is 1 2 minus 2 times 1 is zero let's replace this last row with the last row minus four times this row so we get zero minus four times out of zero four minus four times one is is zero so notice we have a row with zeros here we have two pivot entries two pivot entries two pivot entries or two rows with pivot entries and we also have two pivot columns two pivot columns right there so the rank of this guy right here the rank of one two three four five six the rank of that is equal to two which is not equal to our codomain it is not equal to r3 it's not equal to r3 so s is not s is not on - or not surjective then so it's one of the two conditions for invertibility so we definitely know that s is not s is not invertible hopefully that's helpful now in the next video we're going to focus on the second condition for invertibility and that's being one-to-one