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let's say I have some linear transformation T from r2 to r2 so if that is r2 and then this is r2 t just maps from any member of r2 to another member of r2 just like that and I'm going to define T it's a linear transformation I'm going to define it when I when I take the transformation of some member of r2 it's equivalent to multiplying it by this matrix by the matrix 1 minus 3 minus 1 3 so just to kind of understand this transformation a little bit more let's think about all of the values that it can take on in our codomain so let's set it equal to so if we say that 1 minus 3 minus 1 3 times any vector any vector in our domain so x1 x2 it's going to be equal to some other vector in our codomain let me call that vector BB is a member of r2 so it's going to be equal to b1 b2 all I'm doing here is I'm essentially if I call this right here I call that matrix a I'm just trying to find out what are all of the possible values for ax is equal to B I'm trying to find out all of the possible all of the possible bees in this case all the possible bees so if I were to solve this equation for any particular B what I would do is I would put this in reduced row echelon form well first I would make an Augmented matrix of it so I would put a 1 a minus 3 a minus 1 and a 3 and I wouldn't it with the member of our codomain we're trying to be equal to so b1 b2 and then I would put in reduced row echelon form so how can we do that how can we put in reduced row echelon form I'll keep my first row the same 1 minus 3 and then b1 and then my second row I'll replace it with my second row plus my first row so minus 1 plus 1 is 0 3 plus minus 3 is 0 and then B 2 plus b1 is well I could just write that as b1 plus b2 so this is only going to have solutions I have an interesting situation here we've seen it before I have a row of zeros I have a row of zeros and the only way that this is going to actually have solutions is if this thing right here is also going to be equal to zero so the only only members be that are of members of RM that have have solutions are the ones are the ones where if you add up their two terms so let's say you know B is equal to B is equal to B 1 and B 2 are the ones where that the two terms added up have to be equal to 0 B 1 plus B 2 have to be equal to 0 or another way you could write it is that B 2 has to be equal to minus B 1 so if we would actually draw our codomain let's do it we always stay in the abstract but sometimes it's useful to draw an actual example so let's say that our codomain is r2 let me draw some axes right here let's say that this is my this is my B one axis and this is my B two axis I could have called that x and y but I call that B 1 and B 2 what are all of the members of my co domain that have a solution that have my mapping well B 2 has to be equal to minus B 1 B 2 s to be equal to minus B 1 so it's going to look like this it's going to be look like this let me draw it's going to be aligned with a slope of negative 1 this is this is all the B's all the B's that have solution that have a solution right because if you're not on this line if you're a member of your codomain this is the codomain right here our two our two is also our domain but let me make it very clear that this is the co domain that I've drawn this is what we're mapping into it's very clear that if we're not on this line if you pick somebody whose two terms don't add up to equaling zero or aren't the negative each other if you pick someone over here in our codomain and then you try to solve the equation over here you're going to have a zero is going to equal some nonzero number here and you're not going to have a solution you're not going and we touched on this in the last video and so in this case the image we could say that this right here is the image of our transformation is the image of our transformation or even another way of thinking about it obviously if all of if all of if all of our two is this is our codomain let me draw a domain if our domain looks like this if all of our two so if you take any member of our two and it's always mapping onto something onto that line clearly each point on that line is going to be mapped to by more than one vector so we're not dealing with an onto transformation and we saw that in the last video in order for something to be on two when you put it in reduced row-echelon form you cannot have all zeros in one of the rows or another way to say it in reduced row-echelon form every row has to have a pivot entry but let's focus on the B's that actually do have a solution so let's focus on these B's so that when you take a b1 plus b2 it actually is going to be equal to 0 so you know we could have we could have the B I don't know that could be the b5 5 minus 5 or you could have the B well obviously the b0 0 is going to work you could have one negative one maybe that's right there let's focus on those for a second and solve and see how many guys how many members of our domain map into them so if we take this right here and then we apply this equation up here we only have one constraint we're assuming that this is going to be equal to 0 so let's assume let's assume that we're dealing with the bees in our image so let's assume let's assume that we're dealing with something where we can get a solution that b1 plus b2 is equal to 0 then what is our constraints what we'll map to our vector B that word with so if we just take this top equation right here we have 1 times x1 1 times x1 let me switch colors we have 1 times x1 1 times x1 minus 3 times x2 minus 3 times x2 is equal to B is equal to b1 and then this row will give us no constraint because this is just going to be a bunch of 0 so this is the only constraint for a member of our domain that will map to some particular B now that we are picking some particular B that satisfies this constraint so we could write this solution set let me rewrite it as X 1 is equal to B 1 plus 3x 2 or if we wanted to write the entire solution set it would look like this X 1 X 2 is equal to the vector B 1 0 plus let's see X 2 X 1 is B 1 plus 3 times X 2 so plus x 2 x 3 and X 2 is just going to be equal to X 2 we can that's a free variable so X 2 is equal to 0 plus X 2 times 1 so what we're saying here is look this transformation just maps to this line here for all of the vectors in our codomain where they're two entries add up to each other now assuming let's assume that we actually have one of those vectors and so first of all this is definitely not an onto transformation but let's assume that we are dealing with one of those guys so if we are picking a particular one of those guys for a particular B let me write this down for a particular for a particular B that has a solution that has a solution that has a solution to ax is equal to B the solution set the solution set the solution set will be equal to this thing right here it'll be equal to X 1 X 2 equal to the first entry of your bb10 + X 2 times the vector 3 1 and if you think about this if you pick a particular B so let's say we pick our let me let me draw this out because I think it's nice to visualize it all actually maybe I'll draw it like this I don't want to draw our blobs anymore so let me draw my axes so my axes look like this we know that the image of our transformation is the line with a negative 1 slope because the two entries have to be equal to the negative each other let's pick a particular B let's pick a particular B that has a solution so let's say we pick that'd be right there in order for it to have a solution its entries have to be the negative of each other so let's say that's the entry let's say the C entry 5 minus 5 that is our B so what we just showed is that the solution set if we want to say hey what in our domain maps to this guy so let's think about what in our domain maps to this right here let's think about what in our domain maps to this point right here to this particular B so that's going to be all of the all of the X's that satisfy a X is equal to 5 minus 5 and what this is selling it so this is going to be equal to so the solution set is going to be equal to so x1 x2 is going to be equal to b1 it's going to be equal to 5 0 plus x2 plus any scalar multiple any scalar multiple of the vector 3 1 so our solution set is going to be you take the vector 5 0 so maybe the vector 5 0 specifies this position right here and then you're going to add to it multiples of the vector 3 1 the vector 3 1 looks like this 1 2 3 and you go up 1 the vector 3 1 is going to go to look like this it's going to look like this so if you add multiples of this multiples of that could stretch out like that or it could go negative like that to this vector 5 0 you're essentially going to let me see if I can draw this neatly you're going to end up with a solution set that looks like that right there so if you pick if you pick a particular B right there that has a solution we just said that everything on this line everything on this line will map to that point in our solution set everything on that line would map to our solution set and in fact if you picked another point let's say you picked let's say you pick the point -5 5 let's say you picked let me write it this way -5 5 then the solution set that maps to that would actually be you would take - this first term would be the minus 5 would be here and all of these guys all of these guys would map to that what this is all interesting I mean we've been doing a lot of abstract things and I think it might be satisfying that you're actually seeing something kind of more concrete in this example but I'm doing all of this for a reason because I want to understand what the solution set is to a general non non-homogeneous equation like this and to understand a little bit better let's imagine what is the solution set if I were to pick this guy if I were to pick the 0 vector if I pick the vector this is the 0 vector right there then what's going to be the solution set it's going to be the vector so if we say that ax is equal to 0 then our solution set is going to be the vector 0 0 + right 0 0 + x 2 x 3 1 so it's going to be what it's going to be this is just the 0 vector so it's going to be here it's going to be here and it's going to just be multiples of 3 1 so it's going to look like C it's going to look something like that but what's this what is the solution set to the equation of AX equal to 0 this is the null space this by definition is the null space this is the null space of a this right here is the null space of a so notice and this is the big kind of takeaway from from this video is that for any solution you know we're picking B's that actually do have solutions because we're picking them on this line we're picking them in the image of our codomain but any the solution set to any ax is equal to some B where B does have a solution it's essentially equal to a shifted version of the null set this right here this are the null space this right here is the null space that right there is a null space for any real number X to any scalar multiple of three one is the null space I just showed it right there it's going to be that and so all of these other solution sets are just some particular vector some X particular plus the null space plus the null space obviously this vector by itself would also be a solution to ax is equal to B because you could just set X to to be equal to zero so in general and I haven't proven this to your vigor rigorously but hopefully you you kind of get the intuition behind it the solution and I'll do this in the next video just cuz I realize I'm running along on time the assuming assuming a xB a xB has a solution as a solution in the example we just did we as we can assume it has if we pick one of these points here so assuming it has a solution if we pick a point off of our image we're not going to have a solution but assuming a xB has a solution the solution set the solution set is going to be equal to some particular vector is going to be equal to some particular vector some particular vector so you can just think of it as a set with just one vector right there with or combined with or the union of that set with all your null space with your null space with your null space of this matrix right here I haven't proven this to you yet but hopefully you get the intuition why this is true we just solved it for particular cases that do have solutions we say hey it's going to take this form and I just showed you that this is the form of the null space this thing right here is the null space and the reason why we're doing that is because we've been talking about invertibility in order to be invertible you have to be you have to be on 2 & 1/2 1 and for something to be one-two-one one-two-one you have to have at most one solution at most one solution for that maps to a particular vector you might have none but you might have at most one so in order to have at most one solution and the solution set is always going to be equal to this so you're always going to have this solution so in order to have at most one solution your null space can't have anything in it or it can just have the zero vector so it's just going to have to be so that means that your null space the null space of a has to be to be trivial or just to have to be empty or just to just have just have the zero vector I'll do this a little bit more rigorously in the next video but I think when you do it sometimes with the rigger you don't necessarily get the intuition but this is going to be a very interesting takeaway and I think you already understand that this leading up to conditions for invertibility