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# Determining whether a transformation is onto

## Video transcript

Let's say I have a linear transformation T that's a mapping between Rn and Rm. 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 it a little higher to save space-- so we can write that the transformation T applied to some vector x is equal to some matrix times x. And this matrix, 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 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 have been talking about invertibility of functions, and we can just as easily apply it to this transformation, because transformations are just functions. We just use the word transformations when we start talking about maps between the vector spaces or between sets of vectors, but they are essentially the same thing. Everything we've done the last two 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 Rm. So if you take some vector here, x, T will 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? And we learned in the last video that there's two conditions for invertibility. T has to be onto, or the other way, the other word was surjective. That's one condition for invertibility. And then T also has to be 1 to 1. And the fancy word for that was injective, right there. So in this video, I'm going to just focus on this first one. So I'm not going to prove to you whether T is invertibile. We will at least be able to try to figure out whether T is onto, or whether it's surjective. So just as a reminder, what does onto or surjective mean? It means that you take any element in Rm, you take any element in your co-domain, so you give me any element in your co-domain-- let's call that element b; it's going to be a vector-- the fact, if T is surjective or if T is onto, that means that any b that you pick in our co-domain, there's 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 of our transformation is all of Rm. 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. Now, if T has to be onto, that means that Ax, this matrix vector product, has to be equal to any member of our co-domain 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 onto, implies that for any vector b that is a member of Rm-- so any b here-- there exists at least one solution to a times x is equal to b. Where, of course, x is, the vector x is, a member of Rn. It's just another way of saying exactly what I've been saying in the first part of this video. You give me any b in this set, and then there has to be, if we assume that T is onto, or for T to be onto, 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 instead of any. But they are the same idea. But, 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 A times x has to be equal to-- you can construct any member of Rm by taking a product of A and x, where x is a member of Rn, where x is a member here. Now what is this? If x is an arbitrary member of Rn, Let me write like this. We know that matrix A would look like this. It'll be a bunch of column vectors. a1, a2 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. Well, what does this product look like if where 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 x2 times the second column vector of A, all the way to plus xn times the nth column vector of A. That's what this product is. And in order for T to be onto, this combination has to be equal to any vector in Rm. Well, what does this mean? These are just linear combinations of the column vectors of A. So another way to say that is for T to be onto, So for T to be surjective or onto, the column vectors of A have to span Rm, have to span our co-domain. 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. This vector is just a bunch of arbitrary real numbers. So for T to be onto, the span of a1, a2, all the way to an, has to be equal to Rm, has to be equal to your co-domain. That just means that you can achieve any vector in your co-domain with the linear combinations of the column vectors of this. What's the span of a matrices' column vectors? By definition that is the matrices' column space. So we could say, that means that the span of these guys have to be Rn, or another way is that the column space of A-- let me switch colors-- the column space of A has to be equal to Rm. So how do we know if a vector's column space is equal to Rm? So here, maybe it might be instructive to think about, when can we not find solutions to the equation Ax is equal to b? So whenever you're faced with this type of 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, perform a bunch of row operations. You have to perform the entire rows in both cases, on both sides, and we've done this multiple times. 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 R, capital R, to be the reduced row echelon form of A. And we've done many videos on that. That's just, you have a matrix where you have your pivots and the pivot will be the only non-zero entry in its column. But not every column has to have a pivot in it. Maybe, you have a free column or non-pivot column and then they could have a bunch of 0's. And maybe this has a pivot. This would have to be 0 if there is 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 the don't have pivots, but whenever you do have a pivot, they have to be the only non-zero entry in their column. This is reduced row echelon form. So, what we do with any matrix is we keep performing those row operations so that we eventually get it into a reduced row echelon form. And as we do that we are performing those same operations on the right-hand side. We're performing on the entire row of these augmented matrices. So 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 will perform a bunch of operations and this will be 3, 2, 1, or something of that nature. Now, when does this not have a solution? We reviewed this early on. The only time where you don't have a solution, remember, there are three cases, where you have many solutions. and that's the situation where you have free variables. We've talked about that before. You have the case where you have only one unique solution, that's the other case. And then you have the final case where you have no solutions. And when do you have no solutions? What has to happen to have no solution? 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. I don't know what all this stuff looks like maybe there's a 1 here, a bunch of stuff. There is a 1 here and a 0. But if you have a whole row, at least one whole row of 0's, you just have a bunch of 0's like that. And over here you have something that is non-zero. This is the only time that you have no solution. So let's remember what we're even talking about this stuff for. We are saying that our transformation is onto, if its column vectors or if its column space is Rm, if its column vectors span Rm. And what I'm trying to figure out is how do I know that it spans Rm? So, essentially, for it to span Rm, you can give me any b 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? Well, we definitely don't get a solution if we have a bunch of 0's 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 where we have a bunch of 0's. 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 draw it like this. Let me start it like this. So let's say I have my matrix A, and I have my b1, b2, let me write it all the way, and then you have bm. Remember, this is a member of Rm. And we do our reduced row echelon form with this augmented matrix, and A goes to its reduced row echelon form. Let's say it's reduced row echelon form has a row of 0's at the end of it. So it just has a row of 0's right there. Everything else looks like your standard stuff, 1's and 0's. But the last row, let's say it's a bunch of 0's. And when we perform the row operations here on this generalized member of Rm, this last row has some function. Maybe it just looks like 2b1 plus 3b2-- I'm just writing a particular case, it won't always be this-- minus b3. And it will essentially be some function of all of the b's. So let me write it this way. I'm writing a particular case in here, maybe I shouldn't have written a particular case. This will be some function of b1, b2, all the way to bm. Now, clearly if this is non-zero, we don't have a solution. And so if we don't have a solution for some cases of b, then we are definitely not spanning Rm. So, let me write that down. If we don't have a solution for some cases of b, then we don't span Rm. I don't know 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, and we perform a bunch of row operations until we get A, we get this matrix A to 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 b4, or something. And then 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 0's here, the only way that you're going to have a solution is if your vector b, if its entries satisfy this function here on the right, so that this thing equals 0. So it's only going to be true for certain b's. And if this only has a solution for the certain b's that make this equal to 0, then we definitely are not spanning all of Rm. Let me do it visually. So if that is Rm, and if we put-- if this is only 0 for a couple of b's, for, let's say for some handful of b's, then these are the only guys that we can reach by multiplying A times some vector in Rm. And we definitely won't be spanning all of Rm. In order to span all of Rm, when we put this in a reduced row echelon form, we have to always find a solution. And the only way we are always going be finding a solution is if we don't run into this condition where we have a row of 0's. Because when you have a row 0's, then you have to put the constraint that this right-hand side has to be 0. So what's the only reduced row echelon form where you don't have a row of 0's at the end? Well, any row in reduced row echelon form either has to have all 0's or it has to have a pivot entry in every row. So the only way that you span-- so T is onto, 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 of A has a pivot entry in every row. And how many rows does it have? This is an m by n matrix. It has m rows and n columns. So it has a pivot entry in every row. That means that it has to have m pivot entries, right there. Now, what's another way of thinking about that? Remember, earlier on, several videos ago, we thought about how do you figure out-- and this might confuse things a little bit-- how do you figure out the basis for your column space? So the basis 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 of your matrix, and then, essentially-- let me draw it a little bit different heree-- well, you put it in reduced row echelon form. So let's say that's just reduced row echelon form. And you look at which columns have pivot entries. And the corresponding columns in your original matrices, in your original matrix, forms the basis for your column space. So let me draw that out. So, I'll do a particular instance. So, let's say that it has its column vectors a1, a2, all the way to an. 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 this is a 2 there. I'm just picking particular numbers. 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 0's, 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 these guys. But, what's the minimum set you need to have that same span? Well, you look at which one has a corresponding pivot entries or pivot columns. And you say, I have a pivot column here, and I have a pivot column there. So the basis for my column space must be this column in my original matrix, and that column in my original matrix. And then we said, well, how do you define the dimension of it's column space? And you just essentially count the number of vectors you need for your basis, and we call that the rank of A. This is all review. The rank of A was equal to the dimension of the column space of A, which is equal to 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 onto, if and only if, its column space is Rm, which is the case if it has a pivot entry in every row in its reduced row echelon form. Or, since it has m rows, it has to have m pivot entries. So for every row, you have a pivot entry, but every pivot entry corresponds to a pivot column. So if you have m pivot entries, you also have m 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. So this whole video was just a big long way of saying that T is onto. And another way of saying that is that if you have your domain here, which was Rm, and you have your co-domain here, that is Rm, that every member of Rm can be reached by T by some member of Rn. 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. We're not even talking about 1 to 1 yet. So we say that T is onto, if and only if, the rank of its transformation matrix, A, is equal to m. So that was the big takeaway of this video. 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 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 1, 2, 3, 4, 5, 6, times the vector x. So this is clearly a 3 by 2 matrix. Let's see if S is onto. 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 1, 2, 3, 4, 5, 6. Now, let's keep the first row the same, that's 1, 2. Let's replace the second row with the second row minus 3 times the first row. Actually, let's replace it with 3 times the first row minus the second row. So, 3 times 1 minus 3 is 0. 3 times 2 minus 4, that's 6 minus 4, is 2. Now let's replace the third row with 5 times the first row, minus the third row. So 5 times 1 minus 5 is 0. 5 times 2 is 10, minus 6 is 4. Now, let's see if we can get a 1 right here. So I'm going to keep my middle row the same. Or, actually, let's just divide the middle row by 2, or multiply it by 1/2 So, you get 0, 1 and then you have a 0, 4, 1, 2. And now let's try to make these 0, get it into 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 0. Now, let's replace this last row with the last row minus 4 times this row. So we get 0 minus 4, times that, is 0. 4 minus 4, times 1, is 0. So notice, we have a row with 0's here. We have two pivot entries or two rows with pivot entries, and we also have two pivot columns right there. So the rank of this guy right here, of 1, 2, 3, 4, 5, 6. The rank of that is equal to 2, which is not equal to our co-domain. It is not equal to R3. It's not equal to R3, so S is not onto, or not surjective. It's one of the two conditions for invertibility. So, we definitely know that 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 1 to 1.