im(T): Image of a transformation

Let's say that I have some set V that is a subspace in Rn. And just as a reminder, what does it mean? That's just some set, or some subset of Rn where if I take any two members of that subset-- so let say I take the members a and b-- they're both members my subspace. By the fact that this is a subspace, we then know that the addition of these two vectors, or a plus b, is also in my subspace. And this is our closure under addition. And by the fact that it's a subspace, we also know that if we multiply any member of our subspace by a scalar -- so the fact that those guys are members of our subspace -- we also know that if I pick one of them, let's say a, and I multiply a by some scalar, that this is also going to be a member of our subspace. And we sometimes call this closure under scalar multiplication. And then a somewhat redundant statement is that V, well it must contain the zero vector. And that's true of all subspaces. V -- let me write it this way -- the zero vector is a member of V. And it would be the zero vector with n components here, because V is a subspace of Rn. And why I say that's redundant, because if I say that any multiple of these vectors is also in V, I could just set the scalar to be equal to 0. So this statement kind of takes the statement into account. But in a lot of textbooks, they will always write, oh and the zero vector has to be a member of V. Although, that's kind of redundant with the closure under scalar multiplication. Fair enough. Now, let's say that I also have some transformation T. It is a mapping, a function, from Rn to Rm. What I want to understand, in this video is, I have a subspace right here, V. I want to understand whether the transformation of the subspace -- and what did we call that? We called that the image of our subspace, or our subset, either way. The image of V under T. In the last video, just to kind of help you visualize it. How did that work or -- we had some subset of Rn that looked like this. It was a triangle that looked something like that. And that was in Rn, this was actually in R2, it was a triangle that looked something like that. And we figured out it's image under T. So we went from R2 to R2. and we had our transformation. And it ended up looking something like this. If I remember it properly. It ended up looking like a -- gee, I don't remember it fully, but it was like a triangle that was skewed like this, rotated. So it was a -- actually I think it was more like -- I think that's right. It was rotated a bit clockwise like that and it was skewed. But the exact particulars of that last video aren't what matter. What matters is that you are able to visualize what an image under transformation means. It means you take some subset of R2, all of the vectors that define this triangle right here. That's some subset of R2. You transform all of them, and then you get some subset in your codomain. You could call this the image, because the transformation of that triangle, or if we call this s, it's equal to the transformation of s. Or you could say it's the image of-- you can just call it the set s, but maybe it helps you to visualize-- call it the image of this triangle under T. Or maybe even a neater way of thinking about it is, this triangle-- that skewed, rotated triangle-- this one is the image of this right triangle under T. I think that might make a little bit of visual sense to you. And just as a bit of reminder, in that last video these triangles, these weren't subspaces. And just as you could take scalar multiples of some of the vectors that are members of this triangle, and you'll find that they're not going to be in that triangle. So this wasn't a subspace, this was just a subset of R2. All subsets are not subspaces, but all subspaces are definitely subsets. Although something can be a subset of itself. I don't want to wander off too much. But this just helps you visualize what we mean by an image. It means all of the vectors that are mapped to, from the members of your subset. So I want to know whether the image of V under T is a subspace. So in order for it to be a subspace, if I take the transformation -- let me find two members of T. Well clearly if I take the transformation of any members of V, I'm getting members of the image. Right? So I can write this. Clearly the transformation of a and the transformations of b, these are both of members of our images of V under T. These are both members of that right there. So my question to you is what is the transformation of a plus the transformation of b? And the way I have written this, these are two arbitrary members of our image of V under T. Or maybe I should call it T of capital V. These are two arbitrary members. So what is this equal to? Well, we know from our properties, our definition of linear transformations, the sum of the transformations of two vectors is equal to the transformation of the sum of their of vectors. Now, is the transformation of a plus b, is this a member of TV? Is it a member of our image? Well, a plus b is a member of V, and the image contains the transformation of all of the members of V. So the image contains the transformation of this guy. This guy, a plus b is a member of V. So you're taking a transformation of a member of V which, by definition, is in your image of V under T. So this is definitely true. Now, let's ask the next question. If I take a scalar multiple of some member of my image of V under T, or my T of capital V, right there. If I take the sum scalar, what is this equal to? By definition for linear transformation, this is the same thing as a transformation of the scalar times the vector. Now is this going to be a member of our image of V under T? Well we know that ca is definitely in V, right? That's from the definition of a subspace. This is definitely in V. And so, if this is in V, the transformation of this has to be in V's image under T. So this is in -- this is also a member of V. And obviously, you can set this equal to 0. The zero vector is a member of V, so any transformation of -- if you just put a 0 here, you'll get the zero vector. So the zero vector is definitely -- I don't care what this is, if you multiply it times 0, you are going to get the zero vector. So the zero vector is definitely also a member of TV. So we come on the result that T -- the image of V under T, is a subspace. Which is a useful result which we will be able to use later on. But this, I guess, might naturally lead to the question, what if we go -- everything we have been dealing with so far have been subsets, with the case of this triangle, or subspaces, in the case of V. But what if I were to take the image Rn under T, right? This is the image of Rn under T. Let's think about what this means. This means, what do we get when we take any member of Rn, what is the set of all of the vectors? Then when we take the transformation of all of the members of Rn, let me write this. This is equal to the set of the transformation of all of the x's, where each x is a member of Rn. So you take each of the members of Rn and transform them, and you create this new set. This is the image of Rn under T. Well, there's a couple of ways you can think of this. Remember when we defined -- let's see, T is a mapping from Rn to Rm. We defined this as the domain. All of the possible inputs for our transformation. And we define this as the codomain. And remember I told you that the codomain is essentially part of the definition of the function or of the transformation, and it's the space that we map to. It's not necessarily all of the things that we're mapping to. For example, the image of Rn under transformation, maybe it's all of Rm or maybe it's some subset of Rn. The way you can think about it, and I touched on this in that first video, is-- and they'll never, or at least the linear algebra books I looked at, they didn't specify this-- but you can kind of view this as the range of T. These are the actual members of Rm that T maps to. That if you take the image of Rn under T, you are actually finding-- let's say that Rm looks like that. Obviously it will go in every direction. And let's say that when you take-- let me draw Rn right here. And we know that T is a mapping from Rn to Rm. But let's say when you take every element of Rn and you map them into Rm, let's say you get some subset of Rm, let's say you get something that looks like this. So let me see if I can draw this nicely. So you literally map every point here, and it goes to one of these guys. Or one of these guys can be represented as a mapping from one of these members right here. So if you map all of them you get this subset right here. This subset is, this is T the image of Rn, the image of Rn under T. And in the terminology that you don't normally see in linear algebra a lot, you can also kind of consider it its range. The range of T. Now, this has a special name. This is called -- and I don't want you to get confused -- this is called the image of T. Image of T. This might be a little confusing, image of T. So this is sometimes written as just im of T. Now you are a little confused here, you are like, before when we were talking about subsets, we would call this the image of R subset under T. And that is the correct terminology when you're dealing with a subset. But when you take, all of a sudden, the entire n dimensional space, and you're finding that image, we call that the image of the actual transformation. So we can also call this set right here the image of T. And now what is the image of T? Well, we know that we can write any-- and this is literally any-- so T is going from Rn to Rm. We can write T of x-- we can write any linear transformation like this-- as being equal to some matrix, some m by n matrix times a vector. And these vectors obviously are going to be members of Rn-- times sum Rn. And what is this? So what is the image -- let me write it in a bunch of different ways -- what is the image of Rn under T? So we could write that as T -- let me write it this way. We could write that as T of Rn, which is the same thing as the image of T. Notice we're not saying under anything else, because now were saying the image of the actual transformation. Which we could also write as the image of T. Well what are these equal to? This is equal to the set of all the transformations of x. Well all the transformations of x are going to be Ax where x is a member of Rn. So x is going to be an n-tuple, where each element has to be a real number. So what is this? So if we write A-- let me write my matrix A. It's just a bunch of column vectors, a1, a2. It's going to have n of these, right? Because it has n columns. And so a times any x is going to be-- so if I multiply that times any x that's a member of Rn. I multiply x1, x2, all the way to xn. We've seen this multiple, multiple times. This is equal to x1-- the scalar x1, times a1, plus x2 times a2, all the way to plus xn times an. And we're saying we want the set of all of these sums of these column vectors, where x can take on any vector in Rn. Which means that the elements of x can take on any real scalar values. So the set of all of these is essentially all of the linear combinations of the columns of a, right? Because I can set these guys to be equal to any value. So what is that equal to? That is equal to, and we touched on this, or we actually talked about this when we introduced the idea. This is equal to the column space of A. Or we just denoted it sometimes as C of A. So that's a pretty neat result. If you take -- it's almost obvious, I mean it's just I'm playing with words a little bit-- but any linear transformation can be represented as a matrix vector product. And so the image of any linear transformation, which means the subset of its codomain, when you map all of the elements of its domain into its codomain, this is the image of your transformation. This is equivalent to the column space of the matrix that you're transformation could be represented as. And the column space, of course, is the span of all the column vectors of your matrix. This is just all of the linear combinations, or the span, of all of your column vectors, which we do right here. Anyway hope you found that a little interesting, and you will be able to use these results in the future.