 In R2, all lines that pass to the origin are going to be one-dimensional subspaces. We'll worry about what exactly one-dimensional means later on. We have a geometric intuition right now. So for example, if you take any line through the origin, right, that's what forms the one-dimensional subspaces of R2. So you get something like this, this green and yellow line. These are examples of these one-dimensional subspaces. The x-axis, the y-axis, the subspaces themselves, and it doesn't really matter what angle you choose. As long as it passes through the origin, that's all that's going to matter to form a subspace, a one-dimensional subspace of R2. I should mention, of course, that the origin itself does form a subspace. It would be what we call zero-dimensional. It's just a point. And of course, the whole plane also forms a subspace, somewhat improper because it's everything. But every vector space is a subspace of itself, technically speaking, the way we've defined it. And so these would be all of the subspaces of R2. You have R2, you have the zero space, and you have all these lines that pass through the origin. That's going to be it. One could show that. Now, how does one actually argue that if you have some other subset that why can't it be a subspace? Well, remember, to be a subspace, you have to contain the zero vector. You have to be close under addition, and you have to be close under scalars. And so if you took an arbitrary affine set, if it passes through the origin, then it contains the origin. And like we mentioned here, those would be examples of subspaces. If your affine set doesn't pass through the origin, then automatically it's ruled out. So when it comes to subspaces versus affine sets, a flat is a subspace if and only if it contains the zero vector. We'll talk some more about that maybe a little bit later. But what about some other type of set? Let's take, for example, the set W. W is going to be the set of all vectors in the plane whose X and Y coordinates are non-negative. So let me actually scooch up the picture a little bit more right here, right? So we have our X and Y plane. And so we're talking about these points right here. So we want points which are greater, its X coordinates greater than equals zero, and its Y coordinates greater than equals zero. We're talking about the first quadrant, which is sometimes called Q1, right? If you think of the plane Q2, Q3, Q4, right? It's usual orientation. We're talking about Q1. That's what this set W is talking about right here, this first quadrant. Is it a subspace? I claim the answer is going to be no. Now, why is it not a subspace, right? What stops us here? Now, we have to be very careful when we look at this correctly. Now, does Q1, the first quadrant, contain the zero vector? And the answer is going to be yes, right? If you take the vector zero, zero, notice by definition here, the X coordinate has to be non-negative. Zero is okay. And the Y coordinate likewise has to be non-negative. So if you take zero and zero as your X and Y coordinates, these inequalities are satisfied. So it does pass the first test, right? We have these three conditions. The zero vector needs to be contained inside of W. We get that. What about the second condition? The second condition says if you have vectors U and V, which live inside of W, then the sum of the vectors U plus V must also be inside of W. And so if we try to think of that just like an arbitrary example, we might have a point right here. We'll call this one U. And we take some other point, maybe call it V, something like this. By the parallelogram rule to form the sum, we would take copies of U and V, like we see right here. And then the sum of the vectors would be the diagonal of this. You're going to get U plus V right here. And sure enough, that point does live inside of Q1, right? And so does that happen in general, right? So if you have vectors U and V, so you have like U equals, say, U1, U2, and then V equals V1, V2. When you add these things together, algebraically speaking, you're going to end up with U1 plus V1, U2 plus V2. And are these coordinates going to be non-negative numbers? So think about the X-coordinates. Because the X-coordinates are non-negative, U1 and V1 are numbers that are greater than equal to zero. So if you add non-negatives together, you're going to get something that's non-negative. So U1 plus V1 is going to be greater than or equal to zero in this situation still, right? What about the Y-coordinate? If we focus on this for a second, well, the same arguments in play right here. If Y1 is greater than or equal to zero, that means, sorry, if Y is greater than or equal to zero, that means U2 has to be greater than or equal to zero, V2 has to be greater than or equal to zero. If you take these two positives, or maybe they're zero, right? Yeah, together, you're going to get something that's greater than or equal to zero still. It does turn out that when you focus on axiom two here for a subspace, it passes that condition. The sum of two vectors inside the first quadrant will be a vector in the first quadrant. So so far, we've showed that it has condition one, it has condition two, but I still claim it's not a subspace. So what has to fail, the thing that has to fail is going to have to be the third condition. So let's, assuming my claim's even correct, right? So let me kind of erase what's on the screen a little bit. Imagine you have your vector, say V is this friend over here. Now if we scale V by any positive number, that's just going to elongate V some more, right? So we just make V get longer if it's bigger than one. Maybe if it's smaller than one, it gets a little bit shorter like this. If it's equal to zero itself, that'll just give you the zero vector. All of those vectors are inside of the Q1, aren't they? Well, the thing is, it has to be true for any scalar. In particular, if you take the scalar negative one, you're going to get negative V over here. And notice that negative V is actually living in Q3. That's not in Q1. It's not in the third quadrant anymore. And so this is where the violation happens. That for this, for this set of vectors, it does contain the zero vector, it is closed under addition, but it doesn't satisfy the scalar property closed under scalars. So if U belongs to W, then does C U belong to W? And in this situation, we get a big X. It doesn't belong to the set, and therefore it is not a subspace. To be a subspace, all three conditions have to be checked. Now, to show that something's not a subspace, in fact, we could have skipped steps one and two and jumped straight to here. They're like, oh, here's a counter example. And we should be very explicit about our counter example. Say V is the vector 11. Notice that belongs to W. Let's take the scalar negative one, which is a real number. But on the other hand, if you take negative 11, that is equal to negative one, negative one, which does not belong to W. And so because of this counter example right here, we see that condition three fails. It fails. And then we would then conclude that W is not a subspace. So not every subset of a vector space is automatically a subspace. It has to be a subset that itself resembles a vector space. And to do that, we have these three algebraic conditions. This is our litmus test when it comes to a subspace. It must contain the zero vector. It must be closed under addition. It must be closed under scalars. If any one of those three conditions fails, then we do not have a subspace like we saw in this counter example.