 So let's really cement our understanding of this column space. So the whole idea is I want to solve this equation, a system of linear equations, a linear system, and I solve this by this matrix of coefficients, my matrix of unknowns, and this solution matrix. So let's just ask this question of this matrix of coefficients and I have one, three, two and six and I'm asking, you know, what solutions are there in R2? Really this is going to be in the plane, it's only two rows there, so each column vector is two so I can only have two and I'm asking what, you know, what possible solutions can there be in R2? So remember there's my x, there's my b. So what I'm really asking is what linear combinations of these two column vectors can I have, can I possibly create? And what I'm suggesting is only if b lies inside of that, those linear combinations that I could possibly get this. If b lies outside of the linear combinations of these two, I will never get this, well, zero, zero, zero. Anyway, let's just have a look at one, two and three and six. Notice though that three and six is just three times one and two. So this one times three is three, two times three is six. This vector here is already a linear combination of the first one. I have added nothing to this. I have added nothing, it's the exact same vector that I'm dealing with, so always watch out for that. So that means I only have this one vector. If I look one across one, two up, that's the only vector that I have. And that's the vector, that's the vector. And any linear combination of this vector will just lie on this infinite line through that vector. That means that this matrix here can only have solutions b, the right-hand side of my linear equations that are on this line. If I'm given any b that is off of this line, anyway off of this line, I cannot get this solution. That's what we're trying to say. So the linear combination of these two will give me this space that I have to deal with, the subspace in R2 in this instance. If it was three by three it would be bigger, it wouldn't be square, remember. And that is the only way that I could get the solution is this, better be in the subspace of that. So the second thing that flows from this little example is, if s is a set of column vectors of A, so set of column A, then the linear combination is a subspace in R2 and we call that spanned by s. So that's another way to look at it. So imagine I have another vector. There's our first question there actually. What subspace in R2 can I span with A? And my A is here 1, 2 and 1, 3. So that's the other way to look at it. If I'm just given two arbitrary vectors, say A1 and A2, that's my first one, and let's make it V1 and V2, I'm given those two vectors. I'm asking what subspace in R2, because this is clearly in R2, in R3 where there'd be zero there, what subspace in R2 can I create? What subspace in R2 can I create by linear combinations of these? Linear combinations of these two will be a subspace. Just as the linear combination of these two gave me this subspace, and the only subspace here is this line, and yesterday we said that that line is a subspace in R2. So if I look at these 1 and 2 and 1 and 3, this is not a linear combination of the other one. So the subspace, and if we think of this, you know, it's 1, 1, 1, 2 and 1, 3, they lie there. The subspace, linear combinations of that will actually give me the whole of R2. The whole of R2 is the subspace that I can create in R2 of that. So that actually spans, and that's what we call, it spans, if I'm just given two vectors, it spans a space, linear combinations of them span a space. The second question here, if V, this uppercase V, is all vectors perpendicular to that vector 1 to 1. Let's just clean the board. Board is clean, so I'm asking, you know, vectors perpendicular to that. Remember when vectors are perpendicular, so if I have V there, if I have another U, the dot product of them must be 0. If the dot product is 0, so that means anything that I multiply with, remember that multiplication is just, well, what did we make it? Let's make it then U transpose V, so I can write U1, U2, U3, and V1, V2, V3, and in our instance, let's make that 1 to 1. So I have that U1, so remember this, I'm just going to get a scalar, so U1 plus twice U2 plus U3 must equal 0. If that is so, then these two are perpendicular to each other. So let's look something times U1 plus 2 of this plus 2, so I can make this, what can we make this? So if we make that a negative 1, and we make this a 1 and a 1, 1 plus 1 is 1, minus 2 is 0, so this vector, 1 minus 1, 1, this vector will definitely be perpendicular to that vector. If you multiply those together, you're going to get 0. So I'm saying V is all the vectors that are perpendicular to this because I can make other vectors that are perpendicular, suggest a subspace of that. Well, any line through this one vector would be a subspace, so that would always be one of the easy subspaces. Remember, 0, 0, 0 is also, there's another subspace for you, but anything through this line, 1, negative 1, 1, through that vector would be a subspace. If V is all the 2 by 2 symmetric matrices, remember symmetric matrix is, I have values on my main diagonal and all the values across from that, those two and those two, and these two, they are exactly the same. So if that's all 2 by 2 symmetric matrices, so I'll just have two values here, so as long as they're the same, that's symmetric matrix, suggest a subspace. Well, what about just the diagonal matrix of all of these are 0, and I have one value there and I have another value there. The symmetric, the diagonal matrix is a subspace, for instance, of the symmetric matrices, and then the last one there, if V is all polynomials, f of x, such that f prime, prime, prime, so the third derivative of x equals 0. So first of all, let's look at all the polynomials that will have a third derivative, which is 0. So if I have f of x, and that equals ax squared plus bx plus c, its first derivative is going to be, so if prime of x is the bit of calculus here, that's 2ax plus b, and its second derivative of x, that is going to be 2a, and the third derivative of x is going to equal 0. So definitely, these quadratic, to the power 2 polynomials, is the set of all polynomials, just a subspace of that. Well, a subspace of those polynomials just make that even smaller than that, that would be a subspace of those polynomials. So in each of these cases, you can clearly see what we mean by this idea of a vector space, and specifically in the first one, spanning, but we can also say that the subspaces that we created here, they all span this bigger space, and the questions that you might just get is you first have to construct your original space that you are dealing with before you can get to, so these are just all trick questions, like this little calculus trick question here, to do that, but please understand this idea then, space the subspace, how to, if you've given some vectors, you can span a subspace, and those where we had, especially where we had that one linear combination of the other, watch out for those. So I think by now you have a pretty idea of what we are trying to achieve here is when can we get a solution here, most definitely if B is in the subspace created by a linear combination of the column space of A.