 Let A be a matrix whose column vectors are given as A1, A2, up to An, and let's be more specific. Let's say it's an m by n matrix right here. So I want you to recall the definition of the column space which we had introduced previously. The column space of the matrix A, which we denote as col of A. Sometimes we put parentheses around this just to emphasize it here, but col of A. This is the span of the column vectors of the matrix. Now because the column space is the span of vectors, this implies that the column space is a subspace. Now I want you to notice here that it's a subspace of the vector space fm, where m is the number of rows it has. So there will be n columns inside of the matrix, but each column contains m entries, and so we see that the column space is a subspace of fm. That's actually why in the first place we called it the column space as opposed to something else. This was a premonition of this notion of subspaces we had introduced later on. The column space of A is the set of all vectors b, which belong to fm, for which ax equals b is consistent. Basically can b be written as a linear combination of the columns of A. And so I just want to remind us of an example of how we would do such a thing. In this example, take the matrix A to be this 3x3 matrix, 1, 8, 7, 7, 6, 9, and 8, 7, 6, and then take the vector 3, 3, 7, and we're going to work mod 11 in this example right here. We want to determine whether b belongs to the column space of A or not. So to determine whether b is inside the column space of A, we have to determine whether b is a linear combination of the column vectors of A or not. This is the same as solving the linear equation, I should say the matrix equation, ax equals b, that corresponds to a linear system. And that linear system we have set up ready to go right here. We want to solve the associated augmented, well, we solve the linear system associated to this augmented matrix, which we have right here. So let's say we take the pivot in the first position. We're working on 7, but for the most part the arithmetic is going to be mostly the same as if you're working on the real numbers. To get rid of the 7 that's below the 1, we'll take row 2 minus 7 times row 1, and to get rid of the 8, we're going to get row 3 minus 8 times row 1. So doing the first one, we're going to get minus 7. On the next one, this is where we have to be a little bit more careful, right? On the next one, we're going to get 7 times, not 7, I'm sorry, no, that is right. We want to take 7 times 8, 7 times 8, and we're going to subtract that from 6. So we're working mod 11, so we do want to reduce things as we go. So if we take 8 times 7, be aware that's 56, but if you take away a multiple of 11, like 55 being the first one, we're really just subtracting 1 when you're working mod 7. And for the next example, if we take 7 times 7, that's normally going to be 49, right? But 49, if you take away 44, is going to be 5. So we're really just subtracting 5 when we're working mod 11. And then the last one here, if you take 21 times 7, 21 times 7, that's a negative 21. But if you add 22 to that, that's actually equal to 1. And so I'm just going to add 1 right here. And you can see that working mod 11, you'll get 7 minus 7, which is 0, 6 minus 1, which is 5, 9 minus 5, which is 4, and then 3 plus 1, which is 4 as well. Let's do that for the third row as well. We're going to minus 8, that's an easy one. But then we're going to do 8 times 8, which is 64. We could take away, for example, 66, that's a multiple there. And so instead of subtracting 64, we could actually add 2 to, now we do the same arithmetic mod 11. The next one, right, 7 times 6, or 7 times 8, excuse me, that gives us 56. 56, if you take away 55, that gives you a 1, so we're going to subtract a 1 still. And then lastly, if you take 3 times 8, that's 24, take away 22, 2 is congruent to, sorry, 24 is congruent to 2 mod 11, so we're going to subtract 2. And so that's going to give us 8 minus 8, which is a 0, 7 plus 2, which is a 9. The reason we did that is because we got 7 minus a negative 2, which is where the plus 2 came from, 6 minus 1, which is 5, and then 7 minus 2, which is 2. So the arithmetic's a little bit different working mod 11, but we still get the same thing right there. Moving to the pivot position in 2, 2, how do we get rid of the 9 right down here? Be a little bit more careful. We got to take row 3 minus 9 over 5, row 2. Now, whenever you're working mod anything, mod 11, we don't really want fractions. Can we write 9 fifths instead as something else? Like, for example, the numerator 9 can be replaced with any number, which is congruent to 9 mod 11, so we can add multiples of 11 or subtract multiples of 11 from the denominator if it helps. For example, 9 plus 11 is 20, and 20 is divisible as just usual integers. This is going to give you 4 when you're working mod 11, and therefore, instead of saying we're going to subtract 9 fourths, I think it's going to be a little bit easier to think of as we're going to take 4 times row 2. And I want you to confirm that, right? If you take 9 and you subtract from it 4 times 5, this is going to give you 9 minus 20, which is equal to negative 11, which is the same thing as 0 working mod 11, right? And so that subtracting 4 times row 2 will, in fact, do it for us. So now let's do the next ones. If you take 4 times 4, that's a negative 16, but 16, of course, take away 11 is going to be 5, so really, we're going to subtract 5. I always like to reduce along the way if I can, and then the next one's also subtract 5. And so you can see that all of these numbers are going to cancel out, leaving us a row of zeros in the bottom. Now, in order to determine whether we belong to the column space or not, we just have to make sure that this equation, ax equals b, is consistent. And we can determine consistency from echelon form. Now our matrix is an echelon form. We see the pivots in the first and second column. There's no pivot in the third column, but as we do have a row of zeros, that's not a problem. There's no inconsistency. So because the system is consistent, we then see that b does, in fact, belong to the column space of a, which is what we were trying to determine. So we are able to find out whether b belongs to this subspace or not. And in general, if you wanted to see if of, let's say you have some vector space, w, right, some subspace, and we know it's equal to a span of vectors. Let's call those vectors a1, a2, up to, say, an, like so. And you want to know, does b belong to this subspace? Well, if you have a spanning set, what you can do is you can always create a matrix whose column vectors are the spanning set in question right here. And then to solve, or to determine whether b belongs to the subspace or not, we then just have to solve the equation ax equals b. So in fact, determining whether a vector belongs to a subspace or not comes down to solving a linear system of equations. This is, of course, possible only when we have a spanning set for the subspace, but we'll see in the future that that's actually not too steep of a restriction whatsoever.