 Consider an M by N matrix A. We define the null space of the matrix A, which we'll denote MUL of A, the null space. This is gonna be the solution set to the homogeneous system AX equals zero. Now, sometimes this null space is called the kernel by some people, which if you can follow along the series, then kernel's like, oh, that has something to do with linear transformations, right? Well, there is a connection between linear transformations and matrices. In fact, if we define the matrix transformation associated to A, then the kernel of set transformation would be none other than the null space of this matrix as we're defining it right here. So the null space is the solution set to the homogeneous system AX equals zero. We define the nullity of the matrix A, which we'll just call it nullity of A, is gonna be the number of non-pivot columns in A. This is significant because the non-pivot columns will correspond to free variables. And we've seen before that a homogeneous system has non-trivial solutions for the free variables it has. So if we count the free variables in a system, then that essentially is telling us how big the null space, AKA the solution set of the homogeneous system is going to be. Now, in particular, the null space is going to be a subset of FN because it's a collection of vectors in FN. And that's where N here is the number of columns in the matrix A right here. But we can even do better. It's gonna, the null space is a subspace of FN. That's actually why we call it the null space as opposed to like the null set. We know that it's gonna be a subspace of FN here. And we can try to make an argument that the null space is just a span of vectors, which we saw in the previous example. And we'll see an example in this video as well. But it turns out we can argue that the null space is a subspace directly using the definition without any recitation of spans whatsoever. We don't need the geometric argument here. Because to be a subspace, we have to first show that the zero vector belongs to the null space of A. And that's pretty easy to do because if you take zero times A, that's equal to zero. And therefore zero is a solution to the homogeneous system, AX equals zero. So that gives us the null space that contains zero, right? The second condition was, if the vectors X and Y belong to the null space, then we need that their sum X plus Y also belongs to the null space. So let's suppose that X and Y belong to the null space. So if X belongs to the null space, it's a solution to AX equals zero. Then Y is also a solution. So we see that AY equals zero. Now consider the sum here. Well, A times X plus Y, well, since this is matrix multiplication, matrix times a vector, we can distribute here and we get AX plus AY. Well, AX equals zero and AY equals zero and zero plus zero equals zero. So this tells us that X plus Y is a solution that is A times X plus Y is equal to zero. So it's part of the null space. And the third condition to be part of the null space is that if X belongs to the null space, then we need that C, which is a scalar, CX also belongs to the null space. So again, using the assumption that AX equals zero, pick your favorite scalar C and consider the product here. A times CX, well, you can actually factor the scalar out. So you get C times AX, but AX is equal to zero and any scalar multiple of zero is again zero. So CX is contained in the null space and thus checking the three conditions. We've now determined that the null space is in fact a subspace of FN. So I wanna look at an example of a calculated null space, right? And so since the null space is the solution set of the homogeneous system, we can construct a spanning set for the null space. So we wanna represent the null space as the span of some set of vectors. This is similar to what we did before, but we're gonna have that perspective in mind right here. So let's take this three by four matrix, over the real numbers and we'll find a spanning set for the null space. This is gonna be basically coming down to solving the system, the linear system of equations. Now, because we're trying to solve a homogeneous linear system, normally there should be like this right hand side, zero, zero, zero, but I've seen many times already as you do row operations to a column of zeros, that column of zeros will never, ever, ever change. And so I'm just gonna omit it from consideration. It turns out that computing the spanning set for the null space is really the same thing as just determining whether the column vectors are linearly independent or not. It's just we go a little bit further. So I have a one in the pivot position, one, one. So I'm gonna use that. We're gonna get rid of the two below it by taking row two minus two times row one and getting rid of the three in the third row, we're gonna take row three minus three times row one. So we're gonna get minus two, minus six, plus six and then minus eight for row two. That's gonna give us a zero, negative five, ten and negative five and then for row three, we're gonna get minus three, minus nine, plus nine and then minus 12. Three minus three is zero, negative two minus nine is negative 11, 13 plus nine is 22 and one minus 12 is negative 11. That takes care of the first column. For the second column, we have a negative five right here. I can't help but notice that everything in the second row is actually divisible by five. So I'm actually gonna divide the second row by five, negative five in fact, but I also noticed that everything in the third row is divisible by negative 11. So since I'm at it, as well as we'll do that as well. And so dividing the second row by negative five and the third row by negative 11, I actually can see now that the second and third rows are now actually identical. Kinda like the Weasley twins there, we're identical. In which case, we're just gonna take row three minus row two and that'll just cancel out that entirely, giving us now a row of zeros. So that then gives us a matrix which is in echelon form. We can actually identify all of the pivots in the matrix now. There's a pivot column in the first and second, which actually tells us that column three and column four have no pivots, these are gonna be free variables. We're gonna have in fact two free variables in this system, two free variables. What that tells us is that our null space is gonna be spanned by a minimum of two independent vectors. All right? And so to help us here, we're gonna finish by putting this thing in row reduced echelon form. Take row one minus three times row two. So we're gonna get minus three, we're gonna get plus six and then minus three right here. Three minus three is zero. Negative three plus six is three and four minus three is one. So this gives us our REF right here. And again, identifying our pivots. We see a pivot in the first column and the second column. So we're gonna think of this matrix now in REF. We're gonna realize this as a, we're gonna realize it as a linear system. So we get the equation x one plus three x three plus x four equals zero, x two minus two x three plus x four equals zero and then zero equals zero. We don't need that. Solving for the dependent variables, we get x one equals negative three x three minus x four and then x two equals two x three minus x four. And so if we write our general solution, notice we have x one, x two, x three, x four using the identifications we saw earlier for the dependent variables, we get the following. And the next thing we're gonna do is we're gonna separate this vector we see right here based upon the three variables. Notice that these positions depend on x three. So if we separate those from the x fours, we get a negative three x three. We get a two x three for the second position. We just get an x three for the third position and then there's nothing, there's no x threes in the fourth position. We're gonna do a similar thing for the next vector. Take a negative x four, take a negative x four, get a zero this time because there's no x four in the third position and then x four for the fourth position. So we separated the vector according to the x three, x four, we just separate those entirely. And then looking at the first one, everything's divisible by x three. So we factor out the x three that leaves behind negative three, two, one, and zero. And then looking at the second vector, we factor out the x four and that leaves behind a negative one, negative one, zero, one. And so it turns out these two vectors right here then form the spanners of a plane for which the null space is actually equal to that plane. Set say u to the first vector negative three, two, one, zero and v to be the second vector negative one, negative one, zero, one. And then we've now seen that the null space is the span of these two vectors. And so these two vectors is the form a spanning set for the null space which in this case we see is a plane.