 And so for all of the fundamental spaces, we have an algorithm we can use to find a basis for the subspace by row reducing the matrix A. I wanna provide one for the left null space as well. It's gonna be a little bit different. You need a little more information than just the RREF of the matrix, but this will help us out a lot here. So to find a basis for the left null space here, the basis for the left null space of A. What I want you to do is you're gonna take the matrix A and then you're gonna augment it with the appropriately sized identity matrix. So if A is M by N, then you have to augment the M by M identity matrix. And so then row reduce this matrix to the R, well, honestly, any echelon form would work here, but we'll take the RREF of you. But yeah, any echelon form would work right here for you. Now in the process of reducing A into U, the identity matrix will transform into the matrix, we'll call it, sorry, not E, we're gonna call this one, sorry, not epsilon, we're gonna call it E here. And so I will transform into this matrix E right here. Now suppose that the matrix U had a row of zeros. So let's say that U had a row of zeros. And let's say that this is in fact, the i-th row. If this matrix had a row of zeros in the i-th row, we're gonna let epsilon i be the i-th row of E. So this is gonna equal the i-th row of the matrix E. So since the matrix, if we think of elementary row operations as matrix multiplication, we see that the operations we do to A to get U will transform i into E. And so in fact, the matrix E is just the combination of all of the elementary row operations that turned A into U. So if the above row equivalency happens, that means that E times A is equal to U. And so if the row, epsilon i transforms A into U and the i-th row of U is itself a row of zeros, this tells you that epsilon i, if you times it by the matrix A, it'll give you the i-th row of U, which is a row of zeros. And so in fact, what we see here is that whenever you have a row of zeros inside of U, the corresponding row of a row vector on the other side of the line is gonna give us something inside of the left null space of the matrix U, the matrix A. And in fact, because U is an echelon form, if we select all of these epsilon i's, this will give us a basis for the left null space of U. So let's look at a specific example like we had here. So we're gonna take the exact same example we were working with a moment ago. And so if we take A augment i, we're gonna, in this case, we would take i-3, you see that A is right here and here is i-3. I'm gonna skip the row reduction this time and we did it already. And so if you kept track of things, we have a pivot in the first position, in the second position, second column, and you'll notice there is a row of zeros right here. So this tells us that this right here is our epsilon three, negative one fourth, negative one seventh, and one 28th. So this right here gives you something, epsilon three is inside the left null space of A. And in fact, the span of this single vector, negative one fourth, negative one seventh, and one 28th. This vector by itself spans the entire left null space. So this single vector forms a basis for it. Now, if you don't like the fractions here, I mean, because after all, many of us in the viewing audience right now might be ratio phobics. So some of us are afraid of fractions, right? If you don't want to, if you don't want fractions here, you could scale everything in this vector by 28, the least common denominator there, in which case then you can replace the span, the spanning set with this time the vector, I'm actually gonna times everything by negative seven, negative 28, excuse me. So this gives us seven, four, and negative one as a basis for the left null space. And this vector, seven, four, negative one will probably look familiar as this was, none other than, where did it go? This was none other than the vector we were playing with before seven, four, negative one, all right? And so the left null space, it's the set of all vectors which multiply on the left to give you zero. It's the set of all vectors you cannot hit with the linear transformation, the matrix transformation. And we can find a basis by this, by sort of a calculation very similar to how we found inverse matrices before.