 All right, so now we're to the main topic for this lecture right here. That is, we are here to the fundamental theorem of linear algebra. Now, this thing is a really long thing to describe. But bear with me here. The fundamental theorem of linear algebra, it really is a picture. That's how we want to think of it. But let's state the statements quickly. Let's consider an m by n matrix A. And let's take, then any vector v that lives inside of fn, which notice A is an m by n matrix, any vector v and fn can be decomposed uniquely as a sum of vectors. v into two pieces will call it v rho and v null. The reason for the names is that v rho will live inside of the row space. And v null will live inside the null space. As we've seen before, the row space is perpendicular to the null space. That is, the row space is the orthogonal complement of the null space. And so if you consider the linear system A v equals B, then it turns out that v rho is a particular solution of the linear system A x equals B. So notice the difference here that x is a variable here. So it's a bucket waiting for someone to place a vector in it. And we're saying that v is a specific vector you can put in there. In fact, v rho is a particular solution to that one. And it's the solution of shortest length. That is, there's no solution to A x equals B that's shorter than v rho. And v null is going to be the solution is a solution associated to the homogeneous system A x equals 0. Now if we flip to the transpose, it's also true that if you take the vector w, any w and fm there, it'll be decomposed uniquely as a sum of two vectors, w column and w left null. These right here and this up here is just the orthogonal decomposition we talked about before. W column will be inside the column space. W left null will be inside the left null space. By duality, the column space is the orthogonal complements to the left null space. And if we look at any equation A transpose w equals B, then w column will be a particular solution of the system A ty equals B of shortest length again. And the vector w left null will be a solution to the homogeneous system A ty equals 0. And some other things we've observed is that the rank of A plus the nullity of A is equal to n. This is commonly known as the rank nullity theorem. The dual principle there is that the co-rank plus the co-nullity will always equal m, sorry, number of rows is m, n is the number of columns. And the rank of A always equals the co-rank of A, which is the number of pivots in the echelon form. So that's the statement and that was quite verbose. I apologize for that. But really when it comes to the fundamental theorem of linear algebra, really the result is a picture that I want to illustrate you here. So think of the subspace here in green as the row space, and then next to the row space is the null space. Okay, here's the null space. Now what we know about the row space in the null space is that these are orthogonal complements. So the only thing that lives in both the row space and the null space is the zero vector. Otherwise everything between the two is orthogonal to each other. And so if you put them together, you end up with the domain, which is f, n. All right, now let's come to the other side. If we take the co-domain of our transformation, you have over here the column space of A, the column space, and then there's also the left null space, left null space. These two vector spaces are likewise orthogonal complements to each other. They meet at a right angle, and the only thing that lives in both of them is the zero vector. But if you put the two spaces together, you get all of the co-domain fm. And let's see, the dimension of the null space, like you mentioned before, that was our rank and the rank, I'm sorry, the nullity for the null space here, that's n minus p. If we come back to the row space, its dimension is the rank, which would be p. For the column space, its dimension is likewise the co-rank, which is equal to p. So you see the quality there. And then the dimension of the left null space, the co-nullity, this is m minus p. And so I want us to mention how points move around in this picture. So if we were to take a generic vector v that lives inside this space, okay? Well, it can be broken into two pieces. There's one piece, which we called v-row, and there's this other piece, which we called v-null. So if you put these two pieces together, if you add them, v-row plus v-null, you get back v. Now let's say that v maps over to some vector b over here in the column space. It'll live inside the column space. Well, the row vector that is the vector row will map to that exact same thing, and the vector v-null maps to the zero vector. So the v-null parts doing really nothing through the transformation, it just vanishes v and v-row mapped to the exact same place, right? On the other hand, the left null space is entirely missed by this transformation T. A is the matrix representation of this transformation T. There's nothing in the null space that can get hit. The stuff that we are missing is the null space. The only things we can hit is inside of the column space. And so in particular, when you take all of the vectors v, we take all the vectors v, the shortest, the shortest vector that maps to b is gonna be this v-row. And so this is sort of like a minimum and optimization to this linear system of equations here. And so I wanna show you exactly how one can compute this v-row or x-row in this situation. So let's take a system of equations, three negative, three negative two, first row, negative five, four, three, second row, and one negative five, negative two, third row. If this at all looks familiar, this is the same matrix we've been playing with this whole time. But let's take a different vector b over here, three negative four, five. This is not the y that we talked about before. Let's actually find a solution to this system equation that lives inside the row space. So the first thing we're gonna do is we're gonna first solve, we're gonna solve for the general solution. And we've done this before, solve for the general solution. Well, we have our matrix augmented with the column, we row reduce it, we get something of the following form, the following RREF, you can see that here. And so pulling this thing apart, what are some things we can notice? We have two pivot columns, the first and the second. So our third column represents a free variable. So let's start off by setting the free variable equal to zero. If x3 is equal to zero, then this would suggest that x1 is equal to zero, x2 is equal to negative one. And like we said, x3 was zero. So it has to do with this number here and this number here. All right. Then the next part, we have one free variable and the spinner that's associated to this free variable is we're gonna look at these numbers right here and we're gonna take their negation. So this corresponds to the third variable, so we put a one right there. And so we're gonna get positive one third right here and negative one third right here. Now, if you don't like the fractions, you can replace them. Cause after all, since it's just the spanner of the null space of the matrix, you can times everything by three and that doesn't change the span. So maybe I'll call it a different parameter T now. So we get three, I'm sorry, one times everything by three one, negative one and three. So one negative one three gives us a basis for the null space of A that doesn't require any fractions. So this right here is the general solution. All solutions to the system will look like T minus one minus T and three T. Every solution. So in particular, X row is gonna look like this cause that's a particular solution just looking for the shortest one. Now, how can we find X row using this? Cause we have basically to get the pick a choice for T. Well, one possibility is we could take the length of this and we then wanna choose T to be minimum to T to minimize, I should say. But that turns into essentially a calculus problem and optimization problem for maybe from a calculus one or something. So I wanna provide a linear way of trying to solve this. And the idea is gonna be the following right here. Remember that the row space is the orthogonal compliment to the null space. And since we now have a solution, we have a basis to the homo, we have a basis to the null space which is the general solution to the homogenous system of equations. We can take our vector X row and we take the dot product with each and every basis vector for the null space. And that should equal zero. And so if you take what we got for X row a moment ago, T, negative one minus T, three T, and we dot this with the vector one, negative one, three, which was again, a basis elements for the null space there. That should equal zero. And if we do the dot product, we're gonna get one times T, which is a T, you get negative one times negative one minus T. So that's gonna be a one plus T and then we're gonna get a nine T. Right there. And so this will all equal zero. Combining like terms, we get 11 T plus one equals zero. Thus, 11 T equals negative one or T equals negative one, 11s. Now this gives us the parameter. This is the parameter for which we should plug into the general solution. And this will then give us our X row. And so if we plug that in X row there, so we get negative one, 11th for T, we're gonna get negative one plus 11th, right? That's gonna give us negative 10, 11th. And then three times negative 11th, 11th is negative 3, 11th. And so this right here gives us our X row. And so by the calculation we did before, we can see that this is in fact in the row space because it's orthogonal to the null space one, negative one, three. One can also check it's a solution to the system. Just plug in those numbers, negative 11th into the system. Here, negative 10, 11th's here and negative 3, 11th's there. I'll leave it up to you as an exercise to check that this does solve the system. And this is also the shortest vector one could do. You can't get rid of the fractions this time because we're not trying to change the vector. Earlier we were able to times play fractions because we didn't change the span. We can change the span and set though, but we do need to keep this vector right here. This gives us our shortest vector which solves the system of equations. Now, if the null space had a nullity greater than one, then these equate, this right here, there might be more than one equation. You're gonna take X row times each and every basis element to the null space. And that might actually give you a system of equations for which you solve for all the multiple parameters as well. And so that concludes section 4.6. From watching this video I should hope you're ready to go work on the homework. If you have any questions, feel free to post a comment down below. I'll be glad to answer anyone's questions or have a further discussion about the topics of this lecture if you would like. If you liked this video and want to see more like it, feel free to subscribe and let your friends know this can be fun. All right, I will see you next time. Have a great day, bye.