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Introduction to the null space of a matrix

Showing that the Null Space of a Matrix is a valid Subspace. Created by Sal Khan.
Video transcript
Let's review our notions of subspaces again. And then let's see if we can define some interesting subspaces dealing with matrices and vectors. So a subspace-- let's say that I have some subspace-- oh, let me just call it some subspace s. This is a subspace if the following are true-- and this is all a review-- that the 0 vector-- I'll just do it like that-- the 0 vector, is a member of s. So it contains the 0 vector. Then if v1 and v2 are both members of my subspace, then v1 plus v2 is also a member of my subspace. So that's just saying that the subspaces are closed under addition. You can add any of their two members and you'll get another member of the subspace. And then the last requirement, if you remember, is that subspaces are closed under multiplication. So that if c is real number, and it's just a scalar. And if I multiply, and v1 is a member of my subspace, then if I multiply that arbitrary real number times my member of my subspace, v1, I'm going to get another member of the subspace. So it's closed under multiplication. These were all of what a subspace is. This is our definition of a subspace. If you call something a subspace, these need to be true. Now let's see if we can do something interesting with what we understand about matrix vector multiplication. Let's say I have the matrix a-- I'll make it nice and bold-- and it's an m by n matrix. And I'm interested in the following situation; I want to set up the homogeneous equation. And we'll talk about why it's homogeneous. Well, I'll tell you in a second. So let's say we set up the equation. My matrix a times vector x is equal to the 0 vector. This is a homogeneous equation, because we have a 0 there. And I want to ask the question-- I talked about subspaces. If I take all of the x's-- if I take the world, the universe, the set of all of the x's that satisfy this equation, do I have a valid subspace? Let's think about this. I want to take all of the x's that are a member of Rn. Remember, if our matrix a has n columns, then I've only defined this matrix vector multiplication. If x is a member of r, and if x has to have exactly n components, only then is it defined. So let me define a set of all the vectors that are a member of Rn where they satisfy the equation a times my vector x is equal to the 0 vector. So my question is, is this a subspace? Is this a valid subspace? So the first question is, does it contain the 0 vector? Well in order for this to contain the 0 vector, the 0 vector must satisfy this equation. So what is any m by n matrix a times the 0 vector? Let's write out my matrix a-- my matrix a, a[1,1] a[1,2] all the way to a[1,n] and then this, as we go down a column, we go all the way down to a[m,1] and then as we go all the way to the bottom right, we go to a[m,n] and I'm going to multiply that times the 0 vector that has exactly n components. So the 0 vector with n components is 0, 0, and you're going to have n of these. The number of components here has to be the exact same number of the number of columns you have. But when you take this product, this matrix vector product, what do you get? What do we get? Well, this first term up here is going to be a[1,1] time 0, plus a[1,2] times 0, plus each of these terms times 0. And you add them all up. a[1,1] times 0, plus a[1,2] plus a[1,2] times 0, all the way to a[1,n] and times 0. So you get 0. Now this term is going to be a[2,1] times 0, plus a[2,2] times 0, plus a[2,3] times 0, all the way to a[2,n] times 0. That's, obviously, going to be 0. And you're going to keep doing that because all of these are, essentially,-- you can kind of view it as the dot product of-- I haven't defined dot products with row vectors and column vectors, but I think you get the idea-- the sum of each of these elements, multiplied with the corresponding component in this vector. And, of course, you're just always multiplying by 0 and then adding up. So you're going to get nothing but a bunch of 0's. So the 0 vector does satisfied the equation. A times the 0 vector is equal to the 0 vector. And this is a very unconventional notation. I'm just writing it like that, because I don't feel like bolding out my 0's all the time to make you realize that that's a vector. So we meet our first requirement. The 0 vector is a member of the set. So let me define my set here. Let me define it n. And I'll tell you in a second why I'm calling it n. So we now know that the 0 vector is a member of my set n. Now let's say I have two vectors, v1 and v2 that are members-- let me write this. So let's say I have two factors, v1, and v2, that are both members of our set. What does that mean? That means that they both satisfy this equation. So that means that a-- my matrix a-- times vector 1 is equal to 0. This is by definition. I'm saying that they're a member of the set, which means they must satisfy this. And that also means that a times vector 2 is equal to our 0 vector. So in order for this to be closed under addition, a times vector 1 plus vector 2, the sum of these two vectors should also be a member of n. But let's figure out what this is. The sum of these two vectors is this vector right here. This is equal to-- and I haven't proven this to you yet. I haven't made a video where I prove this. But it's very easy to prove just using the definition of matrix vector multiplication, that matrix vector multiplication does display the distributive property. And maybe I'll make a video on that, but literally, you just have to go through the mechanics of each of the terms. This is equal to a[v,1] plus a[v,2] And we know that this is equal to the 0 vector. And this is equal to the 0 vector. And if you add the 0 vector to itself, this whole thing is going to be equal to the 0 vector. So if v1 is a member of n, and v2 is a member of n, which means they both satisfy this equation, then v1 plus v2 is definitely still a member of n. Because when I multiply a times that, I get the 0 vector again. So let me write that result, as well. So we now know that v1 plus v2 is also a member of n. And the last thing we have to show is that it's closed under multiplication. Let's say that v1 is a member of our space that I defined here, where they satisfy this equation. What about c times v1? Is that a member of n? Well let's think about it. What's our matrix a times the vector-- right? I'm just multiplying this times the scale. I'm just going to get another vector. I don't want to write a capital v there. Lowercase v, so it's a vector. What's this equal to? Well, once again, I haven't prove it to you yet, but it's actually a very straightforward thing to do, to show that when you're dealing with scalars, if you have a scalar here, it doesn't matter if you multiply the scalar times the vector before multiplying it times the matrix or multiplying the matrix times the vector, and then doing the scalar. So it's fairly straightforward to prove that this is equal to c times our matrix a-- I'll make that nice and bold, times our vector v. That these two things are equivalent. Maybe I should just churn out the video that does this, but I'll leave it to you. You, literally, just go through the mechanics component by component. And you show this. But clearly, if there's is true, we already know that v1 is a member of our set, which means that a times v1 is equal to the 0 vector. And so that means this will reduce to c times the 0 vector, which is still the 0 vector. So c[v,1] is definitely a member of n. So it's closed under multiplication. And I kind of assumed this right here. But maybe I'll prove that in a different video. But I want to do all this to show that this set n is a valid subspace. This is a valid subspace. It contains a 0 vector. It's close under addition. It's close under multiplication. And we actually have a special name for this. We call this right here, we call n, the null space of a. Or we could write n is equal to-- maybe I shouldn't have written an n. Let me write orange in there. Our orange n is equal to-- the notation is just the null space of a. Or we could write the null space is equal to the orange notation of n, and literally, if I just give you some arbitrary matrix a, and I say, hey, find me n of a, what is that? Literally, your goal is to find the set of all x's that satisfy the equation a times x is equal to 0. And I'm going to do that in the next video.