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# Introduction to the null space of a matrix

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.