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## Null space and column space

Current time:0:00Total duration:25:13

# Null space and column spaceÂ basis

## Video transcript

What I want to do in this video
-- and it'll probably occur over several videos -- is
really integrate everything we know about matrices, and
null spaces, and column spaces, and linear
independence. So I have this matrix
here, this matrix A. And I guess a good place to
start is, let's figure out its column space and
its null space. The column space is actually
super easy to figure out. It's just the span of the
column vectors of A. So we can right from the get-go
write that the column space of our matrix A-- Let
me do it over here. I can write the column space of
my matrix A is equal to the span of the vectors 1, 2, 3. 1, 1, 4. 1, 4, 1. And 1, 3, 2. I'm done. That was pretty straightforward,
a lot easier than finding null spaces. Now this may or may not
be satisfying to you. And there's a lot of
open questions. Is this a basis for the
space, for example? Is this a linear independent
set of vectors? How can we visualize
this space? And I haven't answered
any of those yet. But if someone just says, hey
what's the column space of A? This is the column space of A. And then we can answer some
of those other questions. If this is a linearly
independent set of vectors, then these vectors would
be a basis for the column space of A. We don't know that yet. We don't know whether these
are linearly independent. But we can figure out if they're
linearly independent by looking at the
null space of A. Remember these are linearly
independent if the null space of A only contains
the 0 vector. So let's figure out what
the null space of A is. And remember, we can do a
little shortcut here. The null space of A is equal to
the null space of the row, the reduced row echelon
form of A. And I showed you that when we
first calculated the null space of a vector, because when
you performed these -- essentially if you want to solve
for the null space of A, you create an augmented
matrix. And you put the augmented matrix
in reduced row echelon form, but the 0's
never change. So essentially you're just
taking A and putting it in reduced row echelon form. Let's do that. So I'll keep row one the
same, 1, 1, 1, 1. And then let me replace
row two with, row two minus row one. So what do I get? No, actually I want to
zero this out here. So row two minus,
2 times row one. Actually even better because
I eventually want to get a 1 here. So let me do 2 times row
one, minus row two. So let me say 2 times row
one, and I'm going to minus row two. So 2 times 1 minus 2 is
0, which is exactly what I wanted there. 2 times 1 minus 1 is 1. That's nice to have
right there. 2 times 1 minus 4 is minus 2. 2 times 1 minus 3 is minus 1. All right, now let me see if I
can zero out this guy here. So what can I do? I could do any combination,
anything that essentially zeroes this guy out. But I want to minimize my number
of negative numbers. So let me take this third row,
minus 3 times this first row. So I'm going take minus 3 times
that first row and add it to this third row. So 3 minus 3 times 1 is 0. These are just going to
be a bunch of 3's. 4 minus 3 times 1 is 1. 1 minus 3 times 1 is minus 2. And 2 minus 3 times
1 is minus 1. Now if we want to get this into
reduced row echelon form we need to target that one
there and that one there. And what can we do? So let's keep my middle
row the same. My middle row is not
going to change. 1, 1, minus 2, minus 1. And to get rid of this one up
here I can just replace my first row with my first row
minus my second row. Because then this
won't change. I'll have 1 minus 0 is 1. 1 minus 1 is 0. That's what we wanted. 1 minus minus 2 is 3. That's 1 plus 2. 1 minus minus 1. That's 1 plus 1. That is 2. Fair enough? Now let me do my third row. Let me replace my third row with
my third row subtracted from my first row. They are obviously
the same thing. So if I subtract the third row
from the second row I'm just going to get a bunch of 0's. 0 minus 0 is 0. 1 minus 1 is 0. Minus 2 minus minus 2 is 0. And minus 1 minus minus 1. That's minus 1 plus 1. That's equal to 0. And just like that
we have it now in reduced row echelon form. So this right here is the
reduced row echelon form of A. That straightforward. Now the whole the reason why
we even went through this exercise is we wanted
to figure out the null space of A. And we already know that the
null space of A is equal to the null space of the reduced
row echelon form of A. So if this is the reduce row
echelon form of A, let's figure out its null space. So the null space is the set
of all of vectors in R4, because we have 4
columns here. 1, 2, 3, 4. The null space is the set of
all of vectors that satisfy this equation, where
we're going to have three 0's right here. That's the 0 vector in R3,
because we have three rows right there, and you
can figure it out. This times this has
to equal that 0. That dotted with that
essentially is going to equal that 0. That dotted with that
is equal to that 0. I say essentially because I
didn't define a row vector dot a column vector. I've only defined column vectors
dotted with other column vectors. But we've been over that in a
previous video, where you can say this is a transpose
of a column vector. So let's just take this,
and write a system of equations with this. So we get 1 times x1. So this times this is going
to be equal to that 0. So one times x1, that is x1. Plus 0 times x2. Let me just write that out. Plus 3 times x3. Plus 2 times x4 is
equal to that 0. And then -- I'll do it in
yellow right here -- I have 0 times x1. Plus 1 times x2. Minus 2 times x3. Minus x4 is equal to 0. And then this gives
me no information. 0 times all this
is equal to 0. So it just turns into
0 equals 0. So let's see if we can solve for
our pivot entries, or our pivot variables. What are our pivot entries? This is a pivot entry. That's a pivot entry. That's what reduced row echelon
form is all about, getting these entries that are
1 and they're the only non-zero term in their
respective columns. And that every pivot entry
is to the right of a pivot entry above it. And then the columns that don't
have pivot entries? These columns represent
the free variables. So this column has
no pivot entry. And so when you take the dot
product, this column turned into this column in our
system of equations. So we know that x3 is
a free variable. x3 is free. We can set it equal
to anything. Likewise x4 is a
free variable. X1 and x2 are pivot variables,
because their corresponding columns in our reduced row
echelon form have pivot entries in them. Fair enough. So let's see if we
can simplify this into a form we know. And we've seen this before. So if I solve for x1 --
this 0 I can ignore. That 0 I can ignore -- I could
say that x1 is equal to minus 3x3 minus 2x4. I just subtracted these two
from both sides of the equation and I can say that x2
is equal to 2x3 plus x4. And if we want to write our
solution set now, so if I wanted to find the null space of
A, which is the same thing as the null space of the reduced
row echelon form of A, is equal to all of the vectors
-- let me do a new color. Maybe I'll do blue -- is equal
to all of the vectors x1, x2, x3, x4 that are equal to -- So what are they going
to be equal to? X1 has to be equal to
minus 3x3 minus 2x4. Just to be clear, these are free
variables because I can set these to be anything. And these are pivot variables
because I can't just set them to anything. When I determine what my x3's
and my x4's are, they determine what my x1's and
my x2's have to be. So these are pivoted
variables. These are free variables. I can make this guy pi. And I can make this
guy minus 2. We can set them to anything. So x1 is equal to -- let's see,
let me write it this way -- they're equal to x3 -- let me
do it in a different color -- do x3 like this. So it's equal to x3 times some
vector plus x4 times some other vector. So any solution set in my null
space is going to be a linear combination of these
two vectors. We can figure out what these
two vectors are just from these two constraints
right here. So -- let me do it in a neutral
color -- x1 is equal to minus 3 times x3
minus 2 times x4. Straightforward enough. x2 is equal to 2 times
x3 plus x4. What's x3 equal to? Well x3 is equal to itself. Whatever we set x3 equal to,
that's going to be x3. So x3 is going to be 1 times
x3 plus 0 times x4. It is not going to have
any x4 in it. X3 is going to be kind of
an independent variable. It's going to be free. We can set whatever it is. We set it and then that's
going to be x3 in our solution set. x4 is not going to have
any x3 in it. It's just going to
be 1 times x4. And so our null space is
essentially all of the linear combinations of these
two vectors. This can be any real number. This is just any real number and
x4 is just a member of the real space. So all of these, the set of all
of the valid solutions to Ax is equal to 0 -- where
did I write that. Did I even write that down? No I haven't even written
that anywhere. The set of all Ax is equal to
0, where this is my x, it equals all the linear
combinations of this vector and that vector right there. And we know what all of the
linear combinations mean. It means my null space is equal
to the span of these two guys, the span of minus
3, 2, 1, 0. And minus 2, 1, 0, 1. Now let me ask you a question. Are the columns in A, are they
a linearly independent set? Are they a linearly
independent set? So if we write these vectors
right there, these are the column vectors of A. So let me write that down. So are the column vectors of
A -- so what were they? Let's see. 1, 3, 2. No it's 1, 2, 3. 1, 1, 4. 1, 4, 1. And 1, 3, 2. So this is just the column
vectors of A. I could just write A is just
this much of columns, but my question is, is this a linearly
independent set? And here you might immediately
start thinking, well when we said that something is linearly
independent -- so linearly independence implies
that there's only one solution -- we saw this I think two
videos ago, that there's only one solution -- one solution
to Ax is equal to 0. And that is the 0 solution,
that x is equal to the 0 vector. Or another way to say that is
that the null space of my matrix A is equal to
just the 0 vector. That's what linear independence
implies. And it goes both ways. If my null space is just a 0
vector, then I know it's linearly independent. If my null space includes other
vectors, then I am not linearly independent. Now my null space of A,
what does it include? Is it just the 0 vector? Well, no it includes
every linear combination of these guys. It includes actually an infinite
number of vectors, not just one solution. Obviously 0 vector is contained
here, if you just multiply both of these -- if you
pick 0 for that and that. It's contained, but you can get
a whole set of vectors. So because the null span of A,
the null space, sorry, the null space of A does not just
contain the 0 vector. So it has more than just 0. So what does that mean? Well that means that
there's more than one solution to this. And that means that this is
a linearly dependent set. And what does that mean? At the very beginning of the
video I said, what's the column space of A. And we said, the column space
of A is just the span of the column vectors. I just wrote it out like that. And I said, well it's not clear
whether this is a valid basis for the column
space of A. And what's a basis? A basis is a set of vectors that
span a subspace, and they are also linearly independent. And we just showed that these
guys are not linearly independent. So that means that they are
not a basis for the column space of A. They do span the column space
of A, by definition really. But they're not a basis. They need to be linearly
independent for them to be a basis. So let's see if we can figure
out what a basis for this column space would be. And to do that we just
have to get rid of some redundant vectors. If I can show you that this guy
can be represented by some combination of these two
guys, then I can get rid of that guy. He's not adding any
new information. Same with that guy. Who knows? So let's see if we can figure
this piece of the puzzle out. So we know already that x1, let
me write it this way, that x1 times -- Maybe I'll just kind
of leave you hanging and continue this in
the next video. But we know that x1
times 1, 2, 3. Plus x2 times 1, 1, 4. Plus x3 times 1, 4, 1. Plus x4 times 1, 3, 2. We know that this
is equal to 0. Now if we are able to solve for
x4 in terms of -- let me just think that I can solve
for the vectors that are associated with my
free variables using the other vectors. Let me see if I can do that. And you'll see it's actually
pretty straightforward. So let's say I want
to solve for x4. So if I subtract this from
both sides of this equation, I get what? Let me put it this way, let
me set x3 equal to 0. It was a free variable. I can do that. So if I set x3 is equal to 0,
then what do I get here? If I said x3 equals 0,
this guy disappears. And if I subtract this from both
sides of this equation, I get x1 times 1, 2, 3. Plus x2 times 1, 1, 4. Is equal to -- I'm just
setting x3 equal to 0. That was a free variable. So I'm setting x3 equal to 0. So this whole thing
disappears. So that is equal to minus
x4 times 1, 3, 2. Now I set x3 equal to 0. Let me set x4 to be
equal to minus 1. If x4 is equal to minus
1, what is minus x4? Well then this thing will
just be equal to 1. And I'll have x1
times 1, 2, 3. Plus x2 times 1, 1, 4 will
equal this fourth vector right here. And can I always find
things like this? Well sure I can actually find
the particular ones. If x3 is equal to 0, and x4 is
minus 1 -- Let me copy and paste this that I have up
here -- Let me scroll down a little bit. This is what we got when we
figured out our null space, right there. So if I'm setting -- remember
these are the free variables -- if I set x3 equal to
0 and x4 is equal to minus 1, what is x1? Then this will imply that x1 is
equal to minus 3 times x3, that's just 0, minus
2 times x4. If x4 is minus 1, minus
2 times minus 1, x1 will equal 2. And then what will
x2 be equal to? x2 is equal to 2 times x3,
which is 0, plus x4. So it's equal to minus 1. So I just showed you that if I
set this equal to 2 and this equal to minus 1, I have a
linear combination of this vector and this vector
that can add up to this fourth vector. And you can even verify it. 2 times 1 minus 1
is equal to 1. 2 times 2 minus 1
is equal to 3. 2 times 3 is 6, minus
4 is equal to 2. So it checks out. So I just showed you using,
really, our definitions looking at what were
our free variables versus our pivot variables. We were able to show you, kind
of just very simply solve for this third, this fourth
vector, in terms of these first two. So we know, if we go back to
the set that this fourth vector is really unnecessary,
really not adding anything to the span of the set
of vectors. Because this guy can be written
as a combination of this guy and this guy. Now let's see if this guy, this
third guy, we can do the same exercise. This is also dictated
by a free variable. So let's see if I can write
him as a combination of these first two. Well we'll do the exact
same thing. Instead of setting x3 equal to 0
and x4 equal to minus 1, let us set x4 is equal to
0 because I want to cross that out. And let me set x3 is
equal to minus 1. If x3 is equal to minus
1, what does this equation reduce to? We get x1 times 1, 2, 3. Plus x2 times 1, 1, 4. Is equal to -- if this is
minus 1 times 1, 4, 1. And then we add it to both sides
of this equation, we get plus 1 times 1, 4, 1. And once again we can just
solve for our x1 and x2. If x4 is 0 and x3 is minus
1, then x1 x4 is 0. So x3 is just minus 3,
times x3, so x1 would be equal to 3, right? Minus 3 times minus 1. And what would x2 be equal to? x4 is 0, we can ignore that. x2 would be equal to minus 2. So this would be 3, and then
this would be minus 2. Let's see if it works out. 3 times 1 minus 2 is 1. 3 times 2 minus 2 is 4. 3 times 3 minus 8 is 1. It checks out. So I'm able to write this
vector, that was associated with the free variable,
as a linear combination of these two. So we can get rid of
him from our set. So now I've shown that this guy
can be written as a linear combination of these two. This guy can be written
as a linear combination of these two. So the span of all of those guys
should be equal to the span -- So let me write
it this way. The column space of A,
I can now re-write. Before it was the span of
all of those vectors. It was the span of all of
the column vectors, v1, v2, v3, and v4. Now I just showed you that v3
and v4 can be rewritten in terms of v1 and v2. So they're redundant. So that is equal to the span
of v1 and v2 which are just those two vectors. Vector 1, 2, 3, and
vector 1, 1, 4. Now are any of these
guys redundant? Can I express one of
them as a linear combination of the other? Essentially when I'm talking
about the linear combination of only one other
vector it's just multiplying it by a scalar. Well let's think about that. There are multiple ways you can
show this, but the easiest way is well look, to go from
this entry to that entry I'm just multiplying by 1. But if I multiply this whole
vector times 1, then I'm going to get a 2 here and I'm
going to get a 3 here. So it won't work. If I want to represent this guy
as a scalar multiple of that guy, so any scalar multiple
of 1, 2, 3 is going to be equal to 1c, 2c, 3c. Right? And so we're saying this guy has
to be represented somehow like that, if we say that this
guy is somehow a scalar, somehow can be represented
by that guy. So that would have to
be equal to 1, 1, 4. When you look at this top entry
it implies that c would have to be equal to 1. But when you look at this second
entry you think that c would have to be equal to 1/2. So you get a contradiction. Over here c would have
to be equal to 4/3. So there's no c where
this will work. There's no multiple of c. And you can work
that both ways. So there's no way that you can
represent one of these guys as a linear combination
of the other. And you can actually prove
other ways, maybe more formally, that this is
linearly independent. But given that this is linearly
independent -- I think you're satisfied with that
-- we can then say that the set of vectors 1, 2, 3, and
1, 1, 4, this is a basis for the column span of A. Now I'm going to let you go in
this video because I think I've gone well over time. But what I'm going to do in
the next few videos is now that I've established that this
is a basis for the column span of A, we can attempt
to visualize it. Because we can say that the
column span of A is equal to the span of these two vectors. And we can think about
what the span of those two vectors are. We're going to see that
it's a plane in R3. Span of 1, 1, 4. And this is a quick reminder,
I've said a couple times. When I say it's a basis all I'm
saying is that these guys, they both span the column
space of A. When I had four vectors,
they also spanned the column space of A. But what makes them a basis is
that these guys are linearly independent. There's no extra information, or
redundant vectors that can be represented by other vectors
within the basis. They are linearly independent. Anyway, I'll let
you go for now.