Main content

## Linear algebra

### Course: Linear algebra > Unit 2

Lesson 7: Transpose of a matrix- Transpose of a matrix
- Determinant of transpose
- Transpose of a matrix product
- Transposes of sums and inverses
- Transpose of a vector
- Rowspace and left nullspace
- Visualizations of left nullspace and rowspace
- rank(a) = rank(transpose of a)
- Showing that A-transpose x A is invertible

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Rowspace and left nullspace

Rowspace and Left Nullspace. Created by Sal Khan.

## Want to join the conversation?

- Its unclear to me why the dimC(A) = dimR(A)? Its clear that this is the case--but i've researched this a bit and haven't found the why this is the case. thanks(7 votes)
- When you row reduce A, all the rows left in the rref are linearly independent, otherwise one of the remaining rows would be a multiple of another row, and could be eliminated. But the number of rows in the rref is the same as the number of pivot columns, which is the same as the number of linearly independent column vectors.

The row vectors and column vectors of A are respectively the column vectors and row vectors of A transpose, so, as above, the number of linearly independent rows in A transpose is the same as the number of its linearly independent columns. Therefore Rank(A transpose) = Rank(A).(8 votes)

- I am having some trouble figuring out a couple proof type problems in this area of intro linear algebra. For example, right now I have to prove that for any matrix A if vector u is in RowA and vector v is in NullA, then product of u(transpose) and v is 0. Or that if u belongs to both RowA and NullA, then u equals the zero vector.(4 votes)
- This happens because RowA and NullA are orthogonal subspaces.(1 vote)

- Is "nullspace" the same thing as the span of the "kernel"?(2 votes)
- Yes, and a better way to say it is that the kernel is the nullspace. The span of the kernel and the span of the nullspace are just themselves since they are already subspaces.(2 votes)

- Between16:20and17:45Sal explains how A^tx=0 is equivalent to x^TA=0^t. This sounds very confuse because the left part of the equivalence creates a 2x1 matrix and the second part creates a 1x3 matrix. The left part i just need to equal to zero, but what should I do with the right part?

Also, just to be sure, a transpose of zero matrix is zero right?(1 vote)`A`

is a`n⨉m`

matrix, so`Aᵀx⃗`

is a`m⨉1`

column vector. (since`Aᵀ`

is a`m⨉n`

matrix and`x⃗`

is a`m⨉1`

column vector.) Therefore in the equation`Aᵀx⃗ = 0⃗`

, the`0⃗`

is the zero`m⨉1`

column vector.

Now, the equation`x⃗ᵀA`

is a`1⨉m`

row vector (since`x⃗ᵀ`

is a`1xm`

row vector), and therefore`0⃗ᵀ`

is also the zero`1⨉m`

row vector.(4 votes)

- At about6:50, Sal says that x1 is the pivot entry. But couldn't we have solved for any of the other x's also, and then expressed them in the same way?(2 votes)
- Yes, he could. Depending on which row you put on top, and which column comes first and second, and how you do the elimination, you could make any of the 3 variables the pivot variable. Also notice that in this problem the span of any of the column vectors is the same as span of any other.(1 vote)

- For a system of equations Ax=b, how do you prove that the solution for x lies in the row space of A? I tried pondering over this, in vain.(2 votes)
- x doesn't necessarily lie in the row space. Consider:
`[2 1][x1]=[3]`

[x2]

x1 = 1.5

x2 = 0

The given values for x solve the single equation, but x is not a linear multiple of the [2 1] row.(3 votes)

- If Rank(A)=Rank(A Transpose), then that should mean dim( (C(A) )= dim ( C(A Transpose) ). If we have, say, 2 pivots then that means we have two Lin. Ind. rows. In that case dim ( C(A transpose) )= 2. The two independent rows form a basis for the column space of A transpose.

But if we have 2 pivots, then we also have two Lin. Ind. columns and dim ( C(A) )= 2.The two independent columns form a basis for the column space of A. So dim ( C(A) ) =2= dim ( C(A Transpose) ). Is this reasoning correct?(2 votes)- It's absolutely correct.......The number of columns with pivots = the number of rows with pivots........v.i.z. .....Number of independent columns = Number of independent rows

So, C(A) = C(A Transpose) = Rank(A).(3 votes)

- i have a question about taking the transpose of both sides of a equation as sal did. around 15-16 minutes in the video! I can't really see how it's equivivalent because i tried a general case by myself a 2x3 matrix (a,b,c, d,e,f,) then transpose it and get the 3x2((a,b,c)(d,e,f,))(X1,X2X3)=(000) columnvector3x1 which but when i reversedproduct it because i now transposed both sides of the equation i got (X1X2X3)(abc defg)= (000)1x3

and i'm wondering how can we say that a 2x1=1X3 matrix and what does transposing both sides really mean(2 votes)- It has to be equal because ((A)T)x is just a vector, and you can always transpose a vector.

It might help if you review the previous video, "https://www.khanacademy.org/math/linear-algebra/matrix-transformations/matrix-transpose/v/linear-algebra-transpose-of-a-matrix-product". There we learned that:

The transpose of AB "(AB)T" = (B)T(A)T.

and so, since (A)T and x (a vector) are both matrices,

((A)T)x = (x)T*((A)T)T = (x)T*A.(0 votes)

- I used x2 as the pivot variable and got the basis of c(a) right but not the null space. Can there not be different free variables in one matrix?(1 vote)
- Maybe your null space has the same dimensions (R^2), and is even the same 2 dimensional subspace of R^3, but has different basis vectors. You could check this by taking the cross product of the basis vectors of both versions to see if they're parallel.(1 vote)

- At1:00does he mean the 0 vector in R2?(1 vote)

## Video transcript

I've got this matrix, A, here,
it's a 2 by 3 matrix. And just as a bit of review,
let's figure out its nullspace and its columnspace. So the nullspace of A is the set
of all vectors x that are member of-- let's see we have
3 columns here-- so a member of R3, such that A times the
vector are going to be equal to the 0 vector. So we can just set this up. Let me just-- we just need to
figure out all of the x's that satisfy this in R3. So we take our matrix
A 2, minus 1, minus 3, minus 4, 2, 6. Multiply them times some
arbitrary vector in R3 here. So you get x1, x2, x3. And you set them equal
to the 0 vector. It's going to be the
0 vector in R2. Because we have 2 rows here. You multiply a 2 by 3 matrix
times a vector in R3, you're going to get a 2 by 1 vector
or 2 by 1 matrix. So you're going to get
the 0 vector in R3. And to solve what is essentially
a system of equations-- you get 2 x1 minus
x2 minus 3 x3 is equal to 0 and so on and so forth. We can just set up an
augmented matrix. So we can just set up this
augmented matrix right here. 2 minus 1 minus 3
minus 4, 2, 6. And then augment it with what
we're trying to set it equal to to solve the system. And you know we're going to
perform a bunch of row operations here to put this in
reduced row echelon form. And they're not going to change
the right-hand side of this augmented matrix. And that's essentially the
argument as to why the nullspace of the reduced row
echelon form of A is the same thing as the nullspace of A. But anyway, that's just
a bit of review. So let's perform some row
operations to solve this a little bit better. So, the first thing I might
want to do is divide the first row by 2. So if I divide the first row by
2 I get a 1 minus 1/2 minus 3/2 and then of course
0 divided by 2 is 0. And let's just divide this row
right here by-- I don't know just to simplify things--
let's divide it by 4. So I'm doing two row operations
in one step. And you can do that. I could have done it in
two separate steps. So if we divide it by 4, this
becomes minus 1, 1/2 and then you get 3/2 and then
you get 0. And now, let's keep my
first row the same. I'm going to keep my
first row the same. It's 1 minus 1/2 minus 3/2 and
of course the 0 is the right-hand side. Now let's replace my second
row with my second row plus my first row. So these are just linear
operations on these guys. So negative 1 plus 1 is 0. 1/2 plus minus 1/2 is 0. 3/2 plus minus 3/2 is 0. And of course, 0 plus 0 is 0. So what are we left with? we're left with this
right here. This is another way of saying
that x1-- let me write it this way-- x1-- I guess the easiest
way to think about it is-- you're multiplying the reduced
row echelon form of A now. 1 minus 1/2 minus 3/2. You have a bunch of 0's here. Times x1, x2, x3 is equal
to the R2 0 vector. This is another interpretation
of this augmented matrix. So this is just saying,
this is useless. This is saying 0 times that plus
0 times that plus 0 times that is equal to 0. So it's giving us
no information. But this first row tells us
that-- let me switch colors-- 1 times x1 minus 1/2 times
x2 minus 3/2 times x3 is equal to 0. All of the vectors whose
components satisfy this are in my nullspace. If I want to write it a little
bit differently I could write as, x1 is equal to 1/2
x2 plus 3/2 x3. Or if I wanted to write my
solution set in vector form, I could write that my nullspace is
going to be the set of all the vectors x1, x2, x3 that
satisfy these conditions. That are equal to what? Well, x2 and x3 are
free variables. They're associated with the
non-pivot entries, or the non-pivot columns in our reduced
row echelon form. That is a pivot column
right there. So let me write it this way. It's going to be x2
times something plus x3 times something. Those are my two
free variables. And we have here,
x1 is 1/2 x2. 1/2 times x2 plus
3/2 times x3. x2 is just going to be x2
times 1 plus 0 times x3. x3 is going to be 0 times
x2 plus 1 times x3. So, our nullspace, these
can be any real numbers right here. They're free variables. So our nullspace is essentially
all of the linear combinations of this
guy and that guy. Or another way to write it, the
nullspace of A is equal to the span, which is the same
thing as all the linear combinations of the
span of 1/2, 1, 0. Notice these are
vectors in R3. And that makes sense because the
nullspace is going to be a set of vectors in R3. So it's the span of that. And that right there. So 3/2, 0, and 1. Just like that. And what is the columnspace
of our original matrix, A? So the columnspace of A is equal
to just, the subspace created by all of the linear
combinations of these guys. Or essentially the span
of the column vectors. Is equal to the span
of 2 minus 4 minus 1, 2, minus 3, 6. These are all each
separate vectors. So it's the span of
these 3 vectors. Now, these guys might not
be linearly independent. And actually, when you put
this guy in reduced row echelon form, you know that the
basis vectors for this are the vectors that are associated with our pivot columns. So we have one pivot
column here. It's our first column. So we could say that we could
use this as a basis vector. And it makes sense. Because this guy right here
is minus 2 times this guy. This guy right here is minus
3/2 times that guy. So these two guys can definitely
be represented as linear combinations
of that guy. So it's equal to the span of
just the vector, 2, minus 4. So if you were to ask me, and
this is the basis for our columnspace. So if you wanted to know the
rank-- and this is all a bit of review-- the rank of A is
equal to the number of vectors in our basis for our
columnspace. So it's going to
be equal to 1. Now, everything I just did
is a bit of review. But with the last couple of
videos, we've been dealing with transposes. So let's actually figure out
the same ideas for the transpose of A. So A transpose looks
like this. A transpose is equal to the
matrix 2, minus 1, minus 3 is the first column right there. And then the second column is
going to be minus 4, 2 and 6. That is our transpose. So let's figure out the
nullspace and the columnspace of our transpose. Let me put this in reduced
row echelon form so we can get the nullspace. Let me get the nullspace
of this guy. So we could do the exact
same exercise. Let me write it this way. The nullspace of
A transpose-- A transpose is a 3 by 2 matrix. So it's equal to all of
the vectors, x, that are members of R2. Not R3 anymore-- because now we
are taking the transpose's nullspace-- such that A
transpose times R vectors are equal to the 0 vector in R3. And we can do that the same
exact way we did before. We set up an augmented matrix. We could just put it in reduced
row echelon form and set them all equal to 0. So let's just do that. So if we-- let me just put it
in reduced row echelon form. So let me divide my
first row by 2. Let's divide the
first row by 2. I'm just going to put it in
reduced row echelon form. The first row divided
by 2 is 1 minus 2. And then the second row, let
me divide it-- let me just, I'll just keep it the same--
so minus 1, 2. And then this last row,
let me divide it by 3. So it becomes minus 1 and 2. And now, let me keep my
first row the same. 1 minus 2. And now let me replace my second
row with my second row plus my first row. So minus 1 plus 1 is 0. 2 plus minus 2 is 0. You get some 0's. I'm going to do the same thing
with the third row. Replace it with it plus
the first row. Once again you're going
to get some 0's. So this is the reduced row
echelon form of A transpose. And its nullspace is the same
as A transpose's nullspace. We could say, to find this
nullspace we can find all of the solutions to this equation
times the vectors x1 and x2 is equal to 0, 0, and 0. These aren't vectors. These are just entries
right here. 0, 0, 0. So these two lines give
us no information, but this first one does. So we get 1 times x1-- and
notice, this is the pivot column right here. It's associated. So x1 is going to be
a pivot variable. x2 will be a free variable. And just to be clear that
the first column is our pivot column. So if we go back to A transpose,
it's this first column here that is associated
with the pivot column. So when we talk about its
columnspace, this by itself will span the columnspace. This is all a review of
what we did before. We're just applying it
to the transpose. Let's go back to
our nullspace. So this tells us that 1 times
x1, so x1, minus 2 times x2, is equal to 0. Or we could say that x1 is
equal to 2 times x2. So all of the vectors in R2 that
satisfy these conditions with these entries will be in
the nullspace of A transpose. Let me write it this way. So the nullspace of A is going
to be the set of all the vectors-- let me write it here--
the set of all the vectors, x1, x2, that are member
of R2, clearly, such that x1, x2 is going to be
equal to-- well, our free variable is x2-- so it's
x2 times the vector. So x1 has to be 2 times x2. And obviously x2-- that's a 2--
is going to be 1 times x2. So what is this going to be? Well this is all of the linear
combinations of this vector right here. So we could say it's equal to
the span of our vector 2, 1. Now, that's the nullspace. Sorry, this was the nullspace
of A transpose. I have to be very
careful there. Now what is the columnspace? The columnspace of A tranpose? Well, the columnspace of A
transpose is the set of all vectors spanned by
the columns of A. So you could just say the span
of this column vector and this column vector. But we know, when we put it into
reduced row echelon form, only this column vector
was associated with a pivot column. So this by itself, this
guy is a linear combination of this guy. If you multiply him by
minus 2, you get that guy right there. So it's consistent with
everything we've learned. So it equals the span of just
this guy right here. Of just the vector 2, minus
1, and minus 3. That's just a nice, neat
exercise that we did. Notice that your span here,
it's in R3, but it's just going to be a line in R3. Maybe in the next video I'll
do a more graphical representation of it. But I did this whole exercise to
introduce you to the ideas of the nullspace of your
transpose and the columnspace of your transpose. Think about what the
columnspace of your transpose is. It's the subspace spanned by
that vector-- sorry-- spanned by this vector and
that vector. And it turns out that this guy
is a multiple of that guy. So we could say just
by that guy. But these were the rows of
our original matrix, A. So we could also view this as
the span of the row vectors of our original guy. This is that column that is the
basis for the column span of the R transpose matrix. And of course this guy was a
linear combination of that. So we could also view
the column span of our transpose matrix. It's equivalent to the subspace spanned by these rows. Or we could call that
the row space of A. Let me write that down. So the columnspace of
A transpose-- and this is just general. Let me write this generally. It doesn't just apply
to this example. So the columnspace of the
transpose of any matrix, this is called the rowspace of A. And it's a very natural name. Because if A's got a bunch of
rows, we could call them the transpose of some vectors. So that's first row. You got the second row. All the way to, maybe,
the nth row. Just like that. These are vector transposes. They're really just rows. If you imagine the space
that's spanned by these vectors, by the different rows,
that's essentially the columnspace of the transpose. Because when you transpose
it, each of these guys become columns. So that's what the
rowspace is. Now, the nullspace of our
transpose-- let's write it like this-- it was all
of the vectors x that satisfied this equation. Equals the 0 vector
right there. Now, what happens if we take the
transpose of both sides of this equation? Well, we've learned from our
transpose properties, this is equal to the reverse product of
each of those transposes. So this is going to be equal
to, this is a vector, the vector x transpose. If this is a column vector
before, now it's going to become a row vector. And then, times A transpose
transpose. And that's going to
be equal to the transpose of the 0 vector. Or, we could just write
this like this. We could write this as some
matrix-- well let me just write it like this. Some column vector x-- what's
the transpose of A transpose? Well that's just equal to A. So you take the transpose
of this column vector. You now get a row vector. You could view it as a
matrix if you want. If this was a member of Rn, this
is now going to be an 1 by n matrix. If this was a member of Rn. We kind of switched
the orders. And we multiply it times the
transpose of the transpose. We just get the matrix, A. And we set that equal to the
transpose of the 0 vector. Now this is interesting. We now have it in terms of
our original matrix, A. Now what did the nullspace of
our matrix, A look like? The nullspace where all of the
vectors x that satisfied this equation is equal to 0. So the x was on the right. So the nullspace is all the
x's that satisfy this. The nullspace of our transpose
is all of the x's that satisfy this equation. So let me say the set of all
of the x's such that A transpose times x
is equal to 0. That is the nullspace
of A transpose. Or we could also write this as
the set of all of the x's such that the transpose of our x
times A is equal to the transpose of the 0 vector. And we have another
name for this. This is called the left
nullspace of A. Why is it called the
left nullspace? Because now we have
x on our left. In just a regular nullspace
you have x on the right. But now, if you take the
nullspace of the transpose, using just our transpose
properties, that's equivalent to this transpose vector
right here. Actually let me write this. The transpose right there. This transpose vector
multiplying A from the left-hand side. So all of the x's that satisfy
this is the left nullspace. And it's going to be different
than your nullspace. Notice, your nullspace of A
transpose was the span of this right here. This is also the left
nullspace of A. Now what was just the regular
nullspace of A? The regular nullspace of A was
essentially a plane in R3. That's the nullspace of A. The left nullspace of A
is just a line in R2. Very different things. And if you go to the rowspace,
what is the rowspace of A? The rowspace of A
is a line in R3. The rowspace of A
is a line in R3. Well what is the columnspace
of A? The columnspace of A, right
here, where did I have it? Well, this is the only linearly
independent vector. It was essentially
a line in R2. So they're all very
different things. And we'll study a little bit
more how they're all related. Now there's one thing I
want to relate to you. We figured out that the
rank of this vector right here is 1. Because when you put it in
reduced row echelon form there was one pivot column. And the basis vectors are those
associated with that pivot column. And if you count your basis
vectors, that's your dimension of your space. So the dimension of your
columnspace is 1. And that's the same thing
as your rank. Now what is the rank
of A transpose? The rank of A transpose in the
example, when you put it in reduced row echelon form, you
got one linearly independent column vector. So the basis for our
columnspace was also equal to 1. And in general, that's always
going to be the case. That the rank of A, which
is the dimension of its columnspace, is equal to the
rank of A transpose. And if you think about it,
it makes a lot of sense. To figure out the rank of A, you
essentially figure out how many pivot columns they have.
Or another way to say it is how many pivot entries they
have. When you want to find the rank of your transpose
vector, you're essentially just saying-- and I know this is
maybe getting a little bit confusing-- but when you want
the rank of your transpose vector, you're saying, how
many of these columns are linearly independent? Or which of these are linearly
independent? And that's the same question
as saying, how many of your rows up here are linearly
independent? If you want to know how many
columns in your transpose are linearly independent, that's
equivalent to asking how many rows in your original matrix
are linearly independent. And when you put this matrix in
reduced row echelon form, everything in reduced row
echelon form are just row operations. So they are just linear
combinations of these things up here. Or you could go vice versa. Everything up here is just
linear combinations of your matrix in reduced row
echelon form. So if you only have one pivot
entry, then this guy right here, by himself, or one pivot
row, that guy by himself can represent a basis for
your rowspace. Or, all of your rows can be
represented by a linear combination of your
pivot rows. And because of that, you
just count that. You say, OK there's
one in this case. So the dimension of
my rowspace is 1. And that's the same thing
as the dimension of my transpose's columnspace. I know it's getting all
confusing and it's late in the day for me as well. So that, hopefully, will
convince you that the rank of our transpose is the
same as the rank of our original matrix.