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## Linear algebra

### Course: Linear algebra > Unit 1

Lesson 7: Null space and column space- Matrix vector products
- Introduction to the null space of a matrix
- Null space 2: Calculating the null space of a matrix
- Null space 3: Relation to linear independence
- Column space of a matrix
- Null space and column space basis
- Visualizing a column space as a plane in R3
- Proof: Any subspace basis has same number of elements
- Dimension of the null space or nullity
- Dimension of the column space or rank
- Showing relation between basis cols and pivot cols
- Showing that the candidate basis does span C(A)

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# Dimension of the null space or nullity

Dimension of the Null Space or Nullity. Created by Sal Khan.

## Want to join the conversation?

- If we can have exercises for these concepts as well it will be great. As these concepts are difficult to master.(45 votes)
- why x2 is a free variable ? isnt x2 pivot as well ?(17 votes)
- When you're looking at free variables, pay attention to the pivot columns. The first and third columns hold a pivot. That leaves the second, fourth, and fifth as free variables.

Realistically, you could treat x2 as a non-free, but to do so would mean you'd have to make x1 free in order to satisfy the first row of the matrix. The general practice, though, is to treat the pivot, x1, as the non-free and x2 as the free.

When you solve an augmented matrix, it's solving a system of equations. If you were to place a vector of values into the matrix Sal used in an attempt to solve for values, you would still have to CHOOSE random values for x2, x4, and x5 to be able to solve for x1 and x3. That's where the notion of a "free" variable comes from. The math gives you the freedom to choose any value for those 3 variables and still come up with an answer for the pivots.(22 votes)

- I am confused. In a previous video ("null space and column space") we learned that the "basis" are the column vectors that correspond to the pivot entries in the rref(B). Which for our case would be the column vectors [1,1] and [2,3]. These TWO vectors should be the basis of the column space of B, which is clearly in R^2. How can the THREE vectors in this video be also a "basis"! (and these are in R^5?!). Or is it that a "basis of a column space" is different than a "basis of the null space", for the same matrix?(15 votes)
- In short, you are correct to say that 'a "basis of a column space" is different than a "basis of the null space", for the same matrix."

A*basis*is a a set of vectors related to a particular mathematical 'space' (specifically, to what is known as a*vector space*). A basis must:

1. be linearly independent and

2. span the space.

The 'space' could be, for example: R^2, R^3, R^4, ... , the column space of a matrix, the null space of a matrix, a subspace of any of these, and so on (there are many more examples; vector spaces appear everywhere in mathematics). When someone gives you a basis, you need to ask: "A basis*of what space?*"

In general, if I give you two different spaces, A and B, then a basis for A will be different to a basis for B.

Now, in general, the column space of a given matrix does**not**equal the null space of that matrix. Therefore, in general, a basis for the column space of the matrix will be**different**to a basis for the null space of that matrix.(23 votes)

- Hi sal i just wanna ask question about reducing echelon form,how to know if that matrix is been reduce completely???(7 votes)
- There are three conditions for a matrix to be in RREF

1) The first non-zero entry of a row must be a 1; this entry is called a pivot

2) The pivot for each row must to the right of all the pivots in any rows above.

3) Any columns that contain pivots must have zeros for all other entries except the pivot.

As a result of 2 and 3, any rows that contain all zeros must be at the bottom of the matrix.

Hope that helps.(24 votes)

- Didn't Sal say that if Nul B is not the zero vector, the B is linearly dependent?

Why are we proving that B is linearly INDEPENDENT so that it can form the basis for the null space?(8 votes)- In this case, B is not linearly independent, what Sal is actually proving is that the set of solution vectors for NULL B (Bx=0) are linearly independent, which is a different thing.(11 votes)

- Is the null space the same thing as a solution space?(2 votes)
- The null space of B is the solution space of the problem B
**x**=**0**.(8 votes)

- I've seen an example of a subspace in R^3 only have a dimension of 2. How is this visually possible? Doesn't a vector in R^3 mean it's in the third dimension?(1 vote)
- The dimension only means the number of elements in the basis for the subspace. Think of a plane in ℝ³ that passes through the origin, how many vectors/elements does it take at minimum, such that when you take their span, you get a plane? Well, 2. And that tells us that the basis for a plane has 2 vectors in it. If the dimension is again, the number of elements/vectors in the basis, then the dimension of a plane is 2.

So even though the subspace of ℝ³ has dimension 2, the vectors that create that subspace still have 3 entries, in other words, they still live in ℝ³.(8 votes)

- What would the dimension of the nullspace be if the the only sloution to Ax = 0 is x = 0? Can you consider the null vector as a basis?(4 votes)
- I think that the basis of such a space is the empty set ∅, so the dimension of it would be 0.(2 votes)

- If the basis for a solution space is the zero vector, will the dimension for the solution space also be zero?(4 votes)
- At the end of the video, I think the non-pivot column may not represent a free variable. If a non-pivot column of a matrix only contains zero, it can't represent a free variable. Is it right?(2 votes)
- If a non-pivot column contains only zeroes, then it must be a free variable. It has to be, because then it can be anything and it won't affect the answer at all.(4 votes)

## Video transcript

Let's say I have this matrix
B, here, and I want to know what the null space of B is. And we've done this multiple
times but just as a review, the null space of B
is just all of the x's that are a member. It's all the vector x's that
are member of what? 1, 2, 3, 4, 5 that are members
of r to the fifth, where B, my matrix B, times any of these
vector x's, is equal to 0. That's the definition
of the null space. I'm just trying to find the
solution set to this equation right here. And we've seen before, that the
null set of the reduced row echelon form of B is equal
to the null set of B. So what's the reduced row
echelon form of B? And this is actually almost
trivially easy. Let me just take a couple of
steps right here-- to get a 0 here, let's just replace row
2 with row 2 minus row 1 So what do we get? Row 2 minus row 1. Row 1 doesn't change, it's
just 1, 1, 2, 3, 2. And then row 2 minus row 1. 1 minus 1 is 0. 1 minus 1 is 0. 3 minus 2 is 1. 1 minus 3 is minus 2. 4 minus is 2 is 2. We're almost there. Let's see, so this is a free
variable right here. This is a pivot variable
right here. We have a 1. So let me get rid of that
guy right there. And I can get rid of that guy
right there, by replacing row 1 with row 1 minus
2 times row 2. So now row 2 is going
to be the same. 0, 0, 1 minus 2, 2. And let me replace row 1 with
row 1 minus 2 times row 2. So 1 minus 2 times 0 is 1. 1 minus 2 times 0 is 1. 2 minus 2 times 1 is 0. 3 minus 2 times minus 2. So that's 3 plus
4 is 7, right? 2 times this is minus 4 and
we're subtracting it. And then 2 minus 2 times
2-- that's 2 minus 4-- it's minus 2. So this is the reduced row
echelon form of B is equal to that right there. And then if I wanted to figure
out its null space, I have x1, x2, x3, x4, and x5 equaling--
I'm going to have two 0's right here. Now I can just write this
as just a set of or a system of equations. So let me do that. I get x1. I'm going to write my pivot
variables in a green color. x1 plus 1 times x2, so plus
x2, plus 0 times x3. Plus 7 times x4. Minus 2 times x5 is equal
to that 0 right there. And then I get my--
this is x3, right? 0 times x1 plus 0 times
x2 plus 1 times x3. So I get x3 minus 2 times x4
plus 2 times x5 is equal to that 0 right there. And then if we solve for our
pivot variables, right? These are our free variables. We can set them equal
to anything. If we solve for our pivot
variables what do we get? We get x1 is equal to-- I
should do that in green. The color coding helps. I get x1 is equal to minus x2
minus 7x4 plus 2x5, just subtracted these from both
sides of the equation. And I get x3 is equal to--
we've done this multiple times-- 2x4 minus 2x5. And so if I wanted to write
the solution set in vector form, I could write my solution
set or my null space, really, is-- or all
the possible x's. x1, x2, x3, x4, x5. This is my vector
x, that's in r5. It is equal to a linear
combination of these. So let me write it out. The free variables are x2 times
some vector right there. Plus x-- is x3, no x3 is
not a free variable. Plus x4, that's my next free
variable, times some factor. Plus x5 times some vector. I've run out of space. Plus x5 times some vector. And what are those vectors? Let's see. I don't want to make this too
dirty, so let me see if I can maybe move-- nope that's not
what I wanted to do. Let me just rewrite this. I haven't mastered this pen
tool yet, so let me rewrite this here. So x3 is equal to
2x4 minus 2x5. Let me delete this right
over here so I get some extra space. Cross that out. I think that's good enough. So I can go back to what
I was doing before. x5 times some vector
right here. And now what are
those vectors? We just have to look
at these formulas. x1 is equal to minus
1 times x2. So minus 1 times x2. Minus 7 times x4. Plus 2 times x5. Fair enough. And what is x3 equal to? x3 is equal to 2x4. 2x4, right? It had nothing to do with x2
right here, so it's equal to 2x4 minus 2x5. And then 0 times x2, right? Because it had no x2
term right here. And then what is x2 equal to? Well x2 is just equal
to 1 times x2. And so all of these terms
are 0 right there. And I want you to pay
attention to that. I'll write it right here. x2 is a free variable, so it's
just equal to itself, right? 1 and you write a 0 and a 0. x4 is a free variable. And this is the important
point of this exercise. So it's just equal to
1 times itself. You don't have to throw in any
of the other free variables. And x5 is a free variable. So it just equals 1 times itself
and none of the other free variables. So right here we now say that
all of the solutions of our equation Bx equals 0, or the
reduced row echelon form of B times x is equal to 0,
will take this form. Or they are linear combinations
of these vectors. Let's call this v1,
v2, and v3. These are just random
real numbers. I can pick any combination here
to create this solution set, or to create
our null space. So the null space of A, which is
of course equal to the null space of the reduced row echelon
form of A, is equal to all the possible linear
combinations of these 3 vectors, is equal to the span
of my vector v1, v2, and v3. Just like that. Now, the whole reason I went
through this exercise-- because we've done this multiple
times already-- is to think about whether these guys
form a linear independent set. So my question is are these
guys linearly independent? And the reason why I care is
because if they are linearly independent then they
form a basis for the null space, right? That we know that they span the
null space, but if they're linearly independent,
then that's the 2 constraints for a basis. You have to span the subspace,
and you have to be linearly independent. So let's just inspect these
guys right here. This v1, he has a
1 right here. He has a 1 in the second term
because he corresponds to the free variable x2, which is
the second entry, so we just throw a 1 here. And we have a 0 everywhere
else in all of the other vectors in our spanning set. And that's because for the other
free variables we always wanted to multiply them
times a 0, right? And this is going to
be true of any null space problem we do. For any free variable, if this
free variable represents a second entry, we're going
to have a 1 in the second entry here. And then a 0 for the second
entry for all of the other vectors associated with the
other free variables. So can this guy ever be
represented as a linear combination of this
guy and that guy? Well there's nothing that I can
multiply this 0 by and add to something that I multiply
this 0 by to get a 1 here. It's just going to get 0's. So this guy can't be
representated as a linear combination of these guys. Likewise, this vector
right here has a 1 in the fourth position. Why is it a fourth position? Because the fourth position
corresponds to its corresponding free
variable, x4. So this guy's a 1 here. These other guys
will definitely always have a 0 here. So you can't take any linear
combination of them to get this guy. So this guy can't be represented
as a linear combination of those guys. And last, this x5 guy, right
here, has a 1 here. And these guys have 0's here. So no linear combination of
these 0's can equal this 1. So all of these guys are
linearly independent. You can't construct any of
these vectors with some combination of the other. So they are linearly
independent. So v1, the set v1, v2, and v3
is actually a basis for the null space, for the null space
of-- Oh, you know what, I have to be very careful. For the null space of B. Just for variety, I defined my
initial matrix as matrix B, so let me be very careful here. So the null space of B was equal
to the null space of the reduced row echelon form of B. It's good to switch things up
every once in a while, you start thinking that
every matrix is named A if you don't. And that's equal to the
span of these vectors. So these vectors, and we just
said that they're linearly independent. We just showed that because
there's no way to get that one from these guys, that one from
these guys, or that one from these guys. These guys form a basis for
the null space of B. Now this raises an interesting
question. In the last video, I defined
what dimensionality is. And maybe you missed
it because that video was kind of proofy. But the dimensionality, the
dimension, of a subspace-- I'll redefine it here-- is the
number of elements in a basis for the subspace. And in the last video I took
great pains to show that all bases for any given subspace
will have the same number of elements. So this is well defined. So my question to you now is:
what is the dimension of my null space of B? What is that the dimension
of my null space of B? Well, the dimension is just
the number of vectors in a basis set for B. Well this is a basis set
for B right there. And how many vectors
do I have in it? I have 1, 2 3 vectors. So the dimension of the
null space of B is 3. Or another way to think about
it-- or another name for the dimension of the null space
of B-- is the nullity, the nullity of B. And that is also equal to 3. And let's think about
it, you know I went through all this exercise. But what is the nullity of any
matrix going to be equal to? It's the dimension of
the null space. Well the dimension of the null
space-- you're always going to have as many factors here as
you have free variables. So in general, the nullity of
any matrix of any matrix-- let's say matrix A-- is equal
to the number of I guess you could call it free variable
columns or the number free variables in, well, I guess we
call it in the reduced row echelon form, or I guess we
could say the number of non-pivot columns. The number of non-pivot columns
in the reduced row echelon form of A. Because that's essentially the
number of free variables-- all of those free variables have
an associated, linearly independent vector with
each of them, right? So the number of variables is
the number of vectors you're going to have in your basis
for your null space. And the number of free variables
is essentially the number of non-pivot columns
in your reduced row echelon form, right? This was a non-pivot column,
that's a non-pivot column, that's a non-pivot column. And they're associated
with the free variables x2, x4, and x5. So the nullity of a matrix is
essentially the number of non-pivot columns in the reduced
row echelon form of that matrix. Anyway, hopefully you found
that vaguely useful.