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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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# Showing relation between basis cols and pivot cols

Showing that linear independence of pivot columns implies linear independence of the corresponding columns in the original equation. Created by Sal Khan.

## Want to join the conversation?

- I was confused the different between rank(A) and dim(A), why they are not set in one defination?(4 votes)
- "dim" is a function which inputs a subspace and outputs an integer.

"rank" is a function which inputs a matrix and outputs an integer.

"dim" returns the dimensionality of a subspace. So for example, a subspace which takes the shape of a plane would return 2.

"rank" returns the dimensionality of the subspace generated as the span of column vectors. In general:

Given:`A = [a₁ a₂ ... an]`

(lower-case a's are column vectors)

Then:`rank(A) = dim(span({a₁, a₂, ... an}))`

Also note, "span" inputs a set of vectors and outputs a subspace.(16 votes)

- A*c = R*c = 0 = a1c1 + a2c2 + a4c4 = r1c1 + r2c2 + r4c4 = 0. For the RHS, ci = 0 by linear independence of ri's. Suppose there exists a vector c' s.t. Ac' = 0, but not all ci's ≠ 0. This new vector, c', by definition would be in the null space of A. However, since Rc' ≠ 0, c' would not be in the null space of R. This is a contradiction. Therefore, the only soln. to the left hand side is for ci = 0. I didn't understand this until I made this explanation, so I'll post it here in case it's useful.(7 votes)
- exactly .. I too didn't get the proof initially but just like you worked it out and finally got it. for those who are still struggling try re-watching the video from7:00mark.

there are only 5 steps

1) the pivot columns in reduced row echelon form are linearly independent ( because the ones (ie "0 1 0 0") in each column can't be made from the other columns)

2) the solution space i.e all the solutions to the equation Rx=0 and Ax=0 are the same . (as R is just the reduced form of A)

3) then lets just set the coefficients of R3,R5 (free columns) to zero ,and find a solution to the equation Rx=0. from step 1 we know that the remaining columns are linearly independent and so the only solution to the equation is setting the remaining coefficients also to zero

4)from step 3 we see the solution space doesn't contain any other coefficients to R1,R2,R4 except 0 if we set the coefficients of R3,R5 to 0

5)from step 2, the solution to AX=0 if we set A3,A4 to zero should be [0,0,0,0,0]T as well and from that we can see c1*A1+c2*A2+c3*A4=0 has only one solution. (this is the condition necessary to be linearly independent)

6)the remaining part of the proof is about proving that the set of linearly independent vectors span the null space. which is done in the next video(5 votes)

- I don't understand, at around8:00when Sal is explaining vector [c1 c2 0 c4 0] he said that c1,c2,c4 must be zero showing linear independence, now aren't the third and fifth term also zero? If c1 c2 c4 are zero then the solution would be [0 0 0 0 0], doesn't that imply all column vectors to be linearly independent?(2 votes)
- He says that the 1st, 2nd and 4th c
**must**be 0 for the result to be the 0 vector, and hence are linearly independent.

Whereas, the 3rd and 5th just happen to be zero.. but since these vectors can be made from the other vectors, you would also be able to find a way to get back to 0 even if these happened not to be. (He mentions showing this in the next video).(5 votes)

- why do I feel like a lot of this is just going around in circles?(3 votes)
- cuz it kinda is. He's trying to show the relationship btw the properties as he progresses through.(1 vote)

- General Question; By putting the matrix into REF instead of RREF, would you still obtain the linearly independent columns?(3 votes)
- In my opinion yes because from ref, rref always can be found, and the pivot columns don't change.(1 vote)

- If two square nxn matrices are from rank n how can i show that their sum is also from rank n (without using characteristic polynomial ) ??(2 votes)
- The nxn identity matrix and its negative are both nxn matrices with rank n. But their sum is the zero matrix, which has rank 0. So your hypothesis is false.(2 votes)

- Sal, why is there 0 in 3rd and 5th rows in matrix x, which is multiplied by r? This happens at around3:56(2 votes)
- By reducing the matrix to reduced row echelon form, we found the pivot columns. In R, the pivots are columns 1, 2, and 4. This tells us that the other two columns (3 and 5) can be expressed as linear combinations of the pivot columns. You can picture this by imagining each vector as a line. If, for example, r4 could be expressed as 2*r1, then that simply means that r4 is on the same line as r1. The idea remains the same if it is a more complex combination involving two or more vectors.

Now, if we are looking at the span of the vector space (which we are), then these vectors can simply be omitted. This becomes obvious (I hope) when thinking of the lines (or planes, etc. for higher dimensions). If r4 is on the same line as the other vectors, it is included in the span of those vectors.

So, to solve this equation more simply, we set these free variables to 0. If you remember, the values of the free variables can be set arbitrarily, but the set the values of the other vectors. In this example, if we used a value other than 0, we would have to find linear combinations of the pivot columns to cancel out the values in the non-pivot columns (which can be done! try it if you don't believe me!). So 0 is picked as a value to simplify calculation, but it does not affect the end result.(2 votes)

- can some explain again why C1,C2 and C4 has to be zero to be L.Ind? this is what confuses me about L.dep and L.ind.(2 votes)
- If C1, C2 and C4 did not all equal zero, but some could still be some non zero number to make C1*R1 + C2*R2 + C4*R4 = 0 this would imply one of those R vectors is a multiple of at least one of the other two. Maybe some examples.

starting with vectors of length 2. <1,0> and <0,1> are the most basic linearly independent vectors. ANY other vector with two elements will not be linearly independent to both. What this means is that you could multiply <1,0> and <0,1> each by some number to get any other vector in R2. let me know if that is not clear.

Now lets look at <1,0> being x1 and <0,1> being x2. More related to your question we would set up C1*x1 + C2*x2 = 0. Now, is there any combination of C1 or C2 other than both equaling 0 where we could make this linear combination equal the zero vector. So in this case <0,0> There is not. Then it is this thinking that carries on.

If you have n linearly independent vectors there is no way to make a linear combination of them so that you get the 0 vector in the end other than making them all be multiplied by 0. So in the video the rref vctors were <1,0,0,0>, <0,1,0,0>, <0,0,1,0>. now, could you make a linear combination of them to get the 0 vecotr, int his case <0,0,0,0>

If you are having trouble understanding why these three vectors dictate the whole matrix I can explain that too, just let me know(1 vote)

- Is it right to set R3 & R5 = 0 and prove that other vectors in the set are linearly independent?(2 votes)
- Sal mentioned 'particular solution' at3:35. I don't understand what exactly is the significance of a particular solution and what is the difference between a general solution and a particular solution?(1 vote)

## Video transcript

In the last video we saw a
method of figuring out what the basis for column space is. And we use these
exact examples. Whereas this was matrix A, I
just took it and I put it in reduced row echelon form. And I figured out which of these
columns in my reduced row echelon form of A,
are pivot columns. And it turned out to be the
first one, the second one, and the fourth one. And then the method is, you say
look, the corresponding columns in A-- so the first one,
the second one, and the fourth one-- form my basis
for my column space. And since they form the basis,
and if you want to know the dimension of your basis of your
column space, which is also called the rank,
you just say, well there's three in there. So it has a rank of
one, two, three. In this video I want to discuss
a little bit about why this worked. Why were we able to just take
the corresponding columns? Why did linear independence of
these three guys, imply linear independence of these
three guys? Why was the fact that I can
represent these guys-- this guy right here as a linear
combination of these three, or this guy as a linear combination
of these three-- why does that imply that I can
construct this guy as a linear combination of my
basis vectors? So the first thing that wasn't
too much of a stretch of the imagination in the last video,
was the idea that these pivot vectors are linearly
independent. So r1, r2, and r4. And everything I'm doing, I'm
kind of applying to the special case just so that it's
easier to understand. But it should be
generalizable. In fact, it definitely
is generalizable. That all of the pivot columns
in reduced row echelon form are linearly independent. And that's because the very
nature of reduced row echelon form, is that you are the only
pivot column that has a 1 in that respective row. So the only way to construct
it is with that vector. You can't construct it with
the other pivot columns because they're all going
to have 0 in that row. And when I say it's linearly
independent, I'm just saying the set of pivot columns. So let me say this in general. The set of pivot columns for any
reduced row echelon form matrix is linearly
independent. And it's just a very
straightforward argument. Because each column is
going to have a 1 in a very unique place. All of the other pivot columns
are going to have a 0 in that same place. And so you can't take any linear
combinations to get to that 1 because 0 times anything,
minus or plus 0 times anything, can never
be equal to 1. So I think you can
accept that. Now, that means that the
solution to c1 times r1, plus c2 times r2, plus, let
me say, c4 times r4. The solution to this equation,
because these guys are linearly independent, we know
that this only has one solution, and that's c1, c2,
and c4 is equal to 0. That's the only solution
to that. So another way we could say it
is, if we write r times some vector x-- well I'll just write
it times this particular x-- where I write it as
c1, c2, 0, c4, and 0 is equal to 0. So this will be some special
member of your null space. It's a particular solution
to the equation. This is equal to one, two,
three, four 0's because we have four rows here. Now, if we just expand
this out. If we just multiply 1 times c1,
plus 0 times c2, minus 1 times 0, plus 4 times 0, you'll
get-- or actually a better way to explain it-- this
multiplication right here can be written as-- and we've
seen this multiple times-- c1 times r1, plus c2 times
r2, plus 0 times r3. So we could just ignore that
term, plus c4, c4 times r4, plus 0 times r5. That's r5 right there. All of that equal to 0. So the only solution to this,
because we know that these three columns are linearly
independent-- or the set of just those three columns, those
three pivot columns are linearly independent-- the only
solution here is all of these equal to 0. That's exactly what I
said right up here. So the only solution here, where
if these two are 0, is that these guys also all have
to equal 0, if I already constrain these two. Now, the one thing that we've
done over and over again, we know that the solution set of
this equation, the solution set of Rx is equal to 0, is the
same as the solution set of Ax is equal to 0. Now, how do we know that? Or what do I mean? Well the solution set of this
is just the null space. The solution set is just
the null space of r. It's all of x's that satisfy
this equation. And we know that is equal to the
null space of a, because r is just a in reduced
row echelon form. So this is the null space of
a, is all of the x's that satisfy this equation. Now, the only version of this
that satisfied this equation was when c1, c2, and
c4 are equal to 0. So that tells us that the only
version of this, c1, c2, 0, c4, 0, that satisfies this
equation, or this equation, is when c1, c2, and c4
is equal to 0. Or another way of saying that
it if this is vector a1, a2, a4 right here, if you multiply
this out, you get c1-- let me do it over here, let me do it in
blue-- you get c1 times a1 plus c2 times a2, and then 0
times a3, plus c4 times a4 is equal to 0. Now these guys are going to be
linearly independent, if and only if the only solution to
this equation is they all equal to 0. Well we know that the only
solution to this is that they all equal 0 because anything
that's a solution to this is a solution to this. And the only solution to this
was, if I go ahead and I constrain these two terms to
being equal to 0, the only solution to this is all of
these c's have to be 0. So likewise, if I constrain
these to be 0, the only solution to this is that c1,
c2, and c4 have to be 0. So those guys have to be 0,
which imply that these three vectors, a1, a2, and a4, so that
implies that the set a1, a2, and a4 are linearly
independent. So we're halfway there. We've shown that because the
pivot columns here are linearly independent. We can show and they have
the same solution set. The null space of the reduced
row echelon form is the same as the null space of our
original matrix. We were able to show that the
only solution to c1 times this plus c2 times this plus c4 times
this is when all the constants are 0, which shows
that these three vectors or a set of those three vectors
are definitely linearly independent. Now, the next thing to prove
that they are a basis, is to show that all of the other
column vectors can be represented as multiples
of these three guys. And I realize, just for the sake
of clarity, or maybe not boring you too much, I'll do
that in the next video. So in this one we saw that if
the pivot columns are linearly independent, they always are. All pivot columns, by definition
are linearly independent. Or the set of pivot columns
are always linearly independent when you take away
the non-pivot columns, then the corresponding columns in
your original vector are also linearly independent. In the next one we'll show that
these three guys also span your column space.