Null Space 3: Relation to Linear Independence Understanding how the null space of a matrix relates to the linear independence of its column vectors
Null Space 3: Relation to Linear Independence
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- The neat thing about linear algebra in general is some
- very seemingly simple concepts can be interpreted in a bunch
- of different ways, and can be shown to represent different
- ideas or different problems. And that's what I'm going to
- do in this video.
- I'm going to explore the nullspace, or even better, I'm
- going to explore the relationship-- if I have some
- matrix A times some vector x, and that that is
- equal to the 0 vector.
- And we, of course, saw on the last few videos that the
- nullspace of A, is equal to all of the vectors x in Rn.
- So this will have n components.
- This would have to be an m by n matrix.
- If this was an m by a matrix, I'd say all of
- the vectors in Ra.
- So this number right here has to be the same as that number
- in order for the matrix vector multiplication to be valid.
- But the nullspace of A is all of the vectors in Rn that
- satisfy this equation.
- Where if I take A and I multiply it times any one of
- the vectors in the nullspace, I should get the 0 vector.
- And this is going to have m components, and we've seen
- that in previous-- this is going to have the 0 vectors.
- I'll put it this way, the 0 vector's going to
- be a member of Rm.
- So that's what our nullspace is.
- Let's explore it a little bit.
- We know already that our vector, our matrix, can be
- rewritten like this.
- We could just write it as a set of column vectors.
- I could say this right here, that's v1, then I'm going to
- have v2, and I have n columns.
- So this last column right here is going to be v sub n.
- So if I define my vectors this way, that's the first vector,
- that's the second vector, than I can rewrite my matrix A.
- I could say A is equal to just a bunch of column vectors.
- v1, v2, all the way to vn.
- And multiplying this matrix times a vector x, so times x,
- so times x1, x2, all the way to xn.
- We've seen in the past on the matrix vector product
- definition video that this can be interpreted as, this has
- actually just coming straight out of the definition.
- This is the same thing as x1 times vector 1, times the
- first column, plus x2 times the second column, times that
- column, all the way to, and you just keep adding them up,
- all the way to xn.
- Times the nth column.
- This just comes straight out of our definition of matrix
- vector products.
- Now, if we're saying that Ax is equal to 0, we're looking
- for the solution set to that.
- If we're looking for the solution set to Ax is equal to
- 0, then that means-- is equal to the 0 vector, that that
- means that this sum, we're trying to find a solution set
- of this sum is equaling 0.
- We want to figure out the x1's, x2's, x3's, all the way
- to xn's, that make this equal the 0 vector.
- What are we doing?
- We're taking linear combinations
- of our column vectors.
- We're taking linear combinations of our column
- vectors, and seeing if we can take some linear combination
- and get it to the 0 vector.
- Now, this should start ringing bells in your head.
- This little equation, or this little expression right here,
- should start ringing bells.
- This was part of how we defined what linear
- independence was.
- We said that if this was the definition of linear
- independence, or we proved this fell out of the
- definition of linear independence, and if I have a
- bunch of vectors, v1, v2, all the way to vn, we say that
- they are linearly independent.
- There's kind of the non-mathematical way of
- describing it, I guess this is mathematical as well, is that
- look, none of those vectors can be represented as a
- combination of the other ones.
- And then we show that that means that the only solution
- to this equation would be that x1, x2, all of the
- coefficients on this, has to be equal to 0.
- That this is the only solution.
- Linear independence means that this is the only solution to
- this equation right now.
- If the only way that you get the 0 vector, by taking
- combination of all of these common vectors, the only way
- to do that is to have all of these guys equal 0.
- Then you are linearly independent.
- Likewise, if v1, v2, all the way to vn are linearly
- independent, then the only solution to this is for these
- coefficients to be 0.
- And we saw that in our video on linear independence.
- Now, if all of these coefficients are 0,
- what does that mean?
- That means that our vector x is the 0 vector,
- and only the 0 vector.
- That's the only solution.
- So we have something interesting here.
- If our column vectors are linearly independent, if v1,
- v2, all the way to vn, are linearly independent, then
- that means that the only solution to Ax equals 0, is
- that x has to be equal to 0 vector.
- Or put another way, the solution set of this equation,
- which is really just a nullspace, the nullspace is
- all of the x's that satisfy this equation.
- So that the nullspace of A has to only contain the 0 vector.
- So that's an interesting result.
- If we're linearly independent, then the nullspace of A only
- contains the 0 vector.
- Which is another way of saying that-- let me write this--
- well, I already wrote it down, that x1, x2, all of them, have
- to be equal to 0.
- Now if I were to multiply this equation out and get it into
- reduced row echelon form, what does that mean?
- We saw in a previous video that the nullspace of A is
- equal to the nullspace of the reduced row echelon form of A.
- And that's-- the nullspace of A is 0, because its column
- vectors are each linearly independent, and that means
- that the nullspace of the reduced row echelon form of A
- must also equal the 0 vector.
- And that means that if I take the reduced row echelon form
- of A, times-- maybe I'm being a little redundant-- the
- reduced row echelon form of A, and I multiply that times x,
- or I want to solve this equation, the only solution
- right here is x is equal to the 0 vector.
- And if you think about what that means, if this is the
- only solution, that means that this reduced row echelon form
- has no free variables.
- It literally would just have to look like this.
- So this is x, x1, x2, all the way to xn, the reduced row
- echelon form of A, in order for this to have a unique
- solution, and that unique solution being 0, the reduced
- echelon form is going to have to look like this.
- 1 times x1 plus 0 times all the other ones, so you're
- going to have just a bunch of n0's, and you're going to have
- 1 times x2, plus 0's times everything else.
- And those 1's are going to go all the way down the diagonal,
- so it's going to look like that, and then that is going
- to be equal to the 0 vector.
- And this is going to be a square matrix, where this has
- to be n, and this has to be n.
- How do I know that?
- Because I said that x1, x2, and all of these have to be
- equal to 0.
- So they have to be equal to 0.
- If I just write them as a system of equations, if I
- write x1 is equal to 0, x2 is equal to 0, x3 is equal to 0,
- all the way to xn is equal to 0.
- This system of equations, if I wrote it as an augmented
- matrix, remember this is x1 plus 0, x2 plus 0.
- This as in augmented matrix, and we've done this multiple
- times, it would look like this.
- 1, you just have a bunch of 0's, n0's, and then the 1's
- would just go down the diagonal, and then you'd have
- n0's right there.
- So that's where I'm getting it from.
- If we are linearly independent, the nullspace of
- A is going to be just a 0 vector, and if the nullspace
- of A is just a 0 vector, then the nullspace of the reduced
- row echelon form is only the 0 vector.
- The only solution is all of the x's equal to 0.
- Which means a reduced row echelon form of A has to
- essentially just be 1's down the diaganol, with 0's
- everywhere else.
- So anyway, I just want to make this-- this is kind of a neat
- by-product of an
- interpretation of the nullspace.
- Let me write that.
- Let me summarize our results.
- The nullspace of A, if it just equals 0, then that means, you
- can go both ways, that's true if and only if the column
- vectors of A are linearly independent.
- And all of that's only true-- this is true, I was going to
- do a triangle, it might turn into a square-- if x1, x2, all
- of these have to be equal 0.
- This is the only solution.
- And then that implies that the reduced row echelon, and I
- didn't do it as precisely as I would have liked, but the
- reduced row echelon form of A is essentially going to be a
- square n by n matrix.
- And, by the way, this can only be true if we're dealing with
- an n by n matrix to begin with.
- And maybe I'll do that a little bit more precisely in a
- future video.
- But then the reduced row echelon form of A is going to
- have to look like this, just a bunch of 1's down the
- diagonal, with 0's everywhere else.
- And these all imply each other.
- Now, what if the nullspace of A contains some other vectors?
- Well, then we would have to say that the column vectors of
- A are linearly dependent.
- And if they're linearly dependent, then we wouldn't
- have a reduced row echelon form of A that looked like
- this, you would have something that would have some free
- variables that allows you to create more solutions there.
- But anyway, I just wanted to give you this angle on how you
- can interpret the nullspace and how it relates to linear
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