Null space and column space
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Matrix Vector Products
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Introduction to the Null Space of a Matrix
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Null Space 2: Calculating the null space of a matrix
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Null Space 3: Relation to Linear Independence
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Column Space of a Matrix
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Null Space and Column Space Basis
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Visualizing a Column Space as a Plane in R3
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Proof: Any subspace basis has same number of elements
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Dimension of the Null Space or Nullity
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Dimension of the Column Space or Rank
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Showing relation between basis cols and pivot cols
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Showing that the candidate basis does span C(A)
Introduction to the Null Space of a Matrix Showing that the Null Space of a Matrix is a valid Subspace
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- Let's review our notions of subspaces again.
- And then let's see if we can define some interesting
- subspaces dealing with matrices and vectors.
- So a subspace-- let's say that I have some subspace-- oh, let
- me just call it some subspace s.
- This is a subspace if the following are true-- and this
- is all a review-- that the 0 vector-- I'll just do it like
- that-- the 0 vector, is a member of s.
- So it contains the 0 vector.
- Then if v1 and v2 are both members of my subspace, then
- v1 plus v2 is also a member of my subspace.
- So that's just saying that the subspaces are
- closed under addition.
- You can add any of their two members and you'll get another
- member of the subspace.
- And then the last requirement, if you remember, is that
- subspaces are closed under multiplication.
- So that if c is real number, and it's just a scalar.
- And if I multiply, and v1 is a member of my subspace, then if
- I multiply that arbitrary real number times my member of my
- subspace, v1, I'm going to get another
- member of the subspace.
- So it's closed under multiplication.
- These were all of what a subspace is.
- This is our definition of a subspace.
- If you call something a subspace,
- these need to be true.
- Now let's see if we can do something interesting with
- what we understand about matrix vector multiplication.
- Let's say I have the matrix a-- I'll make it nice and
- bold-- and it's an m by n matrix.
- And I'm interested in the following situation; I want to
- set up the homogeneous equation.
- And we'll talk about why it's homogeneous.
- Well, I'll tell you in a second.
- So let's say we set up the equation.
- My matrix a times vector x is equal to the 0 vector.
- This is a homogeneous equation,
- because we have a 0 there.
- And I want to ask the question-- I
- talked about subspaces.
- If I take all of the x's-- if I take the world, the
- universe, the set of all of the x's that satisfy this
- equation, do I have a valid subspace?
- Let's think about this.
- I want to take all of the x's that are a member of Rn.
- Remember, if our matrix a has n columns, then I've only
- defined this matrix vector multiplication.
- If x is a member of r, and if x has to have exactly n
- components, only then is it defined.
- So let me define a set of all the vectors that are a member
- of Rn where they satisfy the equation a times my vector x
- is equal to the 0 vector.
- So my question is, is this a subspace?
- Is this a valid subspace?
- So the first question is, does it contain the 0 vector?
- Well in order for this to contain the 0 vector, the 0
- vector must satisfy this equation.
- So what is any m by n matrix a times the 0 vector?
- Let's write out my matrix a-- my matrix a, a[1,1]
- a[1,2]
- all the way to a[1,n]
- and then this, as we go down a column, we go all
- the way down to a[m,1]
- and then as we go all the way to the bottom
- right, we go to a[m,n]
- and I'm going to multiply that times the 0 vector that has
- exactly n components.
- So the 0 vector with n components is 0, 0, and you're
- going to have n of these.
- The number of components here has to be the exact same
- number of the number of columns you have. But when you
- take this product, this matrix vector
- product, what do you get?
- What do we get?
- Well, this first term up here is going to be a[1,1]
- time 0, plus a[1,2]
- times 0, plus each of these terms times 0.
- And you add them all up. a[1,1]
- times 0, plus a[1,2]
- plus a[1,2]
- times 0, all the way to a[1,n]
- and times 0.
- So you get 0.
- Now this term is going to be a[2,1]
- times 0, plus a[2,2]
- times 0, plus a[2,3]
- times 0, all the way to a[2,n]
- times 0.
- That's, obviously, going to be 0.
- And you're going to keep doing that because all of these are,
- essentially,-- you can kind of view it as the dot product
- of-- I haven't defined dot products with row vectors and
- column vectors, but I think you get the idea-- the sum of
- each of these elements, multiplied with the
- corresponding component in this vector.
- And, of course, you're just always multiplying by 0 and
- then adding up.
- So you're going to get nothing but a bunch of 0's.
- So the 0 vector does satisfied the equation.
- A times the 0 vector is equal to the 0 vector.
- And this is a very unconventional notation.
- I'm just writing it like that, because I don't feel like
- bolding out my 0's all the time to make you realize that
- that's a vector.
- So we meet our first requirement.
- The 0 vector is a member of the set.
- So let me define my set here.
- Let me define it n.
- And I'll tell you in a second why I'm calling it n.
- So we now know that the 0 vector is a
- member of my set n.
- Now let's say I have two vectors, v1 and v2 that are
- members-- let me write this.
- So let's say I have two factors, v1, and v2, that are
- both members of our set.
- What does that mean?
- That means that they both satisfy this equation.
- So that means that a-- my matrix a-- times vector 1 is
- equal to 0.
- This is by definition.
- I'm saying that they're a member of the set, which means
- they must satisfy this.
- And that also means that a times vector 2 is
- equal to our 0 vector.
- So in order for this to be closed under addition, a times
- vector 1 plus vector 2, the sum of these two vectors
- should also be a member of n.
- But let's figure out what this is.
- The sum of these two vectors is this vector right here.
- This is equal to-- and I haven't
- proven this to you yet.
- I haven't made a video where I prove this.
- But it's very easy to prove just using the definition of
- matrix vector multiplication, that matrix vector
- multiplication does display the distributive property.
- And maybe I'll make a video on that, but literally, you just
- have to go through the mechanics of each of the
- terms.
- This is equal to a[v,1]
- plus a[v,2]
- And we know that this is equal to the 0 vector.
- And this is equal to the 0 vector.
- And if you add the 0 vector to itself, this whole thing is
- going to be equal to the 0 vector.
- So if v1 is a member of n, and v2 is a member of n, which
- means they both satisfy this equation, then v1 plus v2 is
- definitely still a member of n.
- Because when I multiply a times that, I get
- the 0 vector again.
- So let me write that result, as well.
- So we now know that v1 plus v2 is also a member of n.
- And the last thing we have to show is that it's closed under
- multiplication.
- Let's say that v1 is a member of our space that I defined
- here, where they satisfy this equation.
- What about c times v1?
- Is that a member of n?
- Well let's think about it.
- What's our matrix a times the vector-- right?
- I'm just multiplying this times the scale.
- I'm just going to get another vector.
- I don't want to write a capital v there.
- Lowercase v, so it's a vector.
- What's this equal to?
- Well, once again, I haven't prove it to you yet, but it's
- actually a very straightforward thing to do,
- to show that when you're dealing with scalars, if you
- have a scalar here, it doesn't matter if you multiply the
- scalar times the vector before multiplying it times the
- matrix or multiplying the matrix times the vector, and
- then doing the scalar.
- So it's fairly straightforward to prove that this is equal to
- c times our matrix a-- I'll make that nice and bold, times
- our vector v.
- That these two things are equivalent.
- Maybe I should just churn out the video that does this, but
- I'll leave it to you.
- You, literally, just go through the mechanics
- component by component.
- And you show this.
- But clearly, if there's is true, we already know that v1
- is a member of our set, which means that a times v1 is equal
- to the 0 vector.
- And so that means this will reduce to c times the 0
- vector, which is still the 0 vector.
- So c[v,1]
- is definitely a member of n.
- So it's closed under multiplication.
- And I kind of assumed this right here.
- But maybe I'll prove that in a different video.
- But I want to do all this to show that this set n is a
- valid subspace.
- This is a valid subspace.
- It contains a 0 vector.
- It's close under addition.
- It's close under multiplication.
- And we actually have a special name for this.
- We call this right here, we call n, the null space of a.
- Or we could write n is equal to-- maybe I shouldn't have
- written an n.
- Let me write orange in there.
- Our orange n is equal to-- the notation is just the null
- space of a.
- Or we could write the null space is equal to the orange
- notation of n, and literally, if I just give you some
- arbitrary matrix a, and I say, hey, find me n
- of a, what is that?
- Literally, your goal is to find the set of all x's that
- satisfy the equation a times x is equal to 0.
- And I'm going to do that in the next video.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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