NIKOLA KAMBUROV: Hi everyone.
In this video, we're going to explore briefly the concept of
a vector subspace.
This is the problem we're going to do.
We're given these four different subsets of R3, and
we are asked to figure out which of these is, in fact,
the subspace.
So I'll give you a few moments to try to do this on your own,
and then please come back to see whether you were right.
So hi again.
So before we start, let's briefly recall what a vector
subspace is.
Well vector subspace, of course, it's a subset.
In this case, a subset of R3.
But it behaves like a vector space itself, meaning that if
we take linear combination of element in the subset, what we
get is still something in that subset.
And in class, professor Strang showed you a few specific
examples of subspaces, which are related to matrices.
One of them was the null space of a matrix, all the vectors
that the matrix sends to 0, and the column space of a
matrix, which is the span of the column of a matrix.
So let's see if we can apply what we've learned in lecture
to our problem.
Number one, we are given a linear equation relation
between b1, b2, and b3.
Once we have something linear, this is a philosophy that
you'll learn in this class, we can always write this as a
matrix equation.
So in particular, b1 plus b2 minus b3 equals 0, we can just
write it as the matrix 1, 1 negative 1, and b1, b2, b3
equal to 0.
So therefore, b1, b2, and b3 precisely describe the null
space of 1, 1 negative 1.
And thus, what we are given in the first
question is a subspace.
What about the second one?
Again, we are given the relation between b1, b2, b3,
but it's not linear.
b3 is the product of b1 and b2.
So right, your guts tell you that this shouldn't be a
vector space.
But we need to prove why.
Well, see that the vector 1, 1, 1 is inside this subset,
because the third entry is the product 1, is the
product 1 and 1.
If the subset were a subspace itself, then we would have 2,
2, and 2 in it as well.
Right, because any multiple of a vector in the subspace is
inside the subspace.
But is 2, 2, 2 described by this equation?
Well no, because the third entry 2 is not
equal to 2 times 2.
So the example in number 2 is not a subspace.
Let's try 3.
And first, so we are given the subset in question 3 is given
as the linear span of these two vectors, 1 0 negative 1
and 1, 0, 1.
Plus the vector 1, 0, 0.
So let's think about this geometrically.
We know that the linear span of two linear independent
vectors, and these are obviously linearly
independent, is a plane in R3.
So we have the plane, and we add a vector to the point on
this plane.
So what if the vector were lying on the plane?
Well we're not going to change the plane.
We're still going to remain in the plane.
What I'm hinting at is the following thing.
That, in fact, 1, 0, 0 is a linear combination of 1, 0
negative 1 and 1, 0, 1, and it's fairly obvious to see
which linear combination of these two vectors it is.
It's 1/2, 1, 0, negative 1 plus 1/2 times 1, 0, 1.
So we can write the whole relation here in
the following way.
b1, b2, b3 is 1, 0, 0, which we can write this.
Plus 1, 0 negative 1, and c2, 1 0, 1.
Let me continue this here.
So b1, b2 and b3 is precisely c1 plus 1/2 of 1, 0, negative
1 plus c2 plus 1/2 of 1, 0, 1.
So indeed, the points b1, b2, and b3 are described by the
linear span 1, 0, negative 1, and 1, 0, 1.
So it is a vector subspace itself.
And finally, let's look at number 4.
So we have a similar situation.
I mean it's almost the same situation.
We again have the linear span of precisely the same vectors
as in question 3.
But this time we add the vector 0, 1, 0.
Now, 0, 1, 0 is not in the span of these two vectors.
So the argument that we just showed for 3 doesn't work.
But what we know about vector subspace is
the following thing.
Almost treated from the axioms that 0 needs to be inside the
subset in order for it to be a subspace.
It's a necessary condition.
Well is 0 inside this subset?
Meaning can we find coefficients c1 and c2 plus 0,
1, 0 to equal 0 vector?
And the answer is no.
And why is this?
There's a very easy way to see it.
Well we just look at the second entry of the vector is,
and we see that any multiple.
So the linear span of these two vectors will have as its
second entry 0.
And when we add it to 1, we can never get a 0 entry here.
So the subset in 4 is not a subspace.
So I hope this was useful and just getting an intuition
which subsets of R3 are subspaces.
I'll see you guys later.