WEBVTT

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NIKOLA KAMBUROV: Hi everyone.

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In this video, we're
going to explore briefly

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the concept of a
vector subspace.

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This is the problem
we're going to do.

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We're given these four
different subsets of R^3,

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and we are asked to figure out
which of these is, in fact,

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a subspace.

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So I'll give you a few moments
to try to do this on your own,

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and then please come back to
see whether you were right.

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So hi again.

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So before we start,
let's briefly recall

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what a vector subspace is.

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Well vector subspace, of
course, it's a subset.

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In this case, a subset of R^3.

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But it behaves like a
vector space itself,

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meaning that if we take
linear combination of elements

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in the subset, what we get is
still something in that subset.

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And in class, Professor
Strang showed you

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a few specific
examples of subspaces,

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which are related to matrices.

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One of them was the
null space of a matrix,

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all the vectors that
the matrix sends to 0,

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and the column space
of a matrix, which

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is the span of the
column of a matrix.

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So let's see if we can
apply what we've learned

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in lecture to our problem.

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Number one: we are given
a linear equation relation

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between b_1, b_2, and b_3.

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Once we have something
linear-- this is a philosophy

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that you'll learn
in this class--

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we can always write this
as a matrix equation.

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So in particular, b_1 plus
b_2 minus b_3 equals 0,

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we can just write
it as the matrix 1,

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1, negative 1, times
[b 1, b 2,  b 3] equal to 0.

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So therefore, b_1,
b_2, and b_3 precisely

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describe the null space
of 1, 1 negative 1.

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And thus, what we are
given in the first question

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is a subspace.

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What about the second one?

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Again, we are given the
relation between b_1, b_2, b_3.

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but it's not linear.

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b_3 is the product
of b_1 and b_2.

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So right, your guts
tell you that this

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shouldn't be a vector space.

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But we need to prove why.

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Well, see that the vector [1, 1,
 1] is inside this subset,

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because the third entry is
the product-- 1 is the product

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1 and 1.

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If the subset were
a subspace itself,

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then we would have 2,
2, and 2 in it as well.

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Right, because any multiple
of a vector in the subspace

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is inside the subspace.

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But is [2, 2, 2] described
by this equation?

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Well no, because
the third entry, 2,

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is not equal to 2 times 2.

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So the example in number
2 is not a subspace.

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Let's try 3.

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And first-- so we are given--
so the subset in question 3

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is given as the linear span of
these two vectors, [1, 0, -1]

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and [1, 0, 1].

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Plus the vector [1, 0, 0].

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So let's think about
this geometrically.

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We know that the linear span
of two linear independent

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vectors-- and these
are obviously linearly

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independent-- is a plane in R^3.

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So we have the plane,
and we add a vector

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to the point on this plane.

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So what if the vector
were lying on the plane?

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Well we're not going
to change the plane.

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We're still going to
remain in the plane.

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What I'm hinting at is
the following thing.

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That, in fact, [1, 0, 0] is a
linear combination of [1, 0,

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-1] and [1, 0, 1], and
it's fairly obvious

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to see which linear combination
of these two vectors it is.

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It's 1/2 [1, 0, -1] plus
1/2 times [1, 0, 1].

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So we can write the
whole relation here

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in the following way.

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[b 1, b 2, b 3] is
[1, 0, 0], which

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we can write this, plus c_1
[1,  0, -1], and c_2 [1 0, 1].

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Let me continue this here.

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So b_1, b_2, and
b_3 is precisely

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c_1 plus 1/2 of [1, 0, -1]
plus c_2 plus 1/2 of [1, 0, 1].

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So indeed, the points
b_1, b_2, and b_3

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are described by the linear
span [1, 0, -1] and [1, 0, 1].

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So it is a vector
subspace itself.

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And finally, let's
look at number 4.

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So we have a similar situation.

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I mean, it's almost
the same situation.

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We again have the linear span
of precisely the same vectors

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as in question 3.

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But this time we add the
vector [0, 1, 0] to them.

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Now, [0, 1, 0] is not in the
span of these two vectors.

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So the argument that we just
showed for 3 doesn't work.

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But what we know
about vector subspaces

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is the following thing.

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I follows almost
trivially from the axioms

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that 0 needs to be inside
the subset in order

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for it to be a subspace.

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It's a necessary condition.

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Well is 0 inside this subset?

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Meaning can we find coefficients
c_1 and c_2, plus [0, 1, 0]

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to equal zero vector?

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And the answer is no.

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And why is this?

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There's a very
easy way to see it.

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Well we just look at the
second entry of the vectors,

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and we see that any multiple--
so the linear span of these two

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vectors will have as
its second entry 0.

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And when we add it to 1, we can
never get a zero entry here.

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So the subset in 4
is not a subspace.

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So I hope this was useful in
just getting an intuition which

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subsets of R^3 are subspaces.

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I'll see you guys later.