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PROFESSOR: Hi there.

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Welcome back to recitation.

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In lecture, you've been learning
about when vectors are linearly

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independent, when they span
the space, what a basis is,

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what a dimension of
a vector space is.

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And the problem for today
is exactly about that.

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We have a vector
space that is given.

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It's spanned by
these four vectors.

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And we're asked to find the
dimension of that vector space

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and the basis for it.

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Well, why don't you hit pause
on the video, and work on it

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for a while.

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And I'll come back in a little
bit to help you out with it.

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We're back.

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Let's work on it.

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So we need to find the
dimension and the basis.

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Remember what the dimension is?

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It's simply the
number of vectors

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in a basis for the vector space.

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So actually, the
problem is backwards.

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We want to find the
basis for the space

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first, and then
find the dimension.

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I'll write "first"
here and "second" here.

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So we want to find a basis
for the vector space spanned

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by these four vectors.

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So you might be
tempted to just say

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that a basis for
this vector space

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is those four vectors, because
they span the vector space.

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But there's another thing
that a basis has to satisfy.

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And it is the elements
of the basis have

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to be linearly independent.

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We don't know that these are.

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So we have to check.

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How do we check
that four vectors

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are linearly independent?

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Well, there's a couple
of different ways.

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But here's what
we're going to do.

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We're going to put these
vectors as rows of a matrix.

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And then we'll do elimination.

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And when we get to the end,
the rows that have pivots

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are the independent ones.

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So let's do that.

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1, 1, -2, 0, -1; 1, 2, 0, -4, 1;
0, 1, 3, -3, 2; 2, 3, 0, -2, 0.

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By now you must have done
elimination a million times,

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so I'll go a little bit faster.

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1, 1, -2, 0, -1; 0,
1, 0, +2, -4, 2--

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this one's already done-- 1,
3, -3, 2; 2 minus 2, 3 minus 2,

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0 plus 4, -2, and 2.

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One more step.

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1, 1, -2, 0, -1-- all these
are done-- 0, 1, 2, -4, 2; 0,

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1 minus 1 is 0, 3 minus 2 is 1,
-3 plus 4 is 1, 2 minus 2 is 0,

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1 minus 1 is 0,
2, 2 again, and 0.

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Well, you can see
where this is going.

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In the next step, this
row is going to disappear.

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1, 1, -2, 0 -1; 0, 1, 2, -4, 2;
0, 0, 1, 1, 0; 0, 0, 0, 0, 0.

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All right.

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We're done with elimination.

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So let's circle our pivots.

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All right, here are our pivots.

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We have three pivots.

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And so these three rows
are linearly independent.

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And in fact, these rows
still span the same space

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that the initial four rows did.

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Because when you do elimination,
all that you're doing

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is recombining
your rows by doing

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linear combinations of them.

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So, for example, your first
row stayed the same throughout.

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Your second row was replaced
by row 2 minus row 1.

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But it's really still
spanning the same space.

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And it goes on.

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And then the fourth row,
it turns out, was useless.

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You only needed the first three.

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So the elements for a basis--
well, it will be these three.

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So let me write that.

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Basis [1, 1, -2, 0, -1],
2, 2, and [0, 0, 1, 1, 0].

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Could you have used
the first three rows?

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Yes, you could.

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You can't always do that.

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Sometimes in elimination,
you have to switch rows,

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because there's a 0
where a pivot should be.

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When that happens, you
have to use these three,

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or you have to keep
track of which row

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you switched to go back
and use the initial ones.

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But it's really safe
to use these ones.

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And also they're simpler than
the ones that you started with,

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so why not?

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The other question
that we had was:

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what is the dimension
of the vector space?

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Well, this is the easy part.

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The dimension of the
vector space is 1, 2, 3.

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And that solves the problem.

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But there's one more thing
that I want to tell you.

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I told you that there's
a couple of ways

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to find out which of
those four vectors

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are linearly independent.

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And the one that I used
was I put them into rows

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and performed elimination
and picked out the rows

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that have pivots on them.

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Another way to do it is to
write the initial vectors

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as columns of the matrix and
then perform elimination.

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That also works
and, as you know,

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because you're only working with
the transpose of this matrix,

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you get the same
number of pivots.

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Let's go over here where I have
the same-- well, the transpose

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of that matrix-- I
have the same vectors,

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but now written as columns.

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My four initial vectors
written as columns.

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And here I have
performed elimination.

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And this is the final result.
Let me circle the pivots again.

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Here they are,
three, which is going

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to give me three linearly
independent vectors

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and hence, three elements
of the basis, and hence,

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dimension 3 for
the vector space.

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But I could no longer
use these three columns

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as elements of the basis,
because when I did elimination,

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I changed the column space.

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So the column space
of this matrix

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is not the same as
the column space

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of the eliminated version.

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So I cannot use these.

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In fact, if you
look at them, you

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can probably understand
that they're not

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going to span the
same space as these.

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Because all that I
have down here are 0's,

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and I get a lot more than
just 0's in the last two

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entries of the vectors.

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So what I need to
do is-- the pivots

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are in the first, second,
and third columns.

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I need to use these three
columns as my basis.

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I'll just write basis down here.

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And that will work too.

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So see, I have produced
two different bases

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for the same vector space,
which is totally fine.

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You can pick the
basis that you prefer.

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All right.

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We're done.

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Thank you.