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A couple of videos ago, I made
the statement that the rank of

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a matrix A is equal to the
rank of its transpose.

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And I made a bit of a
hand wavy argument.

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It was at the end of the
video, and I was tired.

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It was actually the
end of the day.

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And I thought it'd be worthwhile
to maybe flush this

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out a little bit.

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Because it's an important
take away.

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It'll help us understand
everything we've learned a

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little bit better.

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So, let's understand-- I'm
actually going to start with

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the rank of A transpose.

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The rank of A transpose is equal
to the dimension of the

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column space of A transpose.

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That's the definition
of the rank.

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The dimension of the column
space of A transpose is the

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number of basis vectors for the

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column space of A transpose.

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That's what dimension is.

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For any subspace, you figure out
how many basis vectors you

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need in that subspace,
and you count them,

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and that's your dimension.

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So, it's the number of basis
vectors for the column space

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of A transpose, which is, of
course, the same thing.

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This thing we've seen multiple
times, is the same thing as

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the row space of A.

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Right?

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The columns of A transpose
are the same thing

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as the rows of A.

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It's because You switch the
rows and the columns.

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Now, how can we figure out the
number of basis vectors we

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need for the column space
of A transpose, or the

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row space of A?

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Let's just think about what
the column space of A

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transpose is telling us.

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So, it's equivalent to--
so let's say, let

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me draw A like this.

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That's a matrix A.

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Let's say it's an
m by n matrix.

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Let me just write it as a
bunch of row vectors.

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I could also write it as a bunch
of column vectors, but

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right now let's stick
to the row vectors.

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So we have row one.

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The transpose of
column vectors.

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That's row one, and we're going
to have row two, and

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we're going to go all the
way down to row m.

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Right?

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It's an m by n matrix.

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Each of these vectors are
members of rn, because they're

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going to have n entries in them

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because we have n columns.

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So, that's what A is
going to look like.

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A is going look like that.

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And then A transpose, all
of these rows are

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going to become columns.

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A transpose is going to look
like this. r1, r2,

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all the way to rm.

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And this is of course going
to be an n by m matrix.

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You swap these out.

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So, all these rows are
going to be columns.

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Right?

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And, obviously the column
space-- or maybe not so

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obviously-- the column space of
A transpose is equal to the

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span of r1, r2, all
the way to rm.

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Right?

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It's equal to the span
of these things.

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Or you could equivocally call
it, it's equal to the span of

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the rows of A.

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That's why it's also called
the row space.

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This is equal to the span
of the rows of A.

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These two things
are equivalent.

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Now, these are the span.

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That means this is some subspace
that's all of the

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linear combinations of these
columns, or all of the linear

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combinations of these rows.

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If we want the basis for it, we
want to find a minimum set

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of linearly independent vectors
that we could use to

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construct any of
these columns.

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Or that we could use construct
any of these rows, right here.

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Now, what happens when
we put A into

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reduced row echelon form?

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We do a bunch of row operations
to put it into

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reduced row echelon form.

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Right?

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Do a bunch of row operations
and you eventually get

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something like this.

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You'll get the reduced row
echelon form of A.

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The reduced row echelon form
of A is going to look

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something like this.

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You're going to have some pivot
rows, some rows that

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have pivot entries.

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Let's say that's one of them.

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Let's say that's one of them.

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This will all have 0's
all the way down.

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This one will have 0's.

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Your pivot entry has to
be the only non-zero

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entry in it's column.

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And everything to the left
of it all has to be 0.

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Let's say that this one isn't.

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These are some non-zero
values.

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These are 0.

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We have another pivot
entry over here.

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Everything else is 0.

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Let's say everything else
are non-pivot entries.

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So you come here and you have
a certain number of pivot

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rows, or a certain number
of pivot entries, right?

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And you got there by performing
linear row

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operations on these guys.

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So those linear row operations--
you know, I take

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3 times row two, and I add it
to row one, and that's going

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to become my new row two.

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And you keep doing that and
you get these things here.

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So, these things
here are linear

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combinations of those guys.

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Or another way to do it, you
can reverse those row

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operations.

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I could start with these
guys right here.

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And I could just as easily
perform the reverse row

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operations.

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Any linear operation, you can
perform the reverse of it.

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We've seen that multiple
times.

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You could perform row operations
with these guys to

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get all of these guys.

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Or another way to view it is,
these vectors here, these row

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vectors right here, they span
all of these-- or all of these

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row vectors can be represented
of linear combinations of your

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pivot rows right here.

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Obviously, your non-pivot rows
are going to be all 0's.

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And those are useless.

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But, your pivot rows, if you
take linear combinations of

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them, you can clearly do reverse
row echelon form and

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get back to your matrix.

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So, all of these guys can
be represented as linear

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combinations of them.

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And all of these pivot entries
are by definition-- well,

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almost by definition--
they are linearly

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independent, right?

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Because I've got a 1 here.

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No one else has a 1 there.

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So this guy can definitely not
be represented as a linear

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combination of the other guy.

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So why am I going through
this whole exercise?

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Well, we started off
saying we wanted a

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basis for the row space.

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We wanted some minimum set of
linearly independent vectors

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that spans everything that
these guys can span.

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Well, if all of these guys can
be represented as linear

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combinations of these row
vectors in reduced row echelon

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form-- or these pivot rows in
reduced row echelon form-- and

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these guys are all linearly
independent, then they are a

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reasonable basis.

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So these pivot rows right here,
that's one of them, this

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is the second one, this is the
third one, maybe they're the

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only three.

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This is just my particular
example.

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That would be a suitable basis
for the row space.

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So let me write this down.

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The pivot rows in reduced row
echelon form of A are a basis

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for the row space of A.

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And the row space of A is the
same thing, or the column

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space of A transpose.

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The row space of A is the
same thing as the

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column space of A transpose.

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We've see that multiple times.

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Now, if we want to know the
dimension of your column

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space, we just count the number
of pivot rows you have.

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So you just count the number
of pivot rows.

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So the dimension of your row
space, which is the same thing

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as the column space of A
transpose, is going to be the

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number of pivot rows you have
in reduced row echelon form.

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Or, even simpler, the number
of pivot entries you have

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because every pivot entry
has a pivot row.

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So we can write that the rank of
A transpose is equal to the

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number of pivot entries in
reduced row echelon form of A.

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Right?

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Because every pivot entry
corresponds to a pivot row.

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Those pivot rows are a suitable
basis for the entire

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row space, because every row
could be made with a linear

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combination of these guys.

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And since all these can be, then
anything that these guys

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can construct, these
guys can construct.

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Fair enough.

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Now, what is the rank of A?

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This is the rank of A transpose
that we've been

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dealing with so far.

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The rank of A is equal to
the dimension of the

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column space of A.

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Or, you could say it's the
number of vectors in the basis

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for the column space of A.

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So if we take that same matrix
A that we used above, and we

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instead we write it as a bunch
of column vectors, so c1, c2,

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all the way to cn.

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We have n columns right there.

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The column space is essentially
the subspace

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that's spanned by all of these
characters right here, right?

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Spanned by each of these
column vectors.

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So the column space of A is
equal to the span of c1, c2,

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all the way to cn.

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That's the definition of it.

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But we want to know the number
of basis vectors.

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And we've seen before-- we've
done this multiple times--

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what suitable basis
vectors could be.

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If you put this into reduced row
echelon form, and you have

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some pivot entries and their
corresponding pivot columns,

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so some pivot entries with
their corresponding pivot

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columns just like that.

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Maybe that's like that, and then
maybe this one isn't one,

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and then this one is.

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So you have a certain number
of pivot columns.

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Let me do it with another
color right here.

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When you put A into reduced row
echelon form, we learned

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that the basis vectors, or the
basis columns that form a

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basis for your column space,
are the columns that

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correspond to the
pivot columns.

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So the first column here is a
pivot column, so this guy

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could be a basis vector.

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The second column is, so this
guy could be a pivot vector.

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Or maybe the fourth one right
here, so this guy could be a

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pivot vector.

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So, in general, you just say
hey, if you want to count the

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number basis vectors-- because
we don't even have to know

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what they are to figure
out the rank.

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We just have to know the
number they are.

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Well you say, well for every
pivot column here, we have a

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basis vector over there.

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So we could just count the
number pivot columns.

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But the number of pivot columns
is equivalent to just

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the number of pivot entries we
have. Because every pivot

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entry gets its own column.

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So we could say that the rank of
A is equal to the number of

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pivot entries in the reduced
row echelon form of A.

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And, as you can see very
clearly, that's the exact same

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thing that we deduced was
equivalent to the rank of A

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transpose-- the dimension
of the

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columns space of A transpose.

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Or the dimension of the
row space of A.

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So we can now write
our conclusion.

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The rank of A is definitely the
same thing as the rank of

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A transpose.

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