A couple of videos ago, I made the statement that the rank of
a matrix A is equal to the rank of its transpose.
And I made a bit of a hand wavy argument.
It was at the end of the video, and I was tired.
It was actually the end of the day.
And I thought it'd be worthwhile to maybe flush this
out a little bit.
Because it's an important take away.
It'll help us understand everything we've learned a
little bit better.
So, let's understand-- I'm actually going to start with
the rank of A transpose.
The rank of A transpose is equal to the dimension of the
column space of A transpose.
That's the definition of the rank.
The dimension of the column space of A transpose is the
number of basis vectors for the
column space of A transpose.
That's what dimension is.
For any subspace, you figure out how many basis vectors you
need in that subspace, and you count them,
and that's your dimension.
So, it's the number of basis vectors for the column space
of A transpose, which is, of course, the same thing.
This thing we've seen multiple times, is the same thing as
the row space of A.
Right?
The columns of A transpose are the same thing
as the rows of A.
It's because You switch the rows and the columns.
Now, how can we figure out the number of basis vectors we
need for the column space of A transpose, or the
row space of A?
Let's just think about what the column space of A
transpose is telling us.
So, it's equivalent to-- so let's say, let
me draw A like this.
That's a matrix A.
Let's say it's an m by n matrix.
Let me just write it as a bunch of row vectors.
I could also write it as a bunch of column vectors, but
right now let's stick to the row vectors.
So we have row one.
The transpose of column vectors.
That's row one, and we're going to have row two, and
we're going to go all the way down to row m.
Right?
It's an m by n matrix.
Each of these vectors are members of rn, because they're
going to have n entries in them
because we have n columns.
So, that's what A is going to look like.
A is going look like that.
And then A transpose, all of these rows are
going to become columns.
A transpose is going to look like this. r1, r2,
all the way to rm.
And this is of course going to be an n by m matrix.
You swap these out.
So, all these rows are going to be columns.
Right?
And, obviously the column space-- or maybe not so
obviously-- the column space of A transpose is equal to the
span of r1, r2, all the way to rm.
Right?
It's equal to the span of these things.
Or you could equivocally call it, it's equal to the span of
the rows of A.
That's why it's also called the row space.
This is equal to the span of the rows of A.
These two things are equivalent.
Now, these are the span.
That means this is some subspace that's all of the
linear combinations of these columns, or all of the linear
combinations of these rows.
If we want the basis for it, we want to find a minimum set
of linearly independent vectors that we could use to
construct any of these columns.
Or that we could use construct any of these rows, right here.
Now, what happens when we put A into
reduced row echelon form?
We do a bunch of row operations to put it into
reduced row echelon form.
Right?
Do a bunch of row operations and you eventually get
something like this.
You'll get the reduced row echelon form of A.
The reduced row echelon form of A is going to look
something like this.
You're going to have some pivot rows, some rows that
have pivot entries.
Let's say that's one of them.
This will all have 0's all the way down.
This one will have 0's.
Your pivot entry has to be the only non-zero
entry in it's column.
And everything to the left of it all has to be 0.
Let's say that this one isn't.
These are some non-zero values.
These are 0.
We have another pivot entry over here.
Everything else is 0.
Let's say everything else are non-pivot entries.
So you come here and you have a certain number of pivot
rows, or a certain number of pivot entries, right?
And you got there by performing linear row
operations on these guys.
So those linear row operations-- you know, I take
3 times row two, and I add it to row one, and that's going
to become my new row two.
And you keep doing that and you get these things here.
So, these things here are linear
combinations of those guys.
Or another way to do it, you can reverse those row
operations.
I could start with these guys right here.
And I could just as easily perform the reverse row
operations.
Any linear operation, you can perform the reverse of it.
We've seen that multiple times.
You could perform row operations with these guys to
get all of these guys.
Or another way to view it is, these vectors here, these row
vectors right here, they span all of these-- or all of these
row vectors can be represented of linear combinations of your
pivot rows right here.
Obviously, your non-pivot rows are going to be all 0's.
And those are useless.
But, your pivot rows, if you take linear combinations of
them, you can clearly do reverse row echelon form and
get back to your matrix.
So, all of these guys can be represented as linear
combinations of them.
And all of these pivot entries are by definition-- well,
almost by definition-- they are linearly
independent, right?
Because I've got a 1 here.
No one else has a 1 there.
So this guy can definitely not be represented as a linear
combination of the other guy.
So why am I going through this whole exercise?
Well, we started off saying we wanted a
basis for the row space.
We wanted some minimum set of linearly independent vectors
that spans everything that these guys can span.
Well, if all of these guys can be represented as linear
combinations of these row vectors in reduced row echelon
form-- or these pivot rows in reduced row echelon form-- and
these guys are all linearly independent, then they are a
reasonable basis.
So these pivot rows right here, that's one of them, this
is the second one, this is the third one, maybe they're the
only three.
This is just my particular example.
That would be a suitable basis for the row space.
So let me write this down.
The pivot rows in reduced row echelon form of A are a basis
for the row space of A.
And the row space of A is the same thing, or the column
space of A transpose.
The row space of A is the same thing as the
column space of A transpose.
We've see that multiple times.
Now, if we want to know the dimension of your column
space, we just count the number of pivot rows you have.
So you just count the number of pivot rows.
So the dimension of your row space, which is the same thing
as the column space of A transpose, is going to be the
number of pivot rows you have in reduced row echelon form.
Or, even simpler, the number of pivot entries you have
because every pivot entry has a pivot row.
So we can write that the rank of A transpose is equal to the
number of pivot entries in reduced row echelon form of A.
Right?
Because every pivot entry corresponds to a pivot row.
Those pivot rows are a suitable basis for the entire
row space, because every row could be made with a linear
combination of these guys.
And since all these can be, then anything that these guys
can construct, these guys can construct.
Fair enough.
Now, what is the rank of A?
This is the rank of A transpose that we've been
dealing with so far.
The rank of A is equal to the dimension of the
column space of A.
Or, you could say it's the number of vectors in the basis
for the column space of A.
So if we take that same matrix A that we used above, and we
instead we write it as a bunch of column vectors, so c1, c2,
all the way to cn.
We have n columns right there.
The column space is essentially the subspace
that's spanned by all of these characters right here, right?
Spanned by each of these column vectors.
So the column space of A is equal to the span of c1, c2,
all the way to cn.
That's the definition of it.
But we want to know the number of basis vectors.
And we've seen before-- we've done this multiple times--
what suitable basis vectors could be.
If you put this into reduced row echelon form, and you have
some pivot entries and their corresponding pivot columns,
so some pivot entries with their corresponding pivot
columns just like that.
Maybe that's like that, and then maybe this one isn't one,
and then this one is.
So you have a certain number of pivot columns.
Let me do it with another color right here.
When you put A into reduced row echelon form, we learned
that the basis vectors, or the basis columns that form a
basis for your column space, are the columns that
correspond to the pivot columns.
So the first column here is a pivot column, so this guy
could be a basis vector.
The second column is, so this guy could be a pivot vector.
Or maybe the fourth one right here, so this guy could be a
pivot vector.
So, in general, you just say hey, if you want to count the
number basis vectors-- because we don't even have to know
what they are to figure out the rank.
We just have to know the number they are.
Well you say, well for every pivot column here, we have a
basis vector over there.
So we could just count the number pivot columns.
But the number of pivot columns is equivalent to just
the number of pivot entries we have. Because every pivot
entry gets its own column.
So we could say that the rank of A is equal to the number of
pivot entries in the reduced row echelon form of A.
And, as you can see very clearly, that's the exact same
thing that we deduced was equivalent to the rank of A
transpose-- the dimension of the
columns space of A transpose.
Or the dimension of the row space of A.
So we can now write our conclusion.
The rank of A is definitely the same thing as the rank of
A transpose.