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Current time:0:00Total duration:8:33

In the last video we saw a
method of figuring out what the basis for column space is. And we use these
exact examples. Whereas this was matrix A, I
just took it and I put it in reduced row echelon form. And I figured out which of these
columns in my reduced row echelon form of A,
are pivot columns. And it turned out to be the
first one, the second one, and the fourth one. And then the method is, you say
look, the corresponding columns in A-- so the first one,
the second one, and the fourth one-- form my basis
for my column space. And since they form the basis,
and if you want to know the dimension of your basis of your
column space, which is also called the rank,
you just say, well there's three in there. So it has a rank of
one, two, three. In this video I want to discuss
a little bit about why this worked. Why were we able to just take
the corresponding columns? Why did linear independence of
these three guys, imply linear independence of these
three guys? Why was the fact that I can
represent these guys-- this guy right here as a linear
combination of these three, or this guy as a linear combination
of these three-- why does that imply that I can
construct this guy as a linear combination of my
basis vectors? So the first thing that wasn't
too much of a stretch of the imagination in the last video,
was the idea that these pivot vectors are linearly
independent. So r1, r2, and r4. And everything I'm doing, I'm
kind of applying to the special case just so that it's
easier to understand. But it should be
generalizable. In fact, it definitely
is generalizable. That all of the pivot columns
in reduced row echelon form are linearly independent. And that's because the very
nature of reduced row echelon form, is that you are the only
pivot column that has a 1 in that respective row. So the only way to construct
it is with that vector. You can't construct it with
the other pivot columns because they're all going
to have 0 in that row. And when I say it's linearly
independent, I'm just saying the set of pivot columns. So let me say this in general. The set of pivot columns for any
reduced row echelon form matrix is linearly
independent. And it's just a very
straightforward argument. Because each column is
going to have a 1 in a very unique place. All of the other pivot columns
are going to have a 0 in that same place. And so you can't take any linear
combinations to get to that 1 because 0 times anything,
minus or plus 0 times anything, can never
be equal to 1. So I think you can
accept that. Now, that means that the
solution to c1 times r1, plus c2 times r2, plus, let
me say, c4 times r4. The solution to this equation,
because these guys are linearly independent, we know
that this only has one solution, and that's c1, c2,
and c4 is equal to 0. That's the only solution
to that. So another way we could say it
is, if we write r times some vector x-- well I'll just write
it times this particular x-- where I write it as
c1, c2, 0, c4, and 0 is equal to 0. So this will be some special
member of your null space. It's a particular solution
to the equation. This is equal to one, two,
three, four 0's because we have four rows here. Now, if we just expand
this out. If we just multiply 1 times c1,
plus 0 times c2, minus 1 times 0, plus 4 times 0, you'll
get-- or actually a better way to explain it-- this
multiplication right here can be written as-- and we've
seen this multiple times-- c1 times r1, plus c2 times
r2, plus 0 times r3. So we could just ignore that
term, plus c4, c4 times r4, plus 0 times r5. That's r5 right there. All of that equal to 0. So the only solution to this,
because we know that these three columns are linearly
independent-- or the set of just those three columns, those
three pivot columns are linearly independent-- the only
solution here is all of these equal to 0. That's exactly what I
said right up here. So the only solution here, where
if these two are 0, is that these guys also all have
to equal 0, if I already constrain these two. Now, the one thing that we've
done over and over again, we know that the solution set of
this equation, the solution set of Rx is equal to 0, is the
same as the solution set of Ax is equal to 0. Now, how do we know that? Or what do I mean? Well the solution set of this
is just the null space. The solution set is just
the null space of r. It's all of x's that satisfy
this equation. And we know that is equal to the
null space of a, because r is just a in reduced
row echelon form. So this is the null space of
a, is all of the x's that satisfy this equation. Now, the only version of this
that satisfied this equation was when c1, c2, and
c4 are equal to 0. So that tells us that the only
version of this, c1, c2, 0, c4, 0, that satisfies this
equation, or this equation, is when c1, c2, and c4
is equal to 0. Or another way of saying that
it if this is vector a1, a2, a4 right here, if you multiply
this out, you get c1-- let me do it over here, let me do it in
blue-- you get c1 times a1 plus c2 times a2, and then 0
times a3, plus c4 times a4 is equal to 0. Now these guys are going to be
linearly independent, if and only if the only solution to
this equation is they all equal to 0. Well we know that the only
solution to this is that they all equal 0 because anything
that's a solution to this is a solution to this. And the only solution to this
was, if I go ahead and I constrain these two terms to
being equal to 0, the only solution to this is all of
these c's have to be 0. So likewise, if I constrain
these to be 0, the only solution to this is that c1,
c2, and c4 have to be 0. So those guys have to be 0,
which imply that these three vectors, a1, a2, and a4, so that
implies that the set a1, a2, and a4 are linearly
independent. So we're halfway there. We've shown that because the
pivot columns here are linearly independent. We can show and they have
the same solution set. The null space of the reduced
row echelon form is the same as the null space of our
original matrix. We were able to show that the
only solution to c1 times this plus c2 times this plus c4 times
this is when all the constants are 0, which shows
that these three vectors or a set of those three vectors
are definitely linearly independent. Now, the next thing to prove
that they are a basis, is to show that all of the other
column vectors can be represented as multiples
of these three guys. And I realize, just for the sake
of clarity, or maybe not boring you too much, I'll do
that in the next video. So in this one we saw that if
the pivot columns are linearly independent, they always are. All pivot columns, by definition
are linearly independent. Or the set of pivot columns
are always linearly independent when you take away
the non-pivot columns, then the corresponding columns in
your original vector are also linearly independent. In the next one we'll show that
these three guys also span your column space.