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OK, here's linear algebra
lecture seven.
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I've been talking about vector
spaces and specially the null
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space of a matrix and the column
space of a matrix.
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What's in those spaces.
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Now I want to actually describe
them.
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How do you describe all the
vectors that are in those
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spaces?
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How do you compute these
things?
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So this is the,
turning the idea,
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the definition,
into an algorithm.
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What's the algorithm for
solving A x =0?
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So that's the null space that
I'm interested in.
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So can I take a particular
matrix A and describe the
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natural algorithm,
and I'll execute it for that
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matrix -- here we go.
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So let me take the matrix as an
example.
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So we're definitely talking
rectangular matrices in this
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chapter.
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So I'll make,
I'll have four columns.
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And three rows.
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Two four six eight and three
six eight ten.
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OK.
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If I just look at those
columns, and rows,
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well, I notice right away that
column two is a multiple of
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column one.
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It's in the same direction as
column one.
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It's not independent.
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I'll expect to discover that in
the process.
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Actually, with rows,
I notice that that row plus
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this row gives the third row.
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So the third row is not
independent.
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So, all that should come out of
elimination.
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So now what I --
my algorithm is elimination,
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but extended now to the
rectangular case,
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where we have to continue even
if there's zeros in the pivot
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position, we go on.
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OK, so let me execute
elimination for that matrix.
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My goal is always,
while I'm doing elimination --
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I'm not changing the null
space.
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That's very important,
right?
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I'm solving A x equals zero by
elimination, and when I do these
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operations that you already
know, when I subtract a multiple
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of one equation from another
equation, I'm not changing the
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solutions.
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So I'm not changing the null
space.
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Actually, I changing the column
space, as you'll see.
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So you have to pay attention.
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What does elimination leave
unchanged?
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And the answer is the solutions
to the system are not changed
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because I'm doing the same thing
to -- I'm doing a legitimate
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operations on the equations.
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Of course, on the right hand
side it's always zero,
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and I don't plan to write zero
all the time.
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OK, so I'm really just working
on the left side,
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but the right side is,
is keeping up always zeros.
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OK, so what's elimination?
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Well, you know where the first
pivot is and you know what to do
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with it.
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So can I just take the first
step below here?
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So that pivot row is fine.
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I take two times that row away
from this one and I get zero
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zero.
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That's signaling a difficulty.
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Two, two twos away from the six
leaves me with a two.
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Two twos away from the eight
leaves me with a four.
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And now three of those away
from here is zero,
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again another zero,
three twos away from that eight
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is the two, three twos away from
that ten is a four.
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OK.
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That's the first stage of
elimination.
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I've got the first column
straight.
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So of course I move on to the
second column.
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I look in that position,
I see a zero.
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I look below it,
hoping for a non-zero that I
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can do a row exchange.
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But it's zero below.
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So that's telling me that that
column is -- well,
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what it's really going to be
telling me is that that column
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is a combination of the earlier
columns.
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It's that second column is
dependent on the earlier
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columns.
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But I don't stop to think here.
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In that column there's nothing
to do.
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I go on to the next.
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So here's the next pivot.
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So there's the first pivot and
there's the second pivot,
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and I just keep this
elimination going downwards.
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So, so the next step keeps the
first row, keeps the second row
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with its pivot,
so I've got my two pivots,
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and does elimination to clear
out the column below that pivot.
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So actually you see the
multiplier is one.
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It subtracts row two from row
three and produces a row of
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zeros.
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OK.
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That I would call that matrix
U, right?
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That's our upper -- well,
I can't quite say upper
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triangular.
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Maybe upper -- I don't know --
upper something.
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It's in this so-called echelon
form.
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The word echelon means,
like, staircase form.
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It's the, the non-zeros come in
that staircase form.
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If there was another pivot
here, then the staircase would
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include that.
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But here's a case where we have
two pivots only.
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OK, so actually we've already
discovered the most important
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number about this matrix.
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The number of pivots is two.
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That number we will call the
rank of the matrix.
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So let me put immediately.
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The rank of A -- so I'm telling
you what this word rank means in
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the algorithm.
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It's equal to the number of
pivots.
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And in this case,
two.
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OK, for me that number two is
the crucial number.
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OK, now I go to -- you remember
I'm always solving A x equals
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zero, but now I can solve U x
equals zero, right?
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Same solution,
same null space.
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OK.
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So I could stop here -- why
don't I stop here and do the
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back substitution.
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So now I have to ask you,
how do I describe the
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solutions?
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There are solutions,
right, to A x equals zero.
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I knew there would be.
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I had three equations in four
unknowns.
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I certainly expected some
solutions.
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Now I want to see what are
they.
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OK, here's the critical step.
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I refer to it up here as
separating out the pivot
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variables, the pivot columns,
which are these two.
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Here I have two pivot columns.
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Those, obviously,
they're the columns with the
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pivots.
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So I have two pivot columns.
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And I have the other columns,
I'll call free.
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These are free columns,
OK.
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Why do I use those words?
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Why do I use that word free?
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Because now I want to write,
I want to find the solutions to
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U x equals zero.
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Here is the way I do it.
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These free columns I can assign
any number freely to those --
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to the variables x2 and x4,
the ones that multiply columns
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two and four.
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So I can assign anything I like
to x2 and x4.
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And then I can solve the
equations for x1 and x3.
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Let me say that again.
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If I give -- let me,
let me assign.
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So, so one particular x is to
assign, say, the value one to
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the, to x2, and zero to x4.
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Those are -- that was a free
choice, but it's a convenient
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choice.
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OK.
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Now I want to solve U x equals
zero and find numbers one and
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three, complete the solution.
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OK.
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Can I write down -- let's see.
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Shall we just remember what U x
equals zero represents?
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What are my equations?
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That first equation is x1 plus
just -- I'm just saying what are
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these matrices meaning.
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That's the first equation.
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And the second equation was 2x3
+ 4x4=0.
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Those are my two equations.
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OK.
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Now the point is I can find x1
and x3 by back substitution.
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So we're building on what we
already know.
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The new thing is that there
were some free variables that I
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could give any numbers to.
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And I'm systematically going to
make a choice like this,
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1 and 0.
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Now what is x3?
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Let's, let's go backwards,
back up.
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I look at the last equation.
x3 is zero, from the last
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equation, because,
because x4 we've set x4 to
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zero, and then we get x3 as
zero.
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OK.
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Now we set x2 to be one,
so what is x1?
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Negative two,
right?
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So then I have negative two
plus two, zero and zero,
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correctly giving zero.
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There is a vector in the null
space.
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There is a solution to A x
equals zero.
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In fact, what solution is that?
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That simply says that minus two
times the first column plus one
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times the second column is the
zero column.
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Of course that's right.
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Minus two of that column plus
one of that, or minus two of
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that plus one of that.
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That solution is --
that, that's just what we saw
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immediately, that the second
column is twice as big as the
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first column.
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OK, tell me some more vectors
in the null space.
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I found one.
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Tell me, how to get a bunch
more immediately out of that
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one.
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Just take multiples of it.
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Any multiple of -- I could
multiply this by anything.
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I might as well call it,
I could say,
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C, some multiple of this.
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So let me -- so X could be any
multiple of this one.
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OK, that, that describes now a
line, an infinitely long line in
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four dimensional space.
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But -- which is in the null
space.
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Is that the whole null space?
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No.
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I've got two free variables
here.
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I made this choice,
one and zero,
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for the free variables,
but I could have made another
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choice.
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Let me make the other choice
zero and one.
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You see my system.
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So let me repeat the system.
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This is the algorithm that you,
you just learned to do.
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Do elimination.
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Decide which are the pivot
columns and which are the free
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columns.
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That tells you that,
that variables one and three
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are pivot variables,
two and four are free
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variables.
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Then those free variables,
you assign them -- you give one
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of them the value one and the
others the value zero -- in this
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case, we had a one and a zero --
and complete the solution.
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And you do --
you give the other one the
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value one and zero.
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And now complete the solution.
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So let's complete that
solution.
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I'm looking for a vector in the
null space, and it's absolutely
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going to be different from this
guy, because,
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because any multiple of that
zero is never going to give the
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one.
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So I have somebody new in the
null space, and let me finish it
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out.
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What's x3 here?
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So we're going by back
substitution,
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looking at this equation.
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Now x4 we've changed,
we're doing the other
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possibility, where x2 is zero
and x4 is one.
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So x3 will happen to be minus
two.
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And now what do I get for that
first equation?
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x1 -- let's see.
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Two x3s is minus four plus two
-- do I get a two there?
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Perhaps, yeah.
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Two for x1, minus four,
and two.
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OK.
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That's in the null space.
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What does that say about the
columns?
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That says that two of this
column minus two of this column
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plus this column gives zero,
which it does.
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Two of that minus two of that
and one of that gives the zero
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column.
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OK, now, now I've found another
vector in the null space.
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Now we're ready to tell me the
whole null space.
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What are all the solutions to
Ax=0?
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I've got this guy and when I
have him, what else is,
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goes into the null space along
with that?
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These are my two special
solutions.
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I call them special -- I just
invented that name.
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Special solutions.
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What's special about them is
the special numbers I gave to
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the free variables,
the values zero and one for the
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free variables.
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I could have given the free
variables any values and got
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vectors in the null space.
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But this was a good way to be
sure I got t- got everybody.
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OK, so once I have him,
I also have any multiple,
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right?
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So I could take any multiple of
that and it's in the null space.
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And now what else -- I left a
little space for what?
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What -- a plus sign.
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I can take any combination.
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Here is a line of vectors in
the null space.
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A bunch of solutions.
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Would you rather I say in the
null space or would you rather I
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say, OK, I'm solving Ax=0?
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Well, really I'm solving Ux=0.
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Well, OK, let me put in that
crucial plus sign.
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I'm taking all the combinations
of my two special solutions.
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That's my conclusion there.
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The null space contains,
contains exactly all the
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combinations of the special
solutions.
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And how many special solutions
are there?
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There's one for every free
variable.
284
00:16:44 --> 00:16:47.54
And how many free variables are
there?
285
00:16:47.54 --> 00:16:52
Oh, I mean, we can see all the
whole picture now.
286
00:16:52 --> 00:16:55
If the rank R was two,
this is the,
287
00:16:55 --> 00:17:00
this is the number of pivot
variables, right,
288
00:17:00 --> 00:17:03
because it counted the pivots.
289
00:17:03 --> 00:17:06
So how many free variables?
290
00:17:06 --> 00:17:10
Well, you know it's two,
right?
291
00:17:10 --> 00:17:14
What is it in -- for a matrix
that's m rows,
292
00:17:14 --> 00:17:21
n columns, n variables that
means, with rank r?
293
00:17:21 --> 00:17:27
How many free variables have we
got left?
294
00:17:27 --> 00:17:36
If r of the variables are pivot
variables, we have n-r -- in
295
00:17:36 --> 00:17:42
this case four minus two -- free
variables.
296
00:17:42 --> 00:17:52
Do you see that first of all we
get clean answers here?
297
00:17:52 --> 00:17:56
We get r pivot variables -- so
there really were r equations
298
00:17:56 --> 00:17:57
here.
299
00:17:57 --> 00:18:01.09
There looked like three
equations, but there were really
300
00:18:01.09 --> 00:18:03
only two independent equations.
301
00:18:03 --> 00:18:07
And there were n-r variables
that we could choose freely,
302
00:18:07 --> 00:18:10
and we gave them those special
zero one values,
303
00:18:10 --> 00:18:13
and we got the special
solutions.
304
00:18:13 --> 00:18:14
OK.
305
00:18:14 --> 00:18:20
For me -- we could stop at that
point.
306
00:18:20 --> 00:18:28
That gives you a complete
algorithm for finding all the
307
00:18:28 --> 00:18:33
solutions to A x equals zero.
308
00:18:33 --> 00:18:33
OK.
309
00:18:33 --> 00:18:41
Again, you do elimination --
going onward when a column,
310
00:18:41 --> 00:18:46
when there's nothing to be done
on one column,
311
00:18:46 --> 00:18:48.03
you just continue.
312
00:18:48.03 --> 00:18:52
There's this number r,
the number of pivots,
313
00:18:52 --> 00:18:56
is crucial, and it leaves n-r
free variables,
314
00:18:56 --> 00:19:00
which you give values zero and
one to.
315
00:19:00 --> 00:19:04
I would like to take one more
step.
316
00:19:04 --> 00:19:09
I would like to clean up this
matrix even more.
317
00:19:09 --> 00:19:14
So now I'm going to go to --
this is in its,
318
00:19:14 --> 00:19:20
this is in echelon form,
upper triangular if you like.
319
00:19:20 --> 00:19:26
I want to go one more step to
make it as good as it can be.
320
00:19:26 --> 00:19:32
OK, so now I'm going to speak
about the reduced row echelon
321
00:19:32 --> 00:19:34.3
form.
322
00:19:34.3 --> 00:19:41
OK, so now I'm going to speak
about the matrix R,
323
00:19:41 --> 00:19:46
which is the reduced row
echelon form.
324
00:19:46 --> 00:19:50
So what does that mean?
325
00:19:50 --> 00:19:57.83
That means I just -- I can,
I can work harder on U.
326
00:19:57.83 --> 00:20:04
So let me start,
let me suppose I got as far as
327
00:20:04 --> 00:20:08
U, which was good.
328
00:20:08 --> 00:20:12
Notice how that row of zeros
appeared.
329
00:20:12 --> 00:20:16
I didn't comment on that,
but I should have.
330
00:20:16 --> 00:20:21.08
That row of zeros up here is
because row three was a
331
00:20:21.08 --> 00:20:26
combination of rows one and two,
and elimination discovered that
332
00:20:26 --> 00:20:27
fact.
333
00:20:27 --> 00:20:33
When we get a row of zeros,
that's telling us that the --
334
00:20:33 --> 00:20:40
original row that was there was
a combination of other rows,
335
00:20:40 --> 00:20:44
and elimination knocked it out.
336
00:20:44 --> 00:20:46
OK, so we got this far.
337
00:20:46 --> 00:20:51.1
Now how can I clean that up
further?
338
00:20:51.1 --> 00:20:54.63
I can do, elimination upwards.
339
00:20:54.63 --> 00:20:58
I can get zero above the
pivots.
340
00:20:58 --> 00:21:04
So this reduced row echelon
form has zeros above and below
341
00:21:04 --> 00:21:06.28
the pivots.
342
00:21:06.28 --> 00:21:09
So let me do that.
343
00:21:09 --> 00:21:14
So now I'll subtract one of
this from the row above.
344
00:21:14 --> 00:21:19
That will leave a zero and a
minus two in there.
345
00:21:19 --> 00:21:21
And that's good.
346
00:21:21 --> 00:21:25
OK, and I can clean it up even
one more step.
347
00:21:25 --> 00:21:31
I can make the pivots -- the
pivots I'm going to make equal
348
00:21:31 --> 00:21:37
to one, because I can divide
equation two by the pivot.
349
00:21:37 --> 00:21:41
That won't change the
solutions.
350
00:21:41 --> 00:21:43
So let me do that.
351
00:21:43 --> 00:21:47
And then I really -- I'm ready
to stop.
352
00:21:47 --> 00:21:51
One, two, zero,
minus two, zero,
353
00:21:51 --> 00:21:52
zero, one, two.
354
00:21:52 --> 00:21:59
I divided the second equation
by two, because now I have a one
355
00:21:59 --> 00:22:04
in the pivot and zeros below.
356
00:22:04 --> 00:22:04
OK.
357
00:22:04 --> 00:22:07
This is my matrix R.
358
00:22:07 --> 00:22:13
I guess I'm hoping that you
could now execute the whole
359
00:22:13 --> 00:22:15
algorithm.
360
00:22:15 --> 00:22:22
Matlab will do it immediately
with the command -- reduced row
361
00:22:22 --> 00:22:24
echelon form of A.
362
00:22:24 --> 00:22:31
So if I input that original
matrix A and then I write,
363
00:22:31 --> 00:22:35
then I type that command,
press return,
364
00:22:35 --> 00:22:40
that matrix will appear.
365
00:22:40 --> 00:22:47
That's the reduced row echelon
form, and it's got all the
366
00:22:47 --> 00:22:51
information as clear as can be.
367
00:22:51 --> 00:22:55
What, what information has it
got?
368
00:22:55 --> 00:23:02
Well, of course it immediately
tells me the pivot rows,
369
00:23:02 --> 00:23:07
pivot rows, one and two,
pivot columns,
370
00:23:07 --> 00:23:10
one and three.
371
00:23:10 --> 00:23:15.21
And in fact it's got the
identity matrix in there,
372
00:23:15.21 --> 00:23:15
right?
373
00:23:15 --> 00:23:20
It's, it's got zeros above and
below the pivots,
374
00:23:20 --> 00:23:25
and the pivots are one,
so it's, so it's got a -- so
375
00:23:25 --> 00:23:31
notice the two by two identity
matrix that's sitting in the
376
00:23:31 --> 00:23:36
pivot rows and pivot columns.
it's I in the pivot rows and
377
00:23:36 --> 00:23:38
columns.
378
00:23:38 --> 00:23:42
It's got zero rows below.
379
00:23:42 --> 00:23:49
Those are always indicating
that original rows were,
380
00:23:49 --> 00:23:53
were combinations of other
rows.
381
00:23:53 --> 00:23:58
So we really only had two rows
there.
382
00:23:58 --> 00:24:04
And now it also -- so there's
the identity.
383
00:24:04 --> 00:24:10
Now it's also got its free
columns.
384
00:24:10 --> 00:24:15
And, they're cleaned up as much
as possible.
385
00:24:15 --> 00:24:20
Actually, actually it's now so
cleaned up that the special
386
00:24:20 --> 00:24:26
solutions, I can read off -- you
remember I went through the
387
00:24:26 --> 00:24:31
steps of computing this -- doing
back substitution?
388
00:24:31 --> 00:24:35
Let me, let me,
instead of that system,
389
00:24:35 --> 00:24:38.09
let me take this improved
system.
390
00:24:38.09 --> 00:24:41.71
So I'm going to use these
numbers, right.
391
00:24:41.71 --> 00:24:44
In these equations,
what did I do?
392
00:24:44 --> 00:24:49
I divided this equation by two
and, oh yeah and I had
393
00:24:49 --> 00:24:54.2
subtracted two of this,
which knocked out this guy and
394
00:24:54.2 --> 00:24:57
made that a minus sign.
395
00:24:57 --> 00:25:03
Is that what -- I've now
written Rx equals zero.
396
00:25:03 --> 00:25:08
Now I guess I'm hoping
everybody in this room
397
00:25:08 --> 00:25:15
understands the solutions to the
original A x equals zero,
398
00:25:15 --> 00:25:19
the midway, halfway,
U x equals zero,
399
00:25:19 --> 00:25:26
and the final R x equals zero
are all the same.
400
00:25:26 --> 00:25:32
Because going from one of those
to another one I didn't mess up.
401
00:25:32 --> 00:25:38
I just multiplied equations and
subtracted from other equations,
402
00:25:38 --> 00:25:40
which I'm allowed to do.
403
00:25:40 --> 00:25:40
OK.
404
00:25:40 --> 00:25:46
But my point is that now if I
do this free variables and back
405
00:25:46 --> 00:25:49
substitution,
it's just, the numbers are
406
00:25:49 --> 00:25:51
there.
407
00:25:51 --> 00:25:57
When I let x -- so in this guy,
I let x2 be one and x4 be zero.
408
00:25:57 --> 00:26:00
I, I guess, what I seeing here?
409
00:26:00 --> 00:26:05
Let me, let me sort of separate
this out here.
410
00:26:05 --> 00:26:09
I'm seeing in the pivot,
in the pivot columns,
411
00:26:09 --> 00:26:13
if I, if I put the pivot
columns here,
412
00:26:13 --> 00:26:15
I'm seeing those.
413
00:26:15 --> 00:26:21
And I'm -- in the free columns
I'm seeing -- what I seeing in
414
00:26:21 --> 00:26:23
the free columns?
415
00:26:23 --> 00:26:28
A two, zero in that first free
column, the x2 column,
416
00:26:28 --> 00:26:31
and a minus two,
two in the fourth column,
417
00:26:31 --> 00:26:33
the other free column.
418
00:26:33 --> 00:26:39
And the row of zeros below,
which of course have no -- that
419
00:26:39 --> 00:26:43
equation is zero equals zero.
420
00:26:43 --> 00:26:44
That's satisfied.
421
00:26:44 --> 00:26:46
Here's my point.
422
00:26:46 --> 00:26:48
That when I do back
substitution,
423
00:26:48 --> 00:26:52
these numbers are exactly what
shows up -- oh,
424
00:26:52 --> 00:26:55
their signs get switched.
425
00:26:55 --> 00:26:58
I was going to say those
numbers, two,
426
00:26:58 --> 00:27:02
minus two, zero,
two, can I just circle the --
427
00:27:02 --> 00:27:05
this is the free part of the
matrix.
428
00:27:05 --> 00:27:08
This is the identity part.
429
00:27:08 --> 00:27:12
This is the free part,
maybe I'll call it F.
430
00:27:12 --> 00:27:15
This, of course,
I call I, because it's the
431
00:27:15 --> 00:27:16
identity.
432
00:27:16 --> 00:27:19
The free part is a,
I mean, I'm just doing back
433
00:27:19 --> 00:27:20
substitution.
434
00:27:20 --> 00:27:24
And those free numbers will
show up, with a minus sign,
435
00:27:24 --> 00:27:29
because they pop onto the other
side of the equation --
436
00:27:29 --> 00:27:34
so I see minus two,
zero, and I see two,
437
00:27:34 --> 00:27:35
minus two.
438
00:27:35 --> 00:27:38
So that wasn't magic.
439
00:27:38 --> 00:27:40
It had to happen.
440
00:27:40 --> 00:27:44.87
Let me, show you clearly why it
happened.
441
00:27:44.87 --> 00:27:52
OK, so that's -- this is what
I'm interested in here.
442
00:27:52 --> 00:27:55
And now let me,
let me just,
443
00:27:55 --> 00:27:59
like, do it,
do it for -- let's suppose
444
00:27:59 --> 00:28:06.22
we've, we've got to -- let's
suppose we've got this system
445
00:28:06.22 --> 00:28:09
already in, in rref form.
446
00:28:09 --> 00:28:14
So my matrix R is -- what does
it look like?
447
00:28:14 --> 00:28:20
OK, and I'll -- let me pretend
that the pivot columns come
448
00:28:20 --> 00:28:27
first and then whatever's in the
free columns.
449
00:28:27 --> 00:28:34
And there might be some zero
rows below.
450
00:28:34 --> 00:28:45
There's a typical -- a pretty
typical reduced row echelon
451
00:28:45 --> 00:28:46
form.
452
00:28:46 --> 00:28:50
You see what's typical.
453
00:28:50 --> 00:28:55
It's got -- this is r by r.
454
00:28:55 --> 00:29:01
This is r pivot rows.
455
00:29:01 --> 00:29:04
This is r pivot columns.
456
00:29:04 --> 00:29:08
And here are n-r free columns.
457
00:29:08 --> 00:29:09
OK.
458
00:29:09 --> 00:29:14
Tell me what are the special
solutions?
459
00:29:14 --> 00:29:18
What are the -- what's x?
460
00:29:18 --> 00:29:25
If I want to solve R x equals
zero -- in fact,
461
00:29:25 --> 00:29:33
let me -- I'm really going to,
do the whole --
462
00:29:33 --> 00:29:37
since these -- this is now
block matrices,
463
00:29:37 --> 00:29:42
I might as well do all of the
special solutions at once.
464
00:29:42 --> 00:29:48
So I want to solve R x equals
zero, and I'll have some special
465
00:29:48 --> 00:29:49
solutions.
466
00:29:49 --> 00:29:54
Let me, actually -- can I do
them all at once?
467
00:29:54 --> 00:29:58
I'm going to create a null
space matrix,
468
00:29:58 --> 00:29:59
OK.
469
00:29:59 --> 00:30:00
A matrix.
470
00:30:00 --> 00:30:07
Its, its, its columns are the
special -- the columns are the
471
00:30:07 --> 00:30:09
special solutions.
472
00:30:09 --> 00:30:15
This is, I'm making it sound
harder, it's going to be totally
473
00:30:15 --> 00:30:16
easy.
474
00:30:16 --> 00:30:19
N will be this null space
matrix.
475
00:30:19 --> 00:30:23
I want R N to be the zero
matrix.
476
00:30:23 --> 00:30:30
These columns of N are supposed
to multipl- to get multiplied by
477
00:30:30 --> 00:30:34
R and give zero columns.
478
00:30:34 --> 00:30:36
So what N will do the job?
479
00:30:36 --> 00:30:42
Let me put -- I'm going to put
the identity in the free
480
00:30:42 --> 00:30:47
variable part and then minus F
will show up in the pivot
481
00:30:47 --> 00:30:52
variables, just the way it did
in that example.
482
00:30:52 --> 00:30:55
There we had the identity and
F.
483
00:30:55 --> 00:30:59
Here -- in the special
solution.
484
00:30:59 --> 00:31:05
So these columns are -- there's
the matrix of special solutions.
485
00:31:05 --> 00:31:09
And actually,
there -- so there's a Matlab
486
00:31:09 --> 00:31:13
command or a teaching code
command, NULL -- N equal,
487
00:31:13 --> 00:31:18
so this is the -- produces the
null basis, the null space
488
00:31:18 --> 00:31:22
matrix, NULL of A,
and there it is.
489
00:31:22 --> 00:31:27
And how does that command
actually work?
490
00:31:27 --> 00:31:33.88
It uses Matlab to compute R,
then it picks out the pivot
491
00:31:33.88 --> 00:31:40
variables, the free variables,
puts, puts ones and zeros in
492
00:31:40 --> 00:31:46.15
for the free variables,
and copies out the pivot
493
00:31:46.15 --> 00:31:48
variables.
494
00:31:48 --> 00:31:55
It, it does back substitution,
but back substitution for this
495
00:31:55 --> 00:31:58
system is totally simple.
496
00:31:58 --> 00:32:01
What is this system?
497
00:32:01 --> 00:32:03
R x equals zero.
498
00:32:03 --> 00:32:09
So this is R is I F,
and x is the pivot variables
499
00:32:09 --> 00:32:16
and the free variables,
and it's supposed to give zero.
500
00:32:16 --> 00:32:19
So what does that mean?
501
00:32:19 --> 00:32:24
That means that the pivot
variables plus F times the free
502
00:32:24 --> 00:32:26
variables give zero.
503
00:32:26 --> 00:32:32.56
So let me put F times the free
variables on the other side.
504
00:32:32.56 --> 00:32:36
I get minus F times the free
variables.
505
00:32:36 --> 00:32:41
There's my, equation,
as simple as it can be.
506
00:32:41 --> 00:32:47
That's what back substitution
comes to when I've reduced and
507
00:32:47 --> 00:32:52
reduced and reduced this system
to the, to the best form,
508
00:32:52 --> 00:32:52
OK.
509
00:32:52 --> 00:32:57
And, then if the free
variables, I make this special
510
00:32:57 --> 00:33:02
choice of the identity,
then the pivot variables are
511
00:33:02 --> 00:33:02
minus F.
512
00:33:02 --> 00:33:05
OK, can I do,
another example?
513
00:33:05 --> 00:33:09
Could you do another example?
514
00:33:09 --> 00:33:13
Can I -- let me just take
another matrix and,
515
00:33:13 --> 00:33:18
and let's go through this
algorithm once more,
516
00:33:18 --> 00:33:18
OK.
517
00:33:18 --> 00:33:19
Here we go.
518
00:33:19 --> 00:33:24
Here's a blackboard for another
matrix, OK.
519
00:33:24 --> 00:33:29
So I'll call the matrix A
again, but let me make it --
520
00:33:29 --> 00:33:34
yeah, how big shall we make it
this time?
521
00:33:34 --> 00:33:37
Why don't I do this?
522
00:33:37 --> 00:33:40
Just for the heck of it.
523
00:33:40 --> 00:33:49
Let me take the transpose of
this A and see what happens to
524
00:33:49 --> 00:33:49
that.
525
00:33:49 --> 00:33:55
Two four six eight and three
six eight ten.
526
00:33:55 --> 00:34:03
Before we do the calculations,
tell me what's coming?
527
00:34:03 --> 00:34:10
How many pivot variables do you
expect here?
528
00:34:10 --> 00:34:14
How many columns are going to
have pivots?
529
00:34:14 --> 00:34:18
How many -- we have three
columns in that matrix,
530
00:34:18 --> 00:34:22
but are we going to,
are we going to have three
531
00:34:22 --> 00:34:23
pivots?
532
00:34:23 --> 00:34:28
No, because this third columns
is the sum of the first two
533
00:34:28 --> 00:34:28
columns.
534
00:34:28 --> 00:34:33
I'm totally expecting,
totally expecting that these
535
00:34:33 --> 00:34:38
will be pivot columns --
because they're independent,
536
00:34:38 --> 00:34:42.57
but that this third guy,
the third column,
537
00:34:42.57 --> 00:34:47
which is dependent on the first
two, is going to be a free
538
00:34:47 --> 00:34:48
column.
539
00:34:48 --> 00:34:51
Elimination better discover
that.
540
00:34:51 --> 00:34:55
And elimination will also
straighten out the rows,
541
00:34:55 --> 00:35:00
dependent rows and some
independent rows.
542
00:35:00 --> 00:35:05
What's the, what's the row
reduced echelon form for this?
543
00:35:05 --> 00:35:07
Let's just do it,
OK.
544
00:35:07 --> 00:35:10
So, so that's the first pivot.
545
00:35:10 --> 00:35:15
Two times that away from that
gives me a row of zeros.
546
00:35:15 --> 00:35:21
Two times that away from that
gives me a zero two two.
547
00:35:21 --> 00:35:27
And two times that away from
that gives me a zero four four.
548
00:35:27 --> 00:35:30
OK, first column is straight.
549
00:35:30 --> 00:35:34
First variable is a pivot
variable.
550
00:35:34 --> 00:35:35
No problem.
551
00:35:35 --> 00:35:37
On to the second column.
552
00:35:37 --> 00:35:41
I look at the second pivot,
it's a zero.
553
00:35:41 --> 00:35:44
I look below it.
554
00:35:44 --> 00:35:45
There's a two.
555
00:35:45 --> 00:35:48
OK, I do a row exchange.
556
00:35:48 --> 00:35:50
So this zero is now there.
557
00:35:50 --> 00:35:55
I now have a perfectly good
pivot, and I use it.
558
00:35:55 --> 00:36:00
OK, and I subtract two of that
row away from this row.
559
00:36:00 --> 00:36:04
All right if I do it like that?
560
00:36:04 --> 00:36:06
I've got to the form U now.
561
00:36:06 --> 00:36:09
This was my A.
562
00:36:09 --> 00:36:10
Now there's my U.
563
00:36:10 --> 00:36:13
I can see now -- I have to
stop, right?
564
00:36:13 --> 00:36:16
I would go on to the third
column.
565
00:36:16 --> 00:36:18
I should have tried.
566
00:36:18 --> 00:36:20
I quit, but without trying.
567
00:36:20 --> 00:36:22
I shouldn't have done that.
568
00:36:22 --> 00:36:26.79
On to the third column,
look at the pivot position.
569
00:36:26.79 --> 00:36:29
It's got a zero in it.
570
00:36:29 --> 00:36:31
Look below, all zeros.
571
00:36:31 --> 00:36:34
Now I'm entitled to stop,
OK.
572
00:36:34 --> 00:36:36
So the rank is two again.
573
00:36:36 --> 00:36:39
What about the null space?
574
00:36:39 --> 00:36:44.16
How many special solutions are
there this time?
575
00:36:44.16 --> 00:36:48
How many special solutions for
this matrix?
576
00:36:48 --> 00:36:54
I've got -- and which are the
free variables and which are the
577
00:36:54 --> 00:36:58
pivot variables and so on?
578
00:36:58 --> 00:37:02
Pivot columns,
I've got two pivot columns,
579
00:37:02 --> 00:37:04
and that's no accident.
580
00:37:04 --> 00:37:10
The number of pivot columns for
a matrix A, that's a great fact,
581
00:37:10 --> 00:37:16
that the number of pivot
columns for A and A transpose
582
00:37:16 --> 00:37:18
are the same.
583
00:37:18 --> 00:37:22
And then I have a free column.
584
00:37:22 --> 00:37:24
There's a free column.
585
00:37:24 --> 00:37:30
One free column,
because the count is three
586
00:37:30 --> 00:37:31
minus two.
587
00:37:31 --> 00:37:36
Three minus two gives me one
free column.
588
00:37:36 --> 00:37:42
OK, so now let me solve,
what's in the null space.
589
00:37:42 --> 00:37:46
OK, so how do I -- let's see.
590
00:37:46 --> 00:37:52
These vectors have length
three.
591
00:37:52 --> 00:37:55
They only have three
components.
592
00:37:55 --> 00:38:00
I'm making too much space for
the, to write x.
593
00:38:00 --> 00:38:05
x has just got three
components, and what are they?
594
00:38:05 --> 00:38:08
I'm looking for the null space.
595
00:38:08 --> 00:38:10
OK, so how do I start?
596
00:38:10 --> 00:38:16
I give the free variable some
convenient value.
597
00:38:16 --> 00:38:18.25
And what's that?
598
00:38:18.25 --> 00:38:19.82
I set it to one.
599
00:38:19.82 --> 00:38:22
I set the free variable to one.
600
00:38:22 --> 00:38:28
If I set the free variable to
zero and solve for the pivot
601
00:38:28 --> 00:38:32
variables, I'll get all zeros:
no progress.
602
00:38:32 --> 00:38:37
But by setting the free
variable to one -- you see w- my
603
00:38:37 --> 00:38:44
two equations now are --
my equations are x1+ 2x2+ 3 x3
604
00:38:44 --> 00:38:48
is zero, that's my first
equation.
605
00:38:48 --> 00:38:54
And my second equation is now
2x2+2x3 equals zero.
606
00:38:54 --> 00:38:55
And, OK.
607
00:38:55 --> 00:39:00.31
So if x3 is one,
then x2 is minus one.
608
00:39:00.31 --> 00:39:06
And if x3 is one and x2 is
minus one, then maybe x1 is
609
00:39:06 --> 00:39:07
minus one.
610
00:39:07 --> 00:39:13.74
And actually I go back to check
now.
611
00:39:13.74 --> 00:39:17
I don't, like -- I did a quick
calculation mentally.
612
00:39:17 --> 00:39:20
Can I mentally do a quick
check?
613
00:39:20 --> 00:39:23
That matrix,
that solution x says that minus
614
00:39:23 --> 00:39:27
this column minus this column
plus this one is the zero
615
00:39:27 --> 00:39:28
column.
616
00:39:28 --> 00:39:29
And it is.
617
00:39:29 --> 00:39:33
Minus that minus that plus that
is zero.
618
00:39:33 --> 00:39:35
So that's in the null space.
619
00:39:35 --> 00:39:39
And now you can tell me what
else is in the null space.
620
00:39:39 --> 00:39:42
What's, what's the whole null
space now?
621
00:39:42 --> 00:39:44
I multiply by C,
right.
622
00:39:44 --> 00:39:46
The whole null space is a line.
623
00:39:46 --> 00:39:48
So that's the description.
624
00:39:48 --> 00:39:53
You know, if I ask you on a
homework or a quiz or the final
625
00:39:53 --> 00:39:56
what --
give me, describe,
626
00:39:56 --> 00:40:00
tell me the null space,
find the null space of this
627
00:40:00 --> 00:40:02
matrix, you can take those
steps.
628
00:40:02 --> 00:40:05
And that's the answer I'm
looking for.
629
00:40:05 --> 00:40:10
And I'm looking for that C too,
because that's telling me that
630
00:40:10 --> 00:40:15.42
you're remembering that it's a
whole space and not just one
631
00:40:15.42 --> 00:40:16
vector.
632
00:40:16 --> 00:40:21
Later I will ask you for a
basis for the null space.
633
00:40:21 --> 00:40:23
Then I just want this vector.
634
00:40:23 --> 00:40:28
But if I ask for the whole null
space, then there's the whole
635
00:40:28 --> 00:40:30
line through that vector.
636
00:40:30 --> 00:40:34
OK, now one more natural thing
to do with this example,
637
00:40:34 --> 00:40:38
right, is keep going to the
reduced matrix,
638
00:40:38 --> 00:40:39
R.
639
00:40:39 --> 00:40:42
So can I push onwards to R?
640
00:40:42 --> 00:40:46
That should be quick,
but let's just practice.
641
00:40:46 --> 00:40:48.71
Let me keep going to R.
642
00:40:48.71 --> 00:40:51
OK, so what do I do here?
643
00:40:51 --> 00:40:57.01
I subtract -- I clear out above
the pivot, so I subtract that
644
00:40:57.01 --> 00:41:02
from that, that's one zero one
is left, right?
645
00:41:02 --> 00:41:07
When I subtracted this row from
this it produced a zero above
646
00:41:07 --> 00:41:08
this pivot.
647
00:41:08 --> 00:41:12
And now I want that pivot to be
a one.
648
00:41:12 --> 00:41:16
So for the R matrix,
I'll divide this equation by
649
00:41:16 --> 00:41:21
two, and of course these zero,
zeros are great,
650
00:41:21 --> 00:41:22
they don't change.
651
00:41:22 --> 00:41:23
There's R.
652
00:41:23 --> 00:41:25
That's R.
653
00:41:25 --> 00:41:27
You see what R is?
654
00:41:27 --> 00:41:32
You see the identity matrix
sitting up here?
655
00:41:32 --> 00:41:37
You see the free part F,
the F part here?
656
00:41:37 --> 00:41:40
And you see the zeros below.
657
00:41:40 --> 00:41:43
This is I F zero zero.
658
00:41:43 --> 00:41:45
And what's the x?
659
00:41:45 --> 00:41:52
The x has the identity --
well, it's only a single number
660
00:41:52 --> 00:41:55
one, but it's the identity
matrix in the free,
661
00:41:55 --> 00:41:57
in the free part.
662
00:41:57 --> 00:42:01
And what does it have in the
pivot variables?
663
00:42:01 --> 00:42:03.72
What did back substitution
give?
664
00:42:03.72 --> 00:42:05
It gave minus these guys.
665
00:42:05 --> 00:42:10
You see that what this is is
any multiple of -- this is the
666
00:42:10 --> 00:42:15
identity there,
and this is minus F here.
667
00:42:15 --> 00:42:18
This is our null space matrix N
for this.
668
00:42:18 --> 00:42:24
Our, our null space matrix is
the guy whose columns are the
669
00:42:24 --> 00:42:25
special solutions.
670
00:42:25 --> 00:42:31
So their free variables have
the special values one and,
671
00:42:31 --> 00:42:33
pivot variables have minus F.
672
00:42:33 --> 00:42:37
So do you see,
though, how the minus F just
673
00:42:37 --> 00:42:43
automatically shows up in the
special solutions.
674
00:42:43 --> 00:42:45
That's it really.
675
00:42:45 --> 00:42:51
I don't think there's anything
more I can say about A x equals
676
00:42:51 --> 00:42:52
zero.
677
00:42:52 --> 00:42:58
There's more I can say about A
x equal b, but that'll be on
678
00:42:58 --> 00:42:59
Friday.
679
00:42:59 --> 00:43:04
OK, so that's,
that's the null space.
680
00:43:04 --> 00:43:07
Thanks.