WEBVTT

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OK.

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Good.

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The final class in linear
algebra at MIT this Fall

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is to review the whole course.

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And, you know the best
way I know how to review

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is to take old exams and just
think through the problems.

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So it will be a three-hour
exam next Thursday.

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Nobody will be able to take an
exam before Thursday, anybody

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who needs to take it
in some different way

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after Thursday should
see me next Monday.

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I'll be in my office Monday.

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OK.

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May I just read
out some problems

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and, let me bring the board
down, and let's start.

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OK.

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Here's a question.

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This is about a 3-by-n matrix.

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And we're given --
so we're given --

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given -- A x equals 1
0 0 has no solution.

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And we're also given A x equals
0 1 0 has exactly one solution.

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OK.

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So you can probably
anticipate my first question,

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what can you tell me about m?

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It's an m-by-n matrix
of rank r, as always,

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what can you tell me
about those three numbers?

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So what can you tell me about
m, the number of rows, n,

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the number of columns,
and r, the rank?

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OK.

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See, do you want to
tell me first what m is?

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How many rows in this matrix?

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Must be three, right?

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We can't tell what n
is, but we can certainly

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tell that m is three.

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OK.

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And, what do these
things tell us?

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Let's take them one at a time.

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When I discover that some
equation has no solution,

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that there's some right-hand
side with no answer,

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what does that tell me about
the rank of the matrix?

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It's smaller m.

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Is that right?

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If there is no
solution, that tells me

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that some rows of the matrix
are combinations of other rows.

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Because if I had a
pivot in every row,

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then I would certainly be
able to solve the system.

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I would have particular
solutions and all the good

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So any time that there's a
system with no solutions,

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stuff. that tells me
that r must be below m.

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What about the fact that if,
when there is a solution,

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there's only one?

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What does that tell me?

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Well, normally there
would be one solution,

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and then we could add in
anything in the null space.

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So this is telling me the null
space only has the 0 vector in

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it.

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There's just one
solution, period,

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so what does that tell me?

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The null space has only
the zero vector in it?

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What does that tell me about
the relation of r to n?

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So this one solution
only, that means

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the null space of the matrix
must be just the zero vector,

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and what does that
tell me about r and n?

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They're equal.

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The columns are independent.

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So I've got, now, r equals
n, and r less than m,

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and now I also know m is three.

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So those are really
the facts I know.

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n=r and those numbers
are smaller than three.

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Sorry, yes, yes.

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r is smaller than m, and
n, of course, is also.

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So I guess this summarizes
what we can tell.

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In fact, why not
give me a matrix --

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because I would often ask for
an example of such a matrix --

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can you give me a matrix
A that's an example?

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That shows this possibility?

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Exactly, that
there's no solution

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with that right-hand side, but
there's exactly one solution

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with this right-hand side.

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Anybody want to suggest
a matrix that does that?

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Let's see.

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What do I -- what vector do
I want in the column space?

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I want zero, one, zero,
to be in the column space,

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because I'm able
to solve for that.

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So let's put zero, one,
zero in the column space.

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Actually, I could
stop right there.

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That would be a matrix with
m equal three, three rows,

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and n and r are both one,
rank one, one column,

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and, of course, there's
no solution to that one.

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So that's perfectly
good as it is.

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Or if you, kind of,
have a prejudice

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against matrices that
only have one column,

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I'll accept a second

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column.

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So what could I include
as a second column

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that would just be a different
answer but equally good?

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I could put this vector in the
column space, too, if I wanted.

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That would now be a case
with r=n=2, but, of course,

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three m eq- m is still
three, and this vector is not

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in the column space.

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So you're -- this is just like
prompting us to remember all

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those things, column space,
null space, all that stuff.

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Now, I probably asked
a second question

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about this type of thing.

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OK.

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Oh, I even asked, write
down an example of a

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Ah. matrix that fits
the description.

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Hm.

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I guess I haven't learned
anything in twenty-six years.

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CK.

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Cross out all statements that
are false about any matrix with

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these -- so again, these are
-- this is the preliminary sta-

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these are the facts about my
matrix, this is one example.

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But, of course, by
having an example,

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it will be easy to check some
of these facts, or non-facts.

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Let me, let me write
down some, facts.

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Some possible facts.

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So this is really true or false.

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The determinant --
this is part one,

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the determinant of A transpose
A is the same as the determinant

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of A A transpose.

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Is that true or not?

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Second one, A transpose
A, is invertible.

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Is invertible.

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Third possible fact, A A
transpose is positive definite.

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So you see how, on
an exam question,

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I try to connect the
different parts of the course.

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So, well, I mean,
the simplest way

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would be to try it with that
matrix as a good example,

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but maybe we can
answer, even directly.

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Let me take number two first.

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Because I'm -- you know, I'm
very, very fond of that matrix,

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A transpose A.

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And when is it invertible?

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When is the matrix A
transpose A, invertible?

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The great thing is that I
can tell from the rank of A

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that I don't have to
multiply out A transpose A.

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A transpose A, is invertible --

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well, if A has a null space
other than the zero vector,

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then it -- it's -- no way
it's going to be invertible.

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But the beauty is, if the null
space of A is just the zero

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vector, so the fact
-- the key fact is,

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this is invertible if
r=n, by which I mean,

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independent columns of A.

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In A.

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In the matrix A.

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If r=n -- if the matrix A
has independent columns,

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then this combination,
A transpose A,

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is square and still
that same null space,

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only the zero vector,
independent columns all good,

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and so, what's the true/false?

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Is it -- is this middle one T
or F for this, in this setup?

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Well, we discovered that --

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we discovered that -- that r
was n, from that second fact.

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So this is a true.

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That's a true.

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And, of course, A transpose
A, in this example,

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would probably be -- what
would A transpose A, be,

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for that matrix?

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Can you multiply A
transpose A, and see what

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it looks like for that matrix?

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What shape would it be?

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It will be two by two.

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And what matrix will it be?

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The identity.

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So, it checks out.

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OK, what about A A transpose?

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Well, depending
on the shape of A,

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it could be good or not so good.

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It's always symmetric,
it's always square,

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but what's the size, now?

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This is three by n,
and this is n by three,

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so the result is three by three.

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Is it positive definite?

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I don't think so.

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False.

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If I multiply that by A
transpose, A A transpose,

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what would the rank be?

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It would be the same as
the rank of A, that's --

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it would be just rank two.

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And if it's three-by-three,
and it's only rank two,

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it's certainly not
positive definite.

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So what could I say
about A A transpose,

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if I wanted to, like, say
something true about it?

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It's true that it is
positive semi-definite.

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If I made this semi-definite,
it would always be true, always.

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But if I'm looking
for positive definite,

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then I'm looking at the null
space of whatever's here,

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and, in this case,
it's got a null space.

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So A, A -- eh, shall
we just figure it out,

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here?

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A A transpose, for that
matrix, will be three-by-three.

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If I multiplied
A by A transpose,

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what would the first row be?

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All zeroes, right?

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First row of A A
transpose, could only

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be all zeroes, so it's probably
a one there and a one there,

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or something like that.

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But, I don't even
know if that's right.

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But it's all zeroes
there, so it's certainly

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not positive definite.

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Let me not put anything up
I'm not sh- don't check.

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What about this determinant?

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Oh, well, I guess --

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that's a sort of
tricky question.

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Is it true or
false in this case?

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It's false, apparently, because
A transpose A, is invertible,

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we just got a true for this
one, and we got a false,

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we got a z- we got
a non-invertible one

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for this one.

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So actually, this one
is false, number one.

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That surprises us,
actually, because it's,

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I mean, why was it tricky?

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Because what is true
about determinants?

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This would be true if
those matrices were square.

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If I have two square matrices,
A and any other matrix B,

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could be A transpose, could
be somebody else's matrix.

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Then it would be true that
the determinant of B A

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would equal the
determinant of A B.

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But if the matrices are not
square and it would actually be

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true that it would be equal
-- that this would equal

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the determinant of A times the
determinant of A transpose.

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We could even split up those
two separate determinants.

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And, of course,
those would be equal.

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But only when A is square.

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So that's just, that's a
question that rests on the,

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the falseness rests on the
fact that the matrix isn't

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square in the first place.

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OK, good.

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Let's see.

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Oh, now, even asks more.

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Prove that A transpose
y equals c --

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hah-God, it's -- this
question goes on and on.

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now I ask you about
A transpose y=c.

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So I'm asking you
about the equation --

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about the matrix A transpose.

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And I want you to prove that
it has at least one solution --

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one solution for every c,
every right-hand side c,

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and, in fact -- in fact,
infinitely many solutions

00:15:13.640 --> 00:15:16.630
for every c.

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OK.

00:15:17.910 --> 00:15:20.170
Well, none -- none
of this is difficult,

00:15:20.170 --> 00:15:25.470
but, it's been a little while.

00:15:25.470 --> 00:15:27.900
So we just have to think again.

00:15:27.900 --> 00:15:30.980
When I have a system of
equations -- this is --

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this matrix A transpose is now,
instead of being three by n,

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it's n by three, it's n by m.

00:15:40.970 --> 00:15:43.470
Of course.

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To show that a system has
at least one solution,

00:15:53.900 --> 00:15:56.020
when does this, when
does this system --

00:15:56.020 --> 00:15:57.760
when is the system
always solvable?

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When it has full row rank,
when the rows are independent.

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Here, we have n rows,
and that's the rank.

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So at least one solution,
because the number

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of rows, which is n,
for the transpose,

00:16:22.500 --> 00:16:24.300
is equal to r, the rank.

00:16:27.490 --> 00:16:30.860
This A transpose
had independent rows

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because A had independent
columns, right?

00:16:34.530 --> 00:16:41.480
The original A had independent
columns, when we transpose it,

00:16:41.480 --> 00:16:44.430
it has independent rows, so
there's at least one solution.

00:16:44.430 --> 00:16:46.660
But now, how do I even know
that there are infinitely

00:16:46.660 --> 00:16:48.050
many solutions?

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Oh, what do I --

00:16:49.260 --> 00:16:52.280
I want to know something
about the null space.

00:16:52.280 --> 00:16:57.390
What's the dimension of the
null space of A transpose?

00:16:57.390 --> 00:17:00.100
So the answer has got
to be the dimension

00:17:00.100 --> 00:17:03.850
of the null space of
A transpose, what's

00:17:03.850 --> 00:17:06.119
the general fact?

00:17:06.119 --> 00:17:10.790
If A is an m by n
matrix of rank r,

00:17:10.790 --> 00:17:14.260
what's the dimension
of A transpose?

00:17:14.260 --> 00:17:15.970
The null space of A transpose?

00:17:15.970 --> 00:17:20.470
Do you remember that
little fourth subspace

00:17:20.470 --> 00:17:24.050
that's tagging along
down in our big picture?

00:17:24.050 --> 00:17:26.497
It's dimension was m-r.

00:17:30.780 --> 00:17:32.630
And, that's bigger than zero.

00:17:32.630 --> 00:17:35.610
m is bigger than r.

00:17:35.610 --> 00:17:37.520
So there's a lot
in that null space.

00:17:44.520 --> 00:17:47.790
So there's always one
solution because n i- this

00:17:47.790 --> 00:17:49.230
is speaking about A transpose.

00:17:52.880 --> 00:17:55.730
So for A transpose, the roles
of m and n are reversed,

00:17:55.730 --> 00:17:57.540
of course, so I'm --

00:17:57.540 --> 00:18:01.430
keep in mind that this
board was about A transpose,

00:18:01.430 --> 00:18:04.880
so the roles -- so it's the
null space of a transpose,

00:18:04.880 --> 00:18:08.590
and there are m-r
free variables.

00:18:08.590 --> 00:18:13.630
OK, that's, like,
just some, review.

00:18:13.630 --> 00:18:17.030
Can I take another problem
that's also sort of --

00:18:17.030 --> 00:18:24.710
suppose the matrix A has
three columns, v1, v2, v3.

00:18:24.710 --> 00:18:28.880
Those are the columns
of the matrix.

00:18:28.880 --> 00:18:31.140
All right.

00:18:31.140 --> 00:18:33.500
Question A.

00:18:33.500 --> 00:18:36.827
Solve Ax=v1-v2+v3.

00:18:44.260 --> 00:18:45.670
Tell me what x is.

00:18:53.960 --> 00:18:57.520
Well, there, you're
seeing the most --

00:18:57.520 --> 00:19:04.380
the one absolutely
essential fact about matrix

00:19:04.380 --> 00:19:06.530
multiplication,
how does it work,

00:19:06.530 --> 00:19:10.760
when we do it a column at a
time, the very, very first day,

00:19:10.760 --> 00:19:13.800
way back in September, we
did multiplication a column

00:19:13.800 --> 00:19:14.590
at a time.

00:19:14.590 --> 00:19:16.190
So what's x?

00:19:16.190 --> 00:19:18.070
Just tell me?

00:19:18.070 --> 00:19:19.250
One minus one, one.

00:19:19.250 --> 00:19:19.820
Thanks.

00:19:19.820 --> 00:19:20.740
OK.

00:19:20.740 --> 00:19:22.950
Everybody's got that.

00:19:22.950 --> 00:19:23.550
OK?

00:19:23.550 --> 00:19:26.750
Then the next question is,
suppose that combination is

00:19:26.750 --> 00:19:28.130
zero --

00:19:28.130 --> 00:19:31.560
oh, yes, OK, so
question (b) says --

00:19:31.560 --> 00:19:42.730
part (b) says, suppose
this thing is zero.

00:19:42.730 --> 00:19:45.380
Suppose that's zero.

00:19:45.380 --> 00:19:48.520
Then the solution is not unique.

00:19:48.520 --> 00:19:51.730
Suppose I want true or false.

00:19:51.730 --> 00:19:54.440
-- and a reason.

00:19:54.440 --> 00:20:00.330
Suppose this
combination is zero.

00:20:00.330 --> 00:20:02.170
v1-v2+v3.

00:20:02.170 --> 00:20:06.640
Show that -- what
does that tell me?

00:20:06.640 --> 00:20:08.210
So it's a separate
question, maybe

00:20:08.210 --> 00:20:11.880
I sort of saved time
by writing it that way,

00:20:11.880 --> 00:20:14.620
but it's a totally
separate question.

00:20:14.620 --> 00:20:19.280
If I have a matrix, and I know
that column one minus column

00:20:19.280 --> 00:20:23.160
two plus column
three is zero, what

00:20:23.160 --> 00:20:33.810
does that tell me about whether
the solution is unique or not?

00:20:33.810 --> 00:20:36.420
Is there more than one solution?

00:20:36.420 --> 00:20:40.330
What's uniqueness about?

00:20:40.330 --> 00:20:42.420
Uniqueness is about,
is there anything

00:20:42.420 --> 00:20:44.720
in the null space, right?

00:20:44.720 --> 00:20:46.490
The solution is
unique when there's

00:20:46.490 --> 00:20:49.550
nobody in the null space
except the zero vector.

00:20:49.550 --> 00:20:57.670
And, if that's zero, then this
guy would be in the null space.

00:20:57.670 --> 00:21:08.020
So if this were zero, then this
x is in the null space of A.

00:21:08.020 --> 00:21:18.750
So solutions are never
unique, because I could always

00:21:18.750 --> 00:21:26.890
add that to any solution,
and Ax wouldn't change.

00:21:26.890 --> 00:21:29.610
So it's always that question.

00:21:29.610 --> 00:21:31.760
Is there somebody
in the null space?

00:21:31.760 --> 00:21:33.620
OK.

00:21:33.620 --> 00:21:37.370
Oh, now, here's a totally
different question.

00:21:37.370 --> 00:21:43.770
Suppose those three vectors,
v1, v2, v3, are orthonormal.

00:21:43.770 --> 00:21:48.340
So this isn't going to happen
for orthonormal vectors.

00:21:48.340 --> 00:21:51.680
OK, so part (c),
forget part (b).

00:21:51.680 --> 00:21:52.410
c.

00:21:52.410 --> 00:22:05.040
If v1, v2, v3,
are orthonormal --

00:22:05.040 --> 00:22:08.725
so that I would usually
have called them q1, q2, q3.

00:22:11.540 --> 00:22:16.420
Now, what combination --
oh, here's a nice question,

00:22:16.420 --> 00:22:19.020
if I say so myself --

00:22:19.020 --> 00:22:24.280
what combination of v1
and v2 is closest to v3?

00:22:24.280 --> 00:22:28.120
What point on the
plane of v1 and v2

00:22:28.120 --> 00:22:33.100
is the closest point to v3 if
these vectors are orthonormal?

00:22:33.100 --> 00:22:33.750
So let me --

00:22:33.750 --> 00:22:41.610
I'll start the sentence -- then
the combination something times

00:22:41.610 --> 00:22:51.570
v1 plus something times v2 is
the closest combination to v3?

00:22:51.570 --> 00:22:53.860
And what's the answer?

00:22:53.860 --> 00:22:57.150
What's the closest vector
on that plane to v3?

00:22:57.150 --> 00:23:00.030
Zeroes.

00:23:00.030 --> 00:23:01.480
Right.

00:23:01.480 --> 00:23:07.240
We just imagine the x, y,
z axes, the v1, v2, th- v3

00:23:07.240 --> 00:23:13.380
could be the standard
basis, the x, y, z vectors,

00:23:13.380 --> 00:23:19.090
and, of course, the point on
the xy plane that's closest

00:23:19.090 --> 00:23:24.090
to v3 on the z axis is zero.

00:23:24.090 --> 00:23:28.880
So if we're orthonormal,
then the projection

00:23:28.880 --> 00:23:34.690
of v3 onto that plane
is perpendicular,

00:23:34.690 --> 00:23:36.440
it hits right at zero.

00:23:36.440 --> 00:23:39.850
OK, so that's like a quick --

00:23:39.850 --> 00:23:43.890
you know, an easy question,
but still brings it out.

00:23:43.890 --> 00:23:45.110
OK.

00:23:45.110 --> 00:23:56.720
Let me see what, shall I
write down a Markov matrix,

00:23:56.720 --> 00:24:01.870
and I'll ask you
for its eigenvalues.

00:24:01.870 --> 00:24:02.680
OK.

00:24:02.680 --> 00:24:06.070
Here's a Markov matrix --

00:24:06.070 --> 00:24:14.210
this -- and, tell
me its eigenvalues.

00:24:14.210 --> 00:24:16.020
So here -- I'll
call the matrix A,

00:24:16.020 --> 00:24:19.990
and I'll call this as point
two, point four, point four,

00:24:19.990 --> 00:24:25.480
point four, point four, point
two, point four, point three,

00:24:25.480 --> 00:24:28.391
point three, point four.

00:24:28.391 --> 00:24:28.890
OK.

00:24:33.340 --> 00:24:39.020
Let's see -- it helps out to
notice that column one plus

00:24:39.020 --> 00:24:41.840
column two --

00:24:41.840 --> 00:24:46.870
what's interesting about
column one plus column two?

00:24:46.870 --> 00:24:50.530
It's twice as much
as column three.

00:24:50.530 --> 00:24:53.790
So column one plus column two
equals two times column three.

00:24:53.790 --> 00:24:57.300
I put that in there,
column one plus column two

00:24:57.300 --> 00:24:59.730
equals twice column three.

00:24:59.730 --> 00:25:01.430
That's observation.

00:25:01.430 --> 00:25:02.230
OK.

00:25:02.230 --> 00:25:04.500
Tell me the eigenvalues
of the matrix.

00:25:07.080 --> 00:25:08.455
OK, tell me one eigenvalue?

00:25:11.770 --> 00:25:15.230
Because the matrix is singular.

00:25:15.230 --> 00:25:17.570
Tell me another eigenvalue?

00:25:17.570 --> 00:25:20.320
One, because it's
a Markov matrix,

00:25:20.320 --> 00:25:25.120
the columns add to
the all ones vector,

00:25:25.120 --> 00:25:31.940
and that will be an
eigenvector of A transpose.

00:25:31.940 --> 00:25:33.435
And tell me the
third eigenvalue?

00:25:36.740 --> 00:25:38.770
Let's see, to make
the trace come out

00:25:38.770 --> 00:25:42.570
right, which is point eight,
we need minus point two.

00:25:45.200 --> 00:25:46.270
OK.

00:25:46.270 --> 00:25:52.750
And now, suppose I start
the Markov process.

00:25:52.750 --> 00:25:56.460
Suppose I start with u(0) --

00:25:56.460 --> 00:26:01.490
so I'm going to look at the
powers of A applied to u(0).

00:26:01.490 --> 00:26:04.840
This is uk.

00:26:04.840 --> 00:26:09.930
And there's my matrix, and
I'm going to let u(0) be --

00:26:09.930 --> 00:26:14.860
this is going to
be zero, ten, zero.

00:26:14.860 --> 00:26:21.060
And my question is,
what does that approach?

00:26:21.060 --> 00:26:24.910
If u(0) is equal to
this -- there is u(0).

00:26:24.910 --> 00:26:26.420
Shall I write it in?

00:26:26.420 --> 00:26:27.800
Maybe I'll just write in u(0).

00:26:30.680 --> 00:26:40.430
A to the k, starting with
ten people in state two,

00:26:40.430 --> 00:26:48.210
and every step follows
the Markov rule,

00:26:48.210 --> 00:26:52.920
what does the solution
look like after k steps?

00:26:52.920 --> 00:26:54.980
Let me just ask you that.

00:26:54.980 --> 00:26:58.500
And then, what happens
as k goes to infinity?

00:26:58.500 --> 00:27:00.360
This is a steady-state
question, right?

00:27:00.360 --> 00:27:02.770
I'm looking for
the steady state.

00:27:02.770 --> 00:27:06.090
Actually, the question doesn't
ask for the k step answer,

00:27:06.090 --> 00:27:08.380
it just jumps right
away to infinity --

00:27:08.380 --> 00:27:15.170
but how would I express
the solution after k steps?

00:27:15.170 --> 00:27:23.340
It would be some multiple of
the first eigenvalue to the k-th

00:27:23.340 --> 00:27:26.160
power -- times the
first eigenvector,

00:27:26.160 --> 00:27:30.210
plus some other multiple
of the second eigenvalue,

00:27:30.210 --> 00:27:34.410
times its eigenvector, and
some multiple of the third

00:27:34.410 --> 00:27:38.440
eigenvalue, times
its eigenvector.

00:27:38.440 --> 00:27:39.285
OK.

00:27:39.285 --> 00:27:39.785
Good.

00:27:42.290 --> 00:27:49.140
And these eigenvalues are
zero, one, and minus point two.

00:27:52.810 --> 00:27:55.290
So what happens as
k goes to infinity?

00:27:58.480 --> 00:28:01.550
The only thing that
survives the steady state --

00:28:01.550 --> 00:28:07.860
so at u infinity, this
is gone, this is gone,

00:28:07.860 --> 00:28:16.640
all that's left is c2x2.

00:28:16.640 --> 00:28:18.550
So I'd better find x2.

00:28:18.550 --> 00:28:20.700
I've got to find
that eigenvector

00:28:20.700 --> 00:28:22.710
to complete the answer.

00:28:22.710 --> 00:28:25.260
What's the eigenvector
that corresponds

00:28:25.260 --> 00:28:26.400
to lambda equal one?

00:28:26.400 --> 00:28:29.650
That's the key eigenvector
in any Markov process,

00:28:29.650 --> 00:28:32.430
is that eigenvector.

00:28:32.430 --> 00:28:34.550
Lambda equal one
is an eigenvalue,

00:28:34.550 --> 00:28:38.280
I need its eigenvector
x2, and then

00:28:38.280 --> 00:28:44.480
I need to know how much of it
is in the starting vector u0.

00:28:44.480 --> 00:28:45.240
OK.

00:28:45.240 --> 00:28:47.880
So, how do I find
that eigenvector?

00:28:47.880 --> 00:28:51.500
I guess I subtract one
from the diagonal, right?

00:28:51.500 --> 00:28:56.280
So I have minus point
eight, minus point eight,

00:28:56.280 --> 00:28:59.660
minus point six, and the
rest, of course, is just --

00:28:59.660 --> 00:29:04.360
still point four, point
four, point four, point four,

00:29:04.360 --> 00:29:07.150
point three, point
three, and hopefully,

00:29:07.150 --> 00:29:19.230
that's a singular matrix, so
I'm looking to solve A minus Ix

00:29:19.230 --> 00:29:19.980
equal zero.

00:29:19.980 --> 00:29:22.972
Let's see -- can anybody
spot the solution here?

00:29:22.972 --> 00:29:24.930
I don't know, I didn't
make it easy for myself.

00:29:27.900 --> 00:29:30.630
What do you think there?

00:29:30.630 --> 00:29:39.080
Maybe those first two
entries might be --

00:29:39.080 --> 00:29:41.500
oh, no, what do you think?

00:29:41.500 --> 00:29:44.640
Anybody see it?

00:29:44.640 --> 00:29:46.680
We could use elimination
if we were desperate.

00:29:49.210 --> 00:29:51.810
Are we that desperate?

00:29:51.810 --> 00:29:55.320
Anybody just call out if
you see the vector that's

00:29:55.320 --> 00:29:57.980
in that null space.

00:29:57.980 --> 00:30:00.390
Eh, there better be a
vector in that null space,

00:30:00.390 --> 00:30:03.220
or I'm quitting.

00:30:03.220 --> 00:30:11.555
Uh, ha- OK, well, I guess
we could use elimination.

00:30:15.200 --> 00:30:18.010
I thought maybe somebody might
see it from further away.

00:30:21.100 --> 00:30:23.820
Is there a chance
that these guys are --

00:30:23.820 --> 00:30:28.140
could it be that these two are
equal and this is whatever it

00:30:28.140 --> 00:30:31.500
takes, like, something
like three, three, two?

00:30:31.500 --> 00:30:33.870
Would that possibly work?

00:30:33.870 --> 00:30:37.170
I mean, that's great for this
-- no, it's not that great.

00:30:37.170 --> 00:30:39.810
Three, three, four --

00:30:39.810 --> 00:30:43.930
this is, deeper mathematics
you're watching now.

00:30:43.930 --> 00:30:47.460
Three, three, four, is that --

00:30:47.460 --> 00:30:48.000
it works!

00:30:48.000 --> 00:30:49.030
Don't mess with it!

00:30:49.030 --> 00:30:49.910
It works!

00:30:49.910 --> 00:30:52.420
Uh, yes.

00:30:52.420 --> 00:30:54.140
OK, it works, all right.

00:30:54.140 --> 00:31:01.710
And, yes, OK, and, so that's
x2, three, three, four,

00:31:01.710 --> 00:31:11.870
and, how much of that vector
is in the starting vector?

00:31:11.870 --> 00:31:16.420
Well, we could go through
a complicated process.

00:31:16.420 --> 00:31:19.430
But what's the beauty
of Markov things?

00:31:19.430 --> 00:31:23.460
That the total number
of the total population,

00:31:23.460 --> 00:31:27.870
the sum of these doesn't change.

00:31:27.870 --> 00:31:30.200
That the total number of
people, they're moving around,

00:31:30.200 --> 00:31:35.130
but they don't get born
or die or get dead.

00:31:35.130 --> 00:31:38.520
So there's ten of them at the
start, so there's ten of them

00:31:38.520 --> 00:31:41.570
there, so c2 is
actually one, yes.

00:31:41.570 --> 00:31:45.380
So that would be the
correct solution.

00:31:45.380 --> 00:31:45.880
OK.

00:31:45.880 --> 00:31:48.490
That would be the u infinity.

00:31:48.490 --> 00:31:49.120
OK.

00:31:49.120 --> 00:31:50.890
So I used there,
in that process,

00:31:50.890 --> 00:31:53.190
sort of, the main
facts about Markov

00:31:53.190 --> 00:31:58.160
matrices to, to get
a jump on the answer.

00:31:58.160 --> 00:31:59.090
OK. let's see.

00:31:59.090 --> 00:32:05.810
OK, here's some, kind of
quick, short questions.

00:32:05.810 --> 00:32:09.770
Uh, maybe I'll move over to
this board, and leave that for

00:32:09.770 --> 00:32:11.260
the moment.

00:32:11.260 --> 00:32:16.020
I'm looking for
two-by-two matrices.

00:32:16.020 --> 00:32:19.610
And I'll read out the property
I want, and you give me

00:32:19.610 --> 00:32:23.670
an example, or tell me
there isn't such a matrix.

00:32:23.670 --> 00:32:24.640
All right.

00:32:24.640 --> 00:32:25.180
Here we go.

00:32:25.180 --> 00:32:28.240
First -- so two-by-twos.

00:32:28.240 --> 00:32:36.550
First, I want the
projection onto the line

00:32:36.550 --> 00:32:41.820
through A equals
four minus three.

00:32:46.190 --> 00:32:49.600
So it's a one-dimensional
projection matrix

00:32:49.600 --> 00:32:50.430
I'm looking for.

00:32:53.520 --> 00:32:56.280
And what's the formula for it?

00:32:56.280 --> 00:33:00.950
What's the formula for the
projection matrix P onto a line

00:33:00.950 --> 00:33:03.960
through A. And then we'd just
plug in this particular A.

00:33:03.960 --> 00:33:08.050
Do you remember that formula?

00:33:08.050 --> 00:33:14.150
There's an A and an A
transpose, and normally we

00:33:14.150 --> 00:33:17.030
would have an A transpose
A inverse in the middle,

00:33:17.030 --> 00:33:20.730
but here we've just got numbers,
so we just divide by it.

00:33:20.730 --> 00:33:25.780
And then plug in A
and we've got it.

00:33:25.780 --> 00:33:26.850
So, equals.

00:33:26.850 --> 00:33:28.290
OK.

00:33:28.290 --> 00:33:30.390
You can put in the numbers.

00:33:30.390 --> 00:33:31.790
Trivial, right.

00:33:31.790 --> 00:33:32.370
OK.

00:33:32.370 --> 00:33:33.190
Number two.

00:33:38.440 --> 00:33:39.900
So this is a new problem.

00:33:39.900 --> 00:33:45.910
The matrix with eigenvalue zero
and three and eigenvectors --

00:33:45.910 --> 00:33:49.740
well, let me write these
down. eigenvalue zero,

00:33:49.740 --> 00:33:56.320
eigenvector one, two, eigenvalue
three, eigenvector two, one.

00:33:59.170 --> 00:34:01.970
I'm giving you the
eigenvalues and eigenvectors

00:34:01.970 --> 00:34:04.090
instead of asking for them.

00:34:04.090 --> 00:34:05.485
Now I'm asking for the matrix.

00:34:10.080 --> 00:34:11.590
What's the matrix, then?

00:34:11.590 --> 00:34:12.210
What's A?

00:34:17.310 --> 00:34:19.090
Here was a formula,
then we just put

00:34:19.090 --> 00:34:21.610
in some numbers, what's
the formula here,

00:34:21.610 --> 00:34:26.270
into which we'll just
put the given numbers?

00:34:26.270 --> 00:34:30.840
It's the S lambda
S inverse, right?

00:34:30.840 --> 00:34:35.290
So it's S, which is
this eigenvector matrix,

00:34:35.290 --> 00:34:41.090
it's the lambda, which
is the eigenvalue matrix,

00:34:41.090 --> 00:34:44.790
it's the S inverse, whatever
that turns out to be,

00:34:44.790 --> 00:34:46.369
let me just leave it as inverse.

00:34:49.489 --> 00:34:52.000
That has to be it, right?

00:34:52.000 --> 00:34:54.409
Because if we went in
the other direction,

00:34:54.409 --> 00:35:00.150
that matrix S would diagonalize
A to produce lambda.

00:35:00.150 --> 00:35:02.340
So it's S lambda S inverse.

00:35:02.340 --> 00:35:03.200
Good.

00:35:03.200 --> 00:35:06.150
OK, ready for number three.

00:35:06.150 --> 00:35:13.120
A real matrix that cannot
be factored into A --

00:35:13.120 --> 00:35:17.570
I'm looking for a matrix
A that never could

00:35:17.570 --> 00:35:24.000
equal B transpose B, for any B.

00:35:24.000 --> 00:35:27.840
A two-by-two matrix that could
not be factored in the form B

00:35:27.840 --> 00:35:30.540
transpose B.

00:35:30.540 --> 00:35:33.470
So all you have to do is think,
well, what does B transpose B,

00:35:33.470 --> 00:35:37.060
look like, and then pick
something different.

00:35:37.060 --> 00:35:38.530
What do you suggest?

00:35:42.690 --> 00:35:43.910
Let's see.

00:35:43.910 --> 00:35:46.850
What shall we take for
a matrix that could not

00:35:46.850 --> 00:35:49.430
have this form, B transpose B.

00:35:49.430 --> 00:35:51.770
Well, what do we know
about B transpose B?

00:35:51.770 --> 00:35:54.090
It's always symmetric.

00:35:54.090 --> 00:35:56.280
So just give me any
non-symmetric matrix,

00:35:56.280 --> 00:35:58.460
it couldn't possibly
have that form.

00:35:58.460 --> 00:35:59.160
OK.

00:35:59.160 --> 00:36:02.030
And let me ask the fourth
part of this question --

00:36:02.030 --> 00:36:07.310
a matrix that has
orthogonal eigenvectors,

00:36:07.310 --> 00:36:12.820
but it's not symmetric.

00:36:12.820 --> 00:36:15.900
What matrices have
orthogonal eigenvectors,

00:36:15.900 --> 00:36:17.655
but they're not
symmetric matrices?

00:36:20.560 --> 00:36:28.240
What other families of matrices
have orthogonal eigenvectors?

00:36:28.240 --> 00:36:33.130
We know symmetric matrices
do, but others, also.

00:36:33.130 --> 00:36:40.030
So I'm looking for
orthogonal eigenvectors,

00:36:40.030 --> 00:36:42.540
and, what do you suggest?

00:36:47.220 --> 00:36:51.460
The matrix could
be skew-symmetric.

00:36:51.460 --> 00:36:55.100
It could be an
orthogonal matrix.

00:36:55.100 --> 00:36:59.590
It could be symmetric,
but that was too easy,

00:36:59.590 --> 00:37:01.140
so I ruled that out.

00:37:01.140 --> 00:37:11.050
It could be skew-symmetric
like one minus one, like that.

00:37:11.050 --> 00:37:19.680
Or it could be an orthogonal
matrix like cosine sine,

00:37:19.680 --> 00:37:22.250
minus sine, cosine.

00:37:22.250 --> 00:37:28.240
All those matrices would
have complex orthogonal

00:37:28.240 --> 00:37:30.620
eigenvectors.

00:37:30.620 --> 00:37:35.540
But they would be orthogonal,
and so those examples are fine.

00:37:35.540 --> 00:37:36.270
OK.

00:37:36.270 --> 00:37:45.660
We can continue a little longer
if you would like to, with

00:37:45.660 --> 00:37:47.260
these --

00:37:47.260 --> 00:37:48.620
from this exam.

00:37:48.620 --> 00:37:49.990
From these exams.

00:37:49.990 --> 00:37:50.630
Least squares?

00:37:53.570 --> 00:37:56.180
OK, here's a least
squares problem in which,

00:37:56.180 --> 00:38:00.130
to make life quick,
I've given the answer --

00:38:00.130 --> 00:38:03.240
it's like Jeopardy!, right?

00:38:03.240 --> 00:38:06.240
I just give the answer,
and you give the question.

00:38:06.240 --> 00:38:07.480
OK.

00:38:07.480 --> 00:38:12.540
Whoops, sorry.

00:38:12.540 --> 00:38:17.710
Let's see, can I stay over
here for the next question?

00:38:17.710 --> 00:38:18.210
OK.

00:38:22.300 --> 00:38:22.980
least squares.

00:38:22.980 --> 00:38:28.370
So I'm giving you the problem,
one, one, one, zero, one, two,

00:38:28.370 --> 00:38:35.890
c d equals three, four, one,
and that's b, of course,

00:38:35.890 --> 00:38:37.290
this is Ax=b.

00:38:40.310 --> 00:38:43.040
And the least
squares solution --

00:38:43.040 --> 00:38:46.350
Maybe I put c hat
d hat to emphasize

00:38:46.350 --> 00:38:49.670
it's not the true solution.

00:38:49.670 --> 00:38:54.520
So the least square solution
-- the hats really go here --

00:38:54.520 --> 00:38:58.360
is eleven-thirds and minus one.

00:38:58.360 --> 00:39:01.470
Of course, you could have
figured that out in no time.

00:39:01.470 --> 00:39:05.390
So this year, I'll ask
you to do it, probably.

00:39:05.390 --> 00:39:09.250
But, suppose we're
given the answer,

00:39:09.250 --> 00:39:14.240
then let's just
remember what happened.

00:39:14.240 --> 00:39:16.050
OK, good question.

00:39:16.050 --> 00:39:21.070
What's the projection P of
this vector onto the column

00:39:21.070 --> 00:39:22.110
space of that matrix?

00:39:25.090 --> 00:39:29.350
So I'll write that
question down, one.

00:39:29.350 --> 00:39:30.530
What is P?

00:39:30.530 --> 00:39:31.490
The projection.

00:39:31.490 --> 00:39:41.290
The projection of b onto the
column space of A is what?

00:39:44.860 --> 00:39:50.030
Hopefully, that's what the
least squares problem solved.

00:39:50.030 --> 00:39:51.430
What is it?

00:39:54.430 --> 00:40:02.220
This was the best solution, it's
eleven-thirds times column one,

00:40:02.220 --> 00:40:07.760
plus -- or rather, minus
one times column two.

00:40:07.760 --> 00:40:08.260
Right?

00:40:08.260 --> 00:40:10.280
That's what least squares did.

00:40:10.280 --> 00:40:14.470
It found the combination
of the columns that

00:40:14.470 --> 00:40:16.350
was as close as possible to b.

00:40:16.350 --> 00:40:18.740
That's what least
squares was doing.

00:40:18.740 --> 00:40:20.450
It found the projection.

00:40:20.450 --> 00:40:21.960
OK?

00:40:21.960 --> 00:40:26.440
Secondly, draw the straight
line problem that corresponds to

00:40:26.440 --> 00:40:27.590
this system.

00:40:27.590 --> 00:40:32.060
So I guess that the straight
line fitting a straight line

00:40:32.060 --> 00:40:35.110
problem, we kind of recognize.

00:40:35.110 --> 00:40:37.410
So we recognize,
these are the heights,

00:40:37.410 --> 00:40:41.380
and these are the points,
and so at zero, one, two,

00:40:41.380 --> 00:40:46.420
the heights are three,
and at t equal to one,

00:40:46.420 --> 00:40:50.720
the height is four,
one, two, three, four,

00:40:50.720 --> 00:40:53.640
and at t equal to two,
the height is one.

00:40:56.180 --> 00:41:01.880
So I'm trying to fit
the best straight line

00:41:01.880 --> 00:41:04.470
through those points.

00:41:04.470 --> 00:41:04.970
God.

00:41:07.640 --> 00:41:10.360
I could fit a
triangle very well,

00:41:10.360 --> 00:41:16.130
but, I don't even know which
way the best straight line goes.

00:41:16.130 --> 00:41:19.340
Oh, I do know how it goes,
because there's the answer,yes.

00:41:19.340 --> 00:41:25.860
It has a height eleven-thirds,
and it has slope minus one,

00:41:25.860 --> 00:41:28.590
so it's something like that.

00:41:28.590 --> 00:41:29.370
Great.

00:41:29.370 --> 00:41:29.930
OK.

00:41:29.930 --> 00:41:36.680
Now, finally -- and this
completes the course --

00:41:36.680 --> 00:41:41.140
find a different vector
b, not all zeroes,

00:41:41.140 --> 00:41:45.310
for which the least square
solution would be zero.

00:41:45.310 --> 00:41:50.140
So I want you to
find a different B

00:41:50.140 --> 00:41:54.915
so that the least square
solution changes to all zeroes.

00:42:00.530 --> 00:42:04.360
So tell me what I'm
really looking for here.

00:42:04.360 --> 00:42:08.640
I'm looking for a b where the
best combination of these two

00:42:08.640 --> 00:42:11.780
columns is the zero combination.

00:42:11.780 --> 00:42:15.200
So what kind of a
vector b I looking for?

00:42:15.200 --> 00:42:16.780
I'm looking for
a vector b that's

00:42:16.780 --> 00:42:19.230
orthogonal to those columns.

00:42:19.230 --> 00:42:20.810
It's orthogonal
to those columns,

00:42:20.810 --> 00:42:22.810
it's orthogonal to
the column space,

00:42:22.810 --> 00:42:24.880
the best possible answer is

00:42:24.880 --> 00:42:25.550
zero.

00:42:25.550 --> 00:42:29.870
So a vector b that's orthogonal
to those columns -- let's see,

00:42:29.870 --> 00:42:35.830
maybe one of those minus two
of those, and one of those?

00:42:35.830 --> 00:42:38.460
That would be orthogonal
to those columns,

00:42:38.460 --> 00:42:42.720
and the best vector
would be zero, zero.

00:42:42.720 --> 00:42:43.470
OK.

00:42:43.470 --> 00:42:46.590
So that's as many questions
as I can do in an hour,

00:42:46.590 --> 00:42:50.490
but you get three hours,
and, let me just say,

00:42:50.490 --> 00:42:56.100
as I've said by e-mail, thanks
very much for your patience

00:42:56.100 --> 00:43:00.060
as this series of
lectures was videotaped,

00:43:00.060 --> 00:43:04.490
and, thanks for filling
out these forms,

00:43:04.490 --> 00:43:08.250
maybe just leave them on the
table up there as you go out --

00:43:08.250 --> 00:43:10.901
and above all, thanks
for taking the course.

00:43:10.901 --> 00:43:11.400
Thank you.

00:43:11.400 --> 00:43:12.950
Thanks.