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OK, this is quiz review day.
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The quiz coming up on
Wednesday will before this
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lecture the quiz will be this
hour one o'clock Wednesday
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in Walker, top floor of Walker,
closed book, all normal.
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I wrote down what we've
covered in this second part
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of the course, and actually
I'm impressed as I write it.
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so that's chapter
four on orthogonality
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and you're remembering these --
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what this is suggesting,
these are those columns are
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orthonormal vectors, and then
we call that matrix Q and the --
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what's the key -- how do we
state the fact that those v-
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those columns are
orthonormal in terms of Q,
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it means that Q transpose
Q is the identity.
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So that's the matrix
statement of the --
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of the property that the
columns are orthonormal,
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the dot products are
either one or zero,
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and then we computed the
projections onto lines and onto
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subspaces, and we used that
to solve problems Ax=b in --
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in the least square sense,
when there was no solution,
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we found the best solution.
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And then finally this
Graham-Schmidt idea,
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which takes independent vectors
and lines them up, takes --
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subtracts off the
projections of the part
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you've already done, so that
the new part is orthogonal
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and so it takes a basis
to an orthonormal basis.
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And you -- those calculations
involve square roots a lot
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because you're making
things unit vectors,
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but you should know that step.
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OK, for determinants,
the three big --
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the big picture is the
properties of the determinant,
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one to three d- properties
one, two and three, d-
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that define the determinant,
and then four, five,
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six through ten
were consequences.
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Then the big formula that has
n factorial terms, half of them
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have plus signs and half
minus signs, and then
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the cofactor formula.
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So and which led us to a
formula for the inverse.
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And finally, just
so you know what's
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covered in from chapter
three, it's section six point
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one and two, so that's the
basic idea of eigenvalues
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and eigenvectors, the
equation for the eigenvalues,
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the mechanical step, this
is really Ax equal lambda
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x for all n
eigenvectors at once,
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if we have n independent
eigenvectors,
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and then using that to
compute powers of a matrix.
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So you notice the differential
equations not on this list,
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because that's six point three,
that that's for the third quiz.
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OK.
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Shall I what I usually do for
review is to take an old exam
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and just try to pick out
questions that are significant
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and write them quickly
on the board, shall I --
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shall I proceed that way again?
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This -- this exam is really old.
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November nineteen --
nineteen eighty-four,
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so that was before
the Web existed.
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So not only were the
lectures not on the Web,
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nobody even had a
Web page, my God.
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OK, so can I nevertheless
linear algebra
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was still as great as ever.
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So may I and that wasn't meant
to be a joke, OK, all right,
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so let me just take these
questions as they come.
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All right.
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OK.
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So the first question's
about projections.
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It says we're given
the line, the --
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the vector a is the
vector two, one, two,
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and I want to find the
projection matrix P that
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projects onto the
line through a.
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So my picture is, well I'm in
three dimensions, of course,
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so there's a vector, two --
there's the vector a, two, one,
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two, let me draw the
whole line through it,
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and I want to project any
vector b onto that line,
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and I'm looking for
the projection matrix.
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So -- so this -- the projection
matrix is the matrix that I
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multiply b by to get here.
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And I guess this first
part, this is just part one
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a, I'm really asking you the
-- the quick way to answer,
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to find P, is just to
remember what the formula is.
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And -- and we're in -- we're
projecting onto a line,
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so our formula, our -- our
usual formula is AA transpose A
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inverse A transpose, but
now A is just a column,
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one-column matrix, so it'll be
just a, so I'll just call it
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little a, little a transpose,
and this is just a number now,
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one by one, so I can put
it in the denominator,
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so there's our --
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that's really what we
want to remember, and --
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and using that two, one,
two, what will I get?
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I'm dividing by --
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what's the length
squared of that vector?
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So what's a transpose a?
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Looks like nine, and what's
the matrix, well, I'm --
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I'm doing two, one,
two against two,
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one, two, so it's one-ninth of
this matrix four, two, four,
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two, one, two, four, two, four.
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Now the next part asked
about eigenvalues.
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So you see we're --
since we're learning,
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we know a lot more now, we can
make connections between these
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chapters, so what's
the eigenvector,
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what are the eigenvalues of P?
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I could ask what's
the rank of P. What's
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the rank of that matrix?
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Uh -- one.
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Rank is one.
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What's the column space?
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If I apply P to all vectors
then I fill up the column space,
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it's combinations of the
columns, so what's the column
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space?
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Well, it's just this line.
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The column space is the
line through two, one, two.
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And now what's the eigenvalue?
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So, or since that
matrix has rank one,
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so tell me the eigenvalues
of this matrix.
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It's a singular
matrix, so it certainly
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has an eigenvalue zero.
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Actually, the rank
is only one, so that
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means that there
like going to be
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t- a two-dimensional null
space, there'll be at least two,
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this lambda equals zero
will be repeated twice
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because I can find two
independent eigenvectors
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with lambda equals zero, and
then of course since it's
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got three eigenvalues,
what's the third one?
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It's one.
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How do I know it's one?
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Either from the trace, which is
nine over nine, which is one,
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or by remembering what --
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what is the eigenvector,
and actually now
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it's going to ask
for the eigenvector,
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so what's the eigenvector
for that eigenvalue?
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What's the eigenvector
for that eigenvalue?
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It's the -- it's the vector that
doesn't move, eigenvalue one,
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so the vector that doesn't move
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is a.
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This a is the --
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is -- is also the eigenvector
with lambda equal one,
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because if I apply the
projection matrix to a, I get
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a again.
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Everybody sees that if I apply
that matrix to a, can I do it
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in little letters, if I
apply that matrix to a,
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then I have a transpose
a canceling a transpose a
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and I get a again.
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So sure enough, Pa equals a.
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And the eigenvalue is one.
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OK.
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Good.
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now, actually it asks
you further to solve this
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difference equation, so
this will be -- this is --
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this is solve u(k+1)=Puk,
starting from u0 equal nine,
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nine, zero.
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And find uk.
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So -- so what's up?
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Shall we find u1 first of all?
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So just to get started.
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So what is u1?
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It's Pu0 of course.
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So if I do the projection of --
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of this vector onto the line,
so this is like my vector b now
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that I'm projecting
onto the line,
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I get a times a transpose
u0 over a transpose a.
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Well, one way or another I
just do this multiplication.
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but maybe this is the
easiest way to do it.
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a transpose, can I
remember what a is on
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this board?
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Two, one, two, so I'm projecting
onto the line through there.
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This is the projection,
it's P times the vector u0,
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so what do I have for a
transpose u0 looks like
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eighteen, looks
like twenty-seven,
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and a transpose a
we figured was nine,
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so it's three a, so that this
is the -- this is the x hat,
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the -- the multiple of
a, in -- in our formulas,
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and of course that's
six, three, six.
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So that's u1.
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Computed out directly.
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That's on the line through a and
it's the closest point to u0,
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and it's just Pu0.
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You just straightforward
multiplication produces that.
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OK.
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Now, what's u2?
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Well, u2 is Pu1, I agree.
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Do I need to compute that again?
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No, because once I'm already on
the line through A, uk will be,
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I could do the
projection k times,
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but it's enough
just to do it once.
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It's the same, it's the
same, six, three, six.
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So this is a case
where I could and --
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and actually on the quiz
if you see one of these,
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which could very well be there,
and it could very well be not
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a projection matrix, then we
would use all the eigenvalues
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and eigenvectors.
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Let's think for a moment,
how do you do those?
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M - the point of this small part
of a question was that when P
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is a projection matrix, so that
P squared equals P and P cubed
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equals P, then -- then we don't
need to get into the mechanics
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of all knowing all the other
eigenvalues and eigenvectors.
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We just can go directly.
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But if P was now some other
matrix, can you just --
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let's just remember from these
very recent lectures how you
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would proceed.
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00:12:52,560 --> 00:12:57,590
from these very recent lectures
we know that uk we would --
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we would expand u0 as a
combination of eigenvectors.
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Let me leave -- yeah, as a
combination of eigenvectors,
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c1x1, some multiple of
the second eigenvector,
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00:13:12,650 --> 00:13:15,470
some multiple of the
third eigenvector,
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and then A to the ku0
would be c1, so this --
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we have to find these numbers
here, that's the work actually.
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00:13:27,270 --> 00:13:30,280
The work is find
the eigenvalues,
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00:13:30,280 --> 00:13:33,500
find the eigenvectors and
find the c-s because they all
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come into the formula.
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00:13:35,320 --> 00:13:50,230
We have -- so -- so to do
this, you can see what you have
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00:13:50,230 --> 00:13:51,280
to compute.
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00:13:51,280 --> 00:13:53,320
You have to compute
the eigenvalues,
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you have to compute
the eigenvectors,
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and then to match u0
you compute the c-s,
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00:13:59,230 --> 00:14:01,210
and then you've got it.
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00:14:01,210 --> 00:14:06,610
So it's -- it's just that's a
formula that shows what pieces
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00:14:06,610 --> 00:14:07,810
we need.
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00:14:07,810 --> 00:14:12,550
And what would actually happen
in the case of this projection
217
00:14:12,550 --> 00:14:13,070
matrix?
218
00:14:13,070 --> 00:14:15,310
If this A is a
projection matrix,
219
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then a couple of
eigenvalues are zero.
220
00:14:19,050 --> 00:14:21,200
That's why we just
throw those away.
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The other eigenvalue was a one,
so that we got the same thing
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00:14:27,780 --> 00:14:30,890
every time, c3x3.
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00:14:30,890 --> 00:14:33,050
From the first
time, second time,
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00:14:33,050 --> 00:14:37,160
third, all iterations pro-
left us with this constant,
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left us right here
at six, three, six.
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But maybe I take --
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I'm taking this chance
to remind you of what
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00:14:43,840 --> 00:14:47,900
to do for other matrices.
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OK.
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So that's part way through.
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OK.
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00:14:53,460 --> 00:14:55,980
The next question in
nineteen eighty-four
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00:14:55,980 --> 00:15:01,030
is fitting a straight
line to points.
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00:15:01,030 --> 00:15:04,712
And actually a straight
line through the origin.
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A straight line
through the origin.
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So can I go to question two?
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00:15:10,120 --> 00:15:15,360
So this is fitting a straight
line to these points, can I --
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00:15:15,360 --> 00:15:26,580
I'll just give you the points
at t=1 the y is four, at t=2,
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00:15:26,580 --> 00:15:30,710
y is five, at t=3, y is eight.
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00:15:34,190 --> 00:15:42,630
So we've got points one, two,
three, four, five, and eight.
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00:15:42,630 --> 00:15:45,720
And I'm trying to fit a
straight line through the origin
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to these three values.
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00:15:48,540 --> 00:15:57,170
OK, so my equation that
I'm allowing myself
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00:15:57,170 --> 00:16:00,950
is just y equal Dt.
245
00:16:00,950 --> 00:16:03,470
So I have only one unknown.
246
00:16:03,470 --> 00:16:04,950
One degree of freedom.
247
00:16:04,950 --> 00:16:07,730
One parameter D.
248
00:16:07,730 --> 00:16:12,510
So I'm expecting to end up
my matrix so my -- my --
249
00:16:12,510 --> 00:16:14,510
when I try to --
250
00:16:14,510 --> 00:16:17,440
when I try to fit a
straight line, that
251
00:16:17,440 --> 00:16:19,660
goes through the origin,
that's because it goes
252
00:16:19,660 --> 00:16:22,570
through the origin, I've
lost the constant c here,
253
00:16:22,570 --> 00:16:27,330
so I have just this should
be a quick calculation.
254
00:16:27,330 --> 00:16:32,590
and I can write down the three
equations that -- that --
255
00:16:32,590 --> 00:16:33,200
that would --
256
00:16:33,200 --> 00:16:36,990
I'd like to solve if the line
went through the points, that's
257
00:16:36,990 --> 00:16:38,080
a good start.
258
00:16:38,080 --> 00:16:40,015
Because that
displays the matrix.
259
00:16:43,410 --> 00:16:45,270
So can I continue that problem?
260
00:16:45,270 --> 00:16:48,760
We would like to solve--
261
00:16:48,760 --> 00:16:54,200
so y is Dt, so I'd like
to solve D times one times
262
00:16:54,200 --> 00:17:00,030
D equals four, two times
D equals five and three
263
00:17:00,030 --> 00:17:02,690
times D equals eight.
264
00:17:02,690 --> 00:17:04,500
That would be perfection.
265
00:17:04,500 --> 00:17:09,150
If I could find such
a D, then the line y
266
00:17:09,150 --> 00:17:13,349
equal Dt would satisfy
all three equations,
267
00:17:13,349 --> 00:17:16,500
would go through all three
points, but it doesn't exist.
268
00:17:16,500 --> 00:17:19,140
So -- so I have to
solve this -- so the --
269
00:17:19,140 --> 00:17:23,680
my matrix is now you can see my
matrix, it just has one column.
270
00:17:23,680 --> 00:17:27,589
Multiplying a scalar D.
271
00:17:27,589 --> 00:17:30,260
And you can see the
right-hand side.
272
00:17:30,260 --> 00:17:31,500
This is my Ax=b.
273
00:17:34,280 --> 00:17:38,860
I don't need three equals signs
now because I've got vectors.
274
00:17:38,860 --> 00:17:39,620
OK.
275
00:17:39,620 --> 00:17:43,610
There's Ax=b and you
take it from there.
276
00:17:43,610 --> 00:17:47,570
You the -- the best x will
be -- will come from --
277
00:17:47,570 --> 00:17:49,510
so what's the --
the key equation?
278
00:17:49,510 --> 00:17:54,960
So this is the A, this is
the Ax hat equal b equation.
279
00:17:54,960 --> 00:17:57,900
Well, Ax=b.
280
00:17:57,900 --> 00:18:00,590
And what's the
equation for x hat?
281
00:18:00,590 --> 00:18:08,490
The best D, so to find
the best D, the best x,
282
00:18:08,490 --> 00:18:14,660
the equation is A
transpose A, the best D,
283
00:18:14,660 --> 00:18:20,900
is A transpose times
the right-hand side.
284
00:18:20,900 --> 00:18:24,340
This is all coming from
projection on a line, our --
285
00:18:24,340 --> 00:18:26,620
our matrix only has one column.
286
00:18:26,620 --> 00:18:30,070
So A transpose A would
be maybe fourteen,
287
00:18:30,070 --> 00:18:35,410
D hat, and A transpose
b I'm getting four, ten,
288
00:18:35,410 --> 00:18:37,520
and twenty-four.
289
00:18:37,520 --> 00:18:38,590
Is that right?
290
00:18:38,590 --> 00:18:40,950
Four, ten and twenty-four.
291
00:18:40,950 --> 00:18:43,000
So thirty-eight.
292
00:18:43,000 --> 00:18:47,530
So that tells me the
best D hat is D hat
293
00:18:47,530 --> 00:18:52,150
is thirty-eight over fourteen.
294
00:18:52,150 --> 00:18:53,310
OK.
295
00:18:53,310 --> 00:18:54,930
Fine.
296
00:18:54,930 --> 00:18:57,620
All right.
297
00:18:57,620 --> 00:18:59,240
so we found the best line.
298
00:18:59,240 --> 00:19:01,830
And now here's a --
here's the next question.
299
00:19:01,830 --> 00:19:07,050
What vector did I just
project onto what line?
300
00:19:07,050 --> 00:19:12,210
See in this section on least
squares here's the key point,
301
00:19:12,210 --> 00:19:12,740
I'm --
302
00:19:12,740 --> 00:19:14,700
I'm asking you to think
of the least squares
303
00:19:14,700 --> 00:19:17,240
problem in two ways.
304
00:19:17,240 --> 00:19:18,890
Two different pictures.
305
00:19:18,890 --> 00:19:20,080
Two different graphs.
306
00:19:20,080 --> 00:19:22,320
One graph is this.
307
00:19:22,320 --> 00:19:28,060
This is a graph in the -- in the
b -- in the tb plane, ty plane.
308
00:19:28,060 --> 00:19:30,900
The -- the -- the line itself.
309
00:19:30,900 --> 00:19:32,940
The other picture I'm
asking you to think of
310
00:19:32,940 --> 00:19:35,050
is like my projection picture.
311
00:19:35,050 --> 00:19:38,610
What -- what projection --
what -- what vector I --
312
00:19:38,610 --> 00:19:42,940
I projecting onto what line or
what subspace when I -- when I
313
00:19:42,940 --> 00:19:44,050
do this?
314
00:19:44,050 --> 00:19:48,700
So the -- my second picture is
a projection picture that --
315
00:19:48,700 --> 00:19:51,280
that sees the whole
thing with vectors.
316
00:19:51,280 --> 00:19:53,970
Here's my vector of course
that I'm projecting.
317
00:19:53,970 --> 00:20:06,800
I'm projecting that vector b
onto the column space of A.
318
00:20:06,800 --> 00:20:15,800
Of if you like -- it's just
a line onto that's the line
319
00:20:15,800 --> 00:20:18,620
it's just a line, of course.
320
00:20:18,620 --> 00:20:21,070
That's what this
calculation is doing.
321
00:20:21,070 --> 00:20:25,595
This is computing the best D,
which is -- this is the x hat.
322
00:20:30,230 --> 00:20:34,090
So -- so seeing it as a
projection means I don't see
323
00:20:34,090 --> 00:20:36,570
the projection in
this figure, right?
324
00:20:36,570 --> 00:20:39,060
In this figure I'm not
projecting those points
325
00:20:39,060 --> 00:20:41,760
onto that line or
anything of the sort.
326
00:20:41,760 --> 00:20:48,430
The projection s-picture for --
for least squares is in the --
327
00:20:48,430 --> 00:20:51,900
in the space where b
lies, the whole vector b,
328
00:20:51,900 --> 00:20:54,870
and the columns of A.
329
00:20:54,870 --> 00:21:02,570
And then the x is
the best combination
330
00:21:02,570 --> 00:21:04,600
that gives the projection.
331
00:21:04,600 --> 00:21:05,100
OK.
332
00:21:05,100 --> 00:21:07,980
So that's a chance
to tell me that.
333
00:21:07,980 --> 00:21:08,760
OK.
334
00:21:08,760 --> 00:21:12,310
I'll go -- OK now
finally in orthogonality
335
00:21:12,310 --> 00:21:14,610
there's the Graham-Schmidt idea.
336
00:21:14,610 --> 00:21:20,230
So that's problem two D here.
337
00:21:20,230 --> 00:21:24,840
It asks me if I have two
vectors, a1 equal one, two,
338
00:21:24,840 --> 00:21:32,650
three, and a2 equal
one, one, one, find
339
00:21:32,650 --> 00:21:37,650
two orthogonal
vectors in that plane.
340
00:21:37,650 --> 00:21:43,480
So those two vectors give
a plane, they give a plane.
341
00:21:43,480 --> 00:21:47,860
Which is of course the --
the column space of the --
342
00:21:47,860 --> 00:21:50,710
of the matrix.
343
00:21:50,710 --> 00:21:55,050
And I'm looking for
an orthogonal basis
344
00:21:55,050 --> 00:21:55,710
for that plane.
345
00:21:55,710 --> 00:21:58,300
So I'm looking for two
orthogonal vectors.
346
00:21:58,300 --> 00:22:02,100
And of course there
are lots of --
347
00:22:02,100 --> 00:22:05,330
I mean, I've got a plane there.
348
00:22:05,330 --> 00:22:08,700
If I get one orthogonal
pair, I can rotate it.
349
00:22:08,700 --> 00:22:10,910
There's not just
one answer here.
350
00:22:10,910 --> 00:22:16,140
But Graham-Schmidt says OK,
start with the first vector,
351
00:22:16,140 --> 00:22:19,840
and let that be --
and keep that one.
352
00:22:19,840 --> 00:22:23,300
And then take the second
one orthogonal to this.
353
00:22:23,300 --> 00:22:27,400
So -- so Graham-Schmidt says
start with this one and then
354
00:22:27,400 --> 00:22:31,220
make a second vector B, can
I call that second vector B,
355
00:22:31,220 --> 00:22:35,970
this is going to be orthogonal
to, so perpendicular to a1.
356
00:22:35,970 --> 00:22:40,650
If I can with my chalk
create the key equation.
357
00:22:40,650 --> 00:22:46,740
This vector B is going
to be this one, one, one,
358
00:22:46,740 --> 00:22:50,720
but that one, one -- one, one,
one is not perpendicular to a1,
359
00:22:50,720 --> 00:22:54,340
so I have to subtract
off its projection,
360
00:22:54,340 --> 00:23:00,820
I have to subtract off the B,
the -- the B trans- ye the B --
361
00:23:00,820 --> 00:23:07,780
the -- the I should say the a1
transpose b over a1 transpose
362
00:23:07,780 --> 00:23:11,400
a1, that multiple of
a1, I've got to remove.
363
00:23:14,990 --> 00:23:17,090
So I just have to
compute what that is,
364
00:23:17,090 --> 00:23:21,440
and I get ano- I get a vector
B that's orthogonal to a1.
365
00:23:21,440 --> 00:23:24,440
It's the -- it's --
366
00:23:24,440 --> 00:23:29,630
it's the original vector
minus its projection.
367
00:23:29,630 --> 00:23:32,110
Oh, so what is --
368
00:23:32,110 --> 00:23:33,490
I mean this to be a2.
369
00:23:36,130 --> 00:23:39,430
So I'm projecting a2
onto the line through a1.
370
00:23:39,430 --> 00:23:40,190
Yeah.
371
00:23:40,190 --> 00:23:42,190
That's the part that I
don't want because that's
372
00:23:42,190 --> 00:23:45,250
in the direction I already
have, so I subtract off
373
00:23:45,250 --> 00:23:48,170
that projection and I
get the part I want,
374
00:23:48,170 --> 00:23:50,240
the orthogonal part.
375
00:23:50,240 --> 00:23:52,540
So that's the
Graham-Schmidt thing
376
00:23:52,540 --> 00:23:54,640
and we can put numbers in.
377
00:23:54,640 --> 00:23:55,370
OK.
378
00:23:55,370 --> 00:24:00,710
one, one, one take away
a1 transpose a2 is six,
379
00:24:00,710 --> 00:24:06,900
a1 transpose a1 is
fourteen,multiplying a1.
380
00:24:06,900 --> 00:24:13,400
And that gives us the
new orthogonal vector B.
381
00:24:13,400 --> 00:24:15,340
Because I only ask for
orthogonal right now,
382
00:24:15,340 --> 00:24:19,200
I don't have to divide
by the length which
383
00:24:19,200 --> 00:24:20,590
will involve a square root.
384
00:24:20,590 --> 00:24:23,210
OK.
385
00:24:23,210 --> 00:24:23,940
Third question.
386
00:24:27,540 --> 00:24:28,677
Third question.
387
00:24:28,677 --> 00:24:29,510
All right, let me --
388
00:24:29,510 --> 00:24:33,270
I'll move this board up.
389
00:24:33,270 --> 00:24:37,930
third question will probably
be about eigenvalues.
390
00:24:37,930 --> 00:24:38,610
OK.
391
00:24:38,610 --> 00:24:41,200
Three.
392
00:24:41,200 --> 00:24:44,020
This is a four-by-four matrix.
393
00:24:44,020 --> 00:24:47,320
Its eigenvalues are lambda
one, lambda two, lambda three,
394
00:24:47,320 --> 00:24:50,310
lambda four.
395
00:24:50,310 --> 00:24:52,930
Question one.
396
00:24:52,930 --> 00:24:55,870
What's the condition
on the lambdas so
397
00:24:55,870 --> 00:24:59,380
that the matrix is invertible?
398
00:24:59,380 --> 00:24:59,990
OK.
399
00:24:59,990 --> 00:25:02,510
So under what conditions
on the lambdas
400
00:25:02,510 --> 00:25:05,900
will the matrix be invertible?
401
00:25:05,900 --> 00:25:08,850
So that's easy.
402
00:25:08,850 --> 00:25:17,780
Invertible if what's the
condition on the lambdas?
403
00:25:17,780 --> 00:25:19,370
None of them are zero.
404
00:25:19,370 --> 00:25:23,900
A zero eigenvalue would mean
something in the null space
405
00:25:23,900 --> 00:25:28,850
would mean a solution to
Ax=0x, but we're invertible,
406
00:25:28,850 --> 00:25:32,170
so none of them is
zero, the product --
407
00:25:32,170 --> 00:25:34,120
however you want to say, no --
408
00:25:34,120 --> 00:25:38,510
no zero eigenvalues.
409
00:25:38,510 --> 00:25:39,170
Good.
410
00:25:39,170 --> 00:25:43,200
OK, what's the
determinant of A inverse?
411
00:25:43,200 --> 00:25:45,555
The determinant of A inverse?
412
00:25:49,280 --> 00:25:52,120
So where is that
going to come from?
413
00:25:52,120 --> 00:25:55,470
Well, if we knew the
eigenvalues of A inverse,
414
00:25:55,470 --> 00:25:59,780
we could multiply them together
to find the determinant.
415
00:25:59,780 --> 00:26:02,110
And we do know the
eigenvalues of A inverse.
416
00:26:02,110 --> 00:26:04,370
What are they?
417
00:26:04,370 --> 00:26:08,580
They're just one
over lambda one times
418
00:26:08,580 --> 00:26:11,470
one over lambda two, that's
the second eigenvalue,
419
00:26:11,470 --> 00:26:13,640
the third eigenvalue and the
420
00:26:13,640 --> 00:26:14,760
fourth.
421
00:26:14,760 --> 00:26:18,770
So the product of the four
eigenvalues of the inverse
422
00:26:18,770 --> 00:26:21,450
will give us the
determinant of the inverse.
423
00:26:21,450 --> 00:26:21,950
Fine.
424
00:26:21,950 --> 00:26:23,110
OK.
425
00:26:23,110 --> 00:26:39,600
And what's the
trace of A plus I?
426
00:26:39,600 --> 00:26:41,170
So what do we know about trace?
427
00:26:44,160 --> 00:26:46,510
It's the sum down
the diagonal, but we
428
00:26:46,510 --> 00:26:48,630
don't know what our matrix is.
429
00:26:48,630 --> 00:26:51,960
The trace is also the
sum of the eigenvalues,
430
00:26:51,960 --> 00:26:55,710
and we do know the
eigenvalues of A plus I.
431
00:26:55,710 --> 00:26:57,410
So we just add them up.
432
00:26:57,410 --> 00:27:02,520
So what -- what's the first
eigenvalue of A plus I?
433
00:27:02,520 --> 00:27:05,610
When the matrix A has
eigenvalues lambda one, two,
434
00:27:05,610 --> 00:27:08,030
three and four,
then the eigenvalues
435
00:27:08,030 --> 00:27:11,610
if I add the identity, that
moves all the eigenvalues
436
00:27:11,610 --> 00:27:17,450
by one, so I just add
up lambda one plus one,
437
00:27:17,450 --> 00:27:22,520
lambda two plus one, and so on,
lambda three plus one, lambda
438
00:27:22,520 --> 00:27:27,180
four plus one, so it's lambda
one plus lambda two plus lambda
439
00:27:27,180 --> 00:27:32,180
three plus lambda
four plus four.
440
00:27:32,180 --> 00:27:34,850
Right.
441
00:27:34,850 --> 00:27:38,110
That movement by the identity
moved all the eigenvalues
442
00:27:38,110 --> 00:27:42,290
by one, so it moved the
whole trace by four.
443
00:27:42,290 --> 00:27:45,690
So it was the trace
of A plus four more.
444
00:27:45,690 --> 00:27:46,880
OK.
445
00:27:46,880 --> 00:27:48,240
Let's see.
446
00:27:48,240 --> 00:27:51,600
We may be finished this
quiz twenty minutes early.
447
00:27:51,600 --> 00:27:52,420
No.
448
00:27:52,420 --> 00:27:54,700
There's another question.
449
00:27:54,700 --> 00:27:57,590
Oh, God, OK.
450
00:27:57,590 --> 00:27:59,270
How did this class ever do it?
451
00:27:59,270 --> 00:28:02,710
Well, you'll see.
you'll be able to do it.
452
00:28:02,710 --> 00:28:03,510
OK.
453
00:28:03,510 --> 00:28:06,950
this has got to be a
determinant question.
454
00:28:06,950 --> 00:28:09,290
All right.
455
00:28:09,290 --> 00:28:13,810
More determinants and cofactors
and big formula question.
456
00:28:13,810 --> 00:28:14,430
OK.
457
00:28:14,430 --> 00:28:18,760
Let me do that.
458
00:28:18,760 --> 00:28:22,820
So it's about a matrix, a --
a whole family of matrices.
459
00:28:22,820 --> 00:28:26,200
Here's the four-by-four one.
460
00:28:26,200 --> 00:28:29,990
The four-by-four one is, and
-- and all the matrices in this
461
00:28:29,990 --> 00:28:35,240
family are tridiagonal with --
462
00:28:35,240 --> 00:28:38,230
with ones.
463
00:28:38,230 --> 00:28:40,110
Otherwise zeroes.
464
00:28:40,110 --> 00:28:43,480
So that's the pattern, and
we've seen this matrix.
465
00:28:43,480 --> 00:28:44,840
OK.
466
00:28:44,840 --> 00:28:47,850
So the -- it's tridiagonal
with ones on the diagonal,
467
00:28:47,850 --> 00:28:52,790
ones above and ones below, and
you see the general formula An,
468
00:28:52,790 --> 00:28:58,780
so I'll use Dn for
the determinant of An.
469
00:28:58,780 --> 00:28:59,950
OK.
470
00:28:59,950 --> 00:29:00,990
All right.
471
00:29:00,990 --> 00:29:02,950
So I'm going to do a --
472
00:29:02,950 --> 00:29:14,740
the first question is use
cofactors to show that Dn is
473
00:29:14,740 --> 00:29:20,680
something times D(n-1) plus
something times D(n-2).
474
00:29:20,680 --> 00:29:23,150
And find those somethings.
475
00:29:23,150 --> 00:29:23,650
OK.
476
00:29:26,310 --> 00:29:31,020
So this -- the fact that it's
tridiagonal with these constant
477
00:29:31,020 --> 00:29:37,290
diagonals means that there
is such a recurrence formula.
478
00:29:37,290 --> 00:29:39,660
And so the first
question is find it.
479
00:29:39,660 --> 00:29:41,810
Well, what's the
recurrence formula?
480
00:29:41,810 --> 00:29:43,550
OK, how does it go?
481
00:29:43,550 --> 00:29:47,390
So I'll use cofactors
along the first row.
482
00:29:47,390 --> 00:29:51,570
So I take that number
times its cofactor.
483
00:29:51,570 --> 00:29:56,730
So it's one times its cofactor
and what is its cofactor?
484
00:29:56,730 --> 00:30:00,120
D(n-1), right, exactly,
the cofactor is this --
485
00:30:00,120 --> 00:30:03,510
is this guy uses up
row one and column one,
486
00:30:03,510 --> 00:30:07,670
so the cofactor is down
here, so it's one of those.
487
00:30:10,230 --> 00:30:12,550
OK, that's the
first cofactor term.
488
00:30:12,550 --> 00:30:17,090
Now the other cofactor
term is this guy.
489
00:30:17,090 --> 00:30:20,740
Which uses up row
one and column two
490
00:30:20,740 --> 00:30:24,870
and what's surprising
about that?
491
00:30:24,870 --> 00:30:30,660
When you use row one and column
two that brings in a minus.
492
00:30:30,660 --> 00:30:32,940
There'll be a minus
because the --
493
00:30:32,940 --> 00:30:38,490
the cofactor is this
determinant times minus one.
494
00:30:38,490 --> 00:30:44,440
The the one-two cofactor is
that determinant with its sign
495
00:30:44,440 --> 00:30:45,460
changed.
496
00:30:45,460 --> 00:30:45,960
OK.
497
00:30:45,960 --> 00:30:47,270
So I have to look
at that determinant
498
00:30:47,270 --> 00:30:48,645
and I have to
remember in my head
499
00:30:48,645 --> 00:30:50,620
a sign is going to get changed.
500
00:30:50,620 --> 00:30:51,160
OK.
501
00:30:51,160 --> 00:30:54,970
Now how do I do
that determinant?
502
00:30:54,970 --> 00:30:58,180
How do I make that one clear?
503
00:30:58,180 --> 00:31:01,810
I -- the -- the neat way to
do is -- is here I see I --
504
00:31:01,810 --> 00:31:05,570
I'll use cofactors
down the first column.
505
00:31:05,570 --> 00:31:08,880
Because the first column is
all zeroes except for that one,
506
00:31:08,880 --> 00:31:13,220
so this one is now --
and what's its cofactor?
507
00:31:13,220 --> 00:31:17,490
Within this three-by-three its
cofactor will be two-by-two,
508
00:31:17,490 --> 00:31:19,080
and what is it?
509
00:31:19,080 --> 00:31:20,970
It's this, right?
510
00:31:20,970 --> 00:31:25,730
So -- so that part is all gone,
so I'm taking that times its
511
00:31:25,730 --> 00:31:29,550
cofactor, then zero times
whatever its cofactor is,
512
00:31:29,550 --> 00:31:34,040
so it's really just one times
and what's this in the general
513
00:31:34,040 --> 00:31:35,510
n-by-n case?
514
00:31:35,510 --> 00:31:39,420
It's Dn minus two.
515
00:31:39,420 --> 00:31:43,660
But now so is this a plus or
sign or a minus sign, it's --
516
00:31:43,660 --> 00:31:48,360
it's just a one, because there's
a one from there and a one from
517
00:31:48,360 --> 00:31:50,050
there.
518
00:31:50,050 --> 00:31:52,930
And is it a plus or a minus?
519
00:31:52,930 --> 00:31:56,010
It's minus I guess because
there was a minus the first time
520
00:31:56,010 --> 00:31:57,970
and then the second
time it's a plus,
521
00:31:57,970 --> 00:32:00,240
so it's overall it's a minus.
522
00:32:00,240 --> 00:32:04,864
So there's my a and b
were one and minus one.
523
00:32:04,864 --> 00:32:05,530
Those constants.
524
00:32:05,530 --> 00:32:09,140
Th- that's the --
that's the recurrence.
525
00:32:09,140 --> 00:32:10,220
OK.
526
00:32:10,220 --> 00:32:19,180
And oh, then it asks you to then
it asks you to solve this thing
527
00:32:19,180 --> 00:32:22,070
first by writing it as a --
528
00:32:22,070 --> 00:32:25,120
as a system.
529
00:32:25,120 --> 00:32:28,200
So now I'd like to
know the solution.
530
00:32:28,200 --> 00:32:30,600
I -- I better know
how it starts, right?
531
00:32:30,600 --> 00:32:34,020
It starts with D1,
what was D1, that's
532
00:32:34,020 --> 00:32:39,550
just the one-by-one case, so
D1 is one, and what is D2?
533
00:32:39,550 --> 00:32:41,880
Just to get us started and
then this would give us
534
00:32:41,880 --> 00:32:45,520
D3, D4, and forever.
535
00:32:45,520 --> 00:32:48,390
D2 is this two-by-two
that I'm seeing here
536
00:32:48,390 --> 00:32:51,620
and that determinant
is obviously zero.
537
00:32:51,620 --> 00:32:57,410
So those little ones will start
the recurrence and then we take
538
00:32:57,410 --> 00:32:58,030
off.
539
00:32:58,030 --> 00:33:02,070
And then the idea is to
write this recurrence as --
540
00:33:02,070 --> 00:33:11,990
as a Dn, D(n-1) is some
matrix times the one before,
541
00:33:11,990 --> 00:33:16,613
the D(n-1), D(n-2).
542
00:33:20,250 --> 00:33:22,270
What's the matrix?
543
00:33:22,270 --> 00:33:26,570
You see, you remember this step
of taking a single second order
544
00:33:26,570 --> 00:33:31,090
equation and by introducing
a vector unknown to make it
545
00:33:31,090 --> 00:33:32,600
into a --
546
00:33:32,600 --> 00:33:35,600
to a first order system.
547
00:33:35,600 --> 00:33:36,200
OK.
548
00:33:36,200 --> 00:33:41,860
So Dn is one of Dn minus one
minus one, I think that --
549
00:33:41,860 --> 00:33:43,530
that goes in the
first row, right?
550
00:33:43,530 --> 00:33:45,640
From the equation above?
551
00:33:45,640 --> 00:33:48,200
And the second one is
this is the same as this,
552
00:33:48,200 --> 00:33:50,010
so one and zero are fine.
553
00:33:53,500 --> 00:33:55,330
So there's the matrix.
554
00:33:55,330 --> 00:33:55,920
OK.
555
00:33:55,920 --> 00:33:58,820
So now how do I proceed?
556
00:33:58,820 --> 00:34:01,590
We can guess what
this examiner's
557
00:34:01,590 --> 00:34:02,845
got in his little mind.
558
00:34:07,740 --> 00:34:09,120
well, find the eigenvalues.
559
00:34:12,760 --> 00:34:19,469
And actually it tells
us that the sixth power
560
00:34:19,469 --> 00:34:23,679
of these eigenvalues
turns out to be one.
561
00:34:23,679 --> 00:34:30,065
Uh, well, can -- can we get the
equation for the eigenvalues?
562
00:34:30,065 --> 00:34:31,940
Let's do it and let's
get a formula for them.
563
00:34:31,940 --> 00:34:32,880
OK.
564
00:34:32,880 --> 00:34:35,260
So what are the eigenvalues?
565
00:34:35,260 --> 00:34:39,590
I look at the -- the matrix,
this determinant one minus
566
00:34:39,590 --> 00:34:44,420
lambda and zero minus lambda,
and these guys are still there,
567
00:34:44,420 --> 00:34:49,400
I compute that determinant, I
get lambda squared minus lambda
568
00:34:49,400 --> 00:34:52,310
and then plus one.
569
00:34:52,310 --> 00:34:54,820
And I set that to zero.
570
00:34:54,820 --> 00:34:55,820
OK.
571
00:34:55,820 --> 00:34:59,540
So we're not Fibonacci here.
572
00:34:59,540 --> 00:35:02,930
We're -- we're not
seeing Fibonacci numbers.
573
00:35:02,930 --> 00:35:07,500
Because the sign -- we
had a sign change there.
574
00:35:07,500 --> 00:35:11,040
And it's not clear right
away whether these --
575
00:35:11,040 --> 00:35:13,380
whether this -- is it clear?
576
00:35:13,380 --> 00:35:18,400
Is this matrix
stable or unstable?
577
00:35:18,400 --> 00:35:21,000
When we take -- when we go
further and further out?
578
00:35:21,000 --> 00:35:23,520
Are these Ds increasing?
579
00:35:23,520 --> 00:35:25,050
Are they going to zero?
580
00:35:25,050 --> 00:35:27,860
Are they bouncing
around periodically?
581
00:35:27,860 --> 00:35:30,600
the answers have to be here.
582
00:35:30,600 --> 00:35:34,840
I would like to know how big
these lambdas are, right?
583
00:35:34,840 --> 00:35:37,880
And the point is probably
these -- let's -- let's see,
584
00:35:37,880 --> 00:35:38,930
what's lambda?
585
00:35:38,930 --> 00:35:42,780
From the quadratic
formula lambda is one,
586
00:35:42,780 --> 00:35:45,390
I switch the sign of
that, plus or minus
587
00:35:45,390 --> 00:35:50,570
the square root of one minus
4ac, I getting a minus three
588
00:35:50,570 --> 00:35:51,970
there?
589
00:35:51,970 --> 00:35:52,710
Over two.
590
00:35:58,860 --> 00:36:00,960
What's up?
591
00:36:00,960 --> 00:36:03,340
They're complex.
592
00:36:03,340 --> 00:36:08,240
The -- the eigenvalues are one
plus square root of three I
593
00:36:08,240 --> 00:36:15,600
over two and one minus square
root of three I over two.
594
00:36:15,600 --> 00:36:18,240
What's the magnitude of lambda?
595
00:36:18,240 --> 00:36:19,765
That's the key
point for stability.
596
00:36:22,560 --> 00:36:26,480
These are two numbers
in the complex plane.
597
00:36:26,480 --> 00:36:30,650
One plus some -- somewhere
here, and its complex conjugate
598
00:36:30,650 --> 00:36:32,820
there.
599
00:36:32,820 --> 00:36:37,580
I want to know how far from
the origin are those numbers.
600
00:36:37,580 --> 00:36:40,840
What's the magnitude of lambda?
601
00:36:40,840 --> 00:36:43,230
And do you see what it is?
602
00:36:43,230 --> 00:36:46,690
Do you recognize this
-- a number like that?
603
00:36:46,690 --> 00:36:50,220
Take the real part squared
and the imaginary part squared
604
00:36:50,220 --> 00:36:51,510
and add.
605
00:36:51,510 --> 00:36:53,840
What do you get?
606
00:36:53,840 --> 00:36:57,010
So the real part
squared is a quarter.
607
00:36:57,010 --> 00:36:59,990
The imaginary part
squared is three-quarters.
608
00:36:59,990 --> 00:37:01,200
They add to one.
609
00:37:01,200 --> 00:37:06,090
That's a number with --
that's on the unit circle.
610
00:37:06,090 --> 00:37:08,540
That's an e to the i theta.
611
00:37:08,540 --> 00:37:10,180
That's a cos(theta)+isin(theta).
612
00:37:10,180 --> 00:37:14,650
And what's theta?
613
00:37:14,650 --> 00:37:19,330
This -- this is like a complex
number that's worth knowing,
614
00:37:19,330 --> 00:37:23,670
it's not totally
obvious but it's nice.
615
00:37:23,670 --> 00:37:27,480
That's -- I should see that
as cos(theta)+isin(theta),
616
00:37:27,480 --> 00:37:30,770
and the angle that would
do that is sixty degrees,
617
00:37:30,770 --> 00:37:32,580
pi over three.
618
00:37:32,580 --> 00:37:36,520
So that's a -- let me
improve my picture.
619
00:37:36,520 --> 00:37:39,990
So those -- that's e
to the i pi over six --
620
00:37:39,990 --> 00:37:40,920
pi over three.
621
00:37:40,920 --> 00:37:46,840
This is -- this number is e
to the i pi over three and e
622
00:37:46,840 --> 00:37:49,840
to the minus i pi over three.
623
00:37:49,840 --> 00:37:54,400
We'll be doing more
complex numbers briefly
624
00:37:54,400 --> 00:37:56,545
but a little more in
the next two days.
625
00:37:59,560 --> 00:38:00,940
next two lectures.
626
00:38:00,940 --> 00:38:05,670
Anyway, the -- so what's
the deal with stability,
627
00:38:05,670 --> 00:38:08,600
what do the Dn-s do?
628
00:38:08,600 --> 00:38:13,570
Well, look, if -- if I take the
sixth power I'm around at one,
629
00:38:13,570 --> 00:38:16,300
the problem actually
told me this.
630
00:38:16,300 --> 00:38:18,950
The sixth power of those
eigenvalues brings me around to
631
00:38:18,950 --> 00:38:23,350
What does that tell you about
the matrix, by the way? one.
632
00:38:23,350 --> 00:38:26,010
Suppose you know -- this
was a great quiz question,
633
00:38:26,010 --> 00:38:29,510
so I should never have just
said it, but popped out.
634
00:38:29,510 --> 00:38:33,600
Suppose lambda one to the sixth
and lambda two to the sixth are
635
00:38:33,600 --> 00:38:36,930
-- are one, which they are.
636
00:38:36,930 --> 00:38:40,090
What does that tell me
about a m- a matrix?
637
00:38:40,090 --> 00:38:43,890
About my matrix A here.
638
00:38:43,890 --> 00:38:47,290
Well, what -- what matrix
is connected with lambda one
639
00:38:47,290 --> 00:38:49,180
to the sixth and lambda
two to the sixth?
640
00:38:49,180 --> 00:38:51,890
It's got to be the
matrix A to the sixth.
641
00:38:51,890 --> 00:38:55,700
So what is A to the
sixth for that matrix?
642
00:38:55,700 --> 00:39:00,170
It's got eigenvalues
one and one.
643
00:39:00,170 --> 00:39:04,720
Because when I take the
sixth power, actually, ye,
644
00:39:04,720 --> 00:39:09,610
if I take the sixth power b- all
the sixth power of that is one
645
00:39:09,610 --> 00:39:12,990
and the sixth power of that is
one, the sixth power of this
646
00:39:12,990 --> 00:39:16,660
is e to the two pi i, that's
one, the sixth power of this
647
00:39:16,660 --> 00:39:19,330
is e to the minus
two pi i, that's one.
648
00:39:19,330 --> 00:39:24,190
So the sixth powers, the -- the
sixth power of that matrix has
649
00:39:24,190 --> 00:39:27,850
eigenvalues one and
one, so what is it?
650
00:39:27,850 --> 00:39:30,400
It's the identity, right.
651
00:39:30,400 --> 00:39:33,940
So if I operate this -- if
I run this thing six times,
652
00:39:33,940 --> 00:39:35,930
I'm back where I was.
653
00:39:35,930 --> 00:39:39,340
The sixth power of that
matrix is the identity.
654
00:39:39,340 --> 00:39:40,550
Good.
655
00:39:40,550 --> 00:39:41,670
OK.
656
00:39:41,670 --> 00:39:44,890
So it'll loop around, it's
-- it doesn't go to zero,
657
00:39:44,890 --> 00:39:49,080
it doesn't blow up, it just
periodically goes around with
658
00:39:49,080 --> 00:39:50,360
period six.
659
00:39:50,360 --> 00:39:51,460
OK.
660
00:39:51,460 --> 00:39:54,440
let's just see if there's a --
661
00:39:54,440 --> 00:39:55,095
all right.
662
00:39:58,520 --> 00:40:00,020
I'll -- let's see.
663
00:40:00,020 --> 00:40:02,250
Could I also look at a --
664
00:40:02,250 --> 00:40:05,155
at a final exam from
nineteen ninety-two.
665
00:40:07,870 --> 00:40:11,570
I think that's yeah, let me
do that on this last board.
666
00:40:14,740 --> 00:40:18,080
It starts -- a lot of the
questions in this exam are
667
00:40:18,080 --> 00:40:20,760
about a family of matrices.
668
00:40:20,760 --> 00:40:25,300
Let me give you the fourth, the
fourth guy in the family is --
669
00:40:25,300 --> 00:40:29,880
has a one, so it's
zeroes on the diagonal,
670
00:40:29,880 --> 00:40:34,130
but these are going one,
two, three and so on.
671
00:40:34,130 --> 00:40:37,420
One, two, three, and so on.
672
00:40:37,420 --> 00:40:42,190
But, for the four-by-four
case I'm stopping at four.
673
00:40:42,190 --> 00:40:47,070
You see the pattern?
674
00:40:47,070 --> 00:40:49,670
It's a family of matrices
which is growing,
675
00:40:49,670 --> 00:40:52,760
and actually the numbers
-- it's symmetric, right,
676
00:40:52,760 --> 00:40:56,490
it's equal to A4 transpose.
677
00:40:56,490 --> 00:41:00,880
And we can ask all sorts of
questions about its null space,
678
00:41:00,880 --> 00:41:07,030
its range, r- its column
space find the projection
679
00:41:07,030 --> 00:41:12,330
matrix onto the column space
of A3, for example, is in here.
680
00:41:12,330 --> 00:41:21,730
So -- so one -- so A3 is zero,
one, zero, one, zero, two,
681
00:41:21,730 --> 00:41:24,100
zero, two, zero.
682
00:41:32,400 --> 00:41:36,900
OK, find the projection
matrix onto the column space.
683
00:41:36,900 --> 00:41:41,540
By the way, is that matrix
singular or invertible?
684
00:41:41,540 --> 00:41:42,310
Singular.
685
00:41:42,310 --> 00:41:45,750
Why do we know it's singular?
686
00:41:45,750 --> 00:41:50,500
I see that column three is
a multiple of column one.
687
00:41:50,500 --> 00:41:53,090
Or we could take
its determinant.
688
00:41:53,090 --> 00:41:55,730
So it's certainly singular.
689
00:41:55,730 --> 00:42:00,200
The projection will be
matrix will be three-by-three
690
00:42:00,200 --> 00:42:03,190
but it will project
onto the column space,
691
00:42:03,190 --> 00:42:05,610
it'll project onto this plane.
692
00:42:05,610 --> 00:42:09,240
The column space of A3, and
I guess I would find it from
693
00:42:09,240 --> 00:42:11,100
the formula AA --
694
00:42:11,100 --> 00:42:15,120
AA transpose A inverse,
I would have to --
695
00:42:15,120 --> 00:42:17,120
I would -- I guess
I would do all this.
696
00:42:20,970 --> 00:42:23,050
There may be a
better way, perhaps
697
00:42:23,050 --> 00:42:25,700
I could think there might
be a slightly quicker way,
698
00:42:25,700 --> 00:42:27,390
but that would come
out pretty fast.
699
00:42:27,390 --> 00:42:28,450
OK.
700
00:42:28,450 --> 00:42:31,260
So that's be the
projection matrix.
701
00:42:31,260 --> 00:42:32,360
Next question.
702
00:42:32,360 --> 00:42:35,010
Find the eigenvalues and
eigenvectors of that matrix.
703
00:42:35,010 --> 00:42:36,020
OK.
704
00:42:36,020 --> 00:42:38,590
There's a three-by-three
matrix, oh, yeah,
705
00:42:38,590 --> 00:42:40,560
so what are its eigenvalues
and eigenvectors,
706
00:42:40,560 --> 00:42:42,840
we haven't done any
three-by-threes.
707
00:42:42,840 --> 00:42:45,070
Let's do one.
708
00:42:45,070 --> 00:42:48,770
I want to find, so how
do I find eigenvalues?
709
00:42:48,770 --> 00:42:53,280
I take the determinant
of A3 minus lambda I.
710
00:42:53,280 --> 00:42:57,200
So this is you just have to --
so I'm subtracting lambda from
711
00:42:57,200 --> 00:43:02,650
the diagonal, and I have a
one, one, zero, zero, two,
712
00:43:02,650 --> 00:43:07,200
two there, and I just have
to find that determinant.
713
00:43:07,200 --> 00:43:11,030
OK, since it's three-by-three
I'll just go for it.
714
00:43:11,030 --> 00:43:16,850
This way gives me minus lambda
cubed and a zero and zero.
715
00:43:16,850 --> 00:43:20,060
Then in this direction which
has the minus sign, that's
716
00:43:20,060 --> 00:43:25,090
a zero, four lambdas, I
mean minus four lambdas,
717
00:43:25,090 --> 00:43:29,110
and minus another lambda, so
that's minus five lambdas,
718
00:43:29,110 --> 00:43:32,550
but that direction
goes with a minus sign,
719
00:43:32,550 --> 00:43:36,850
so I think it's
plus five lambda.
720
00:43:36,850 --> 00:43:40,330
That looks like the determinant
of A3 minus lambda I,
721
00:43:40,330 --> 00:43:43,030
so I set it to zero.
722
00:43:43,030 --> 00:43:45,520
So what are the eigenvalues?
723
00:43:45,520 --> 00:43:48,950
Well, lambda equals zero --
lambda factors out of this,
724
00:43:48,950 --> 00:43:52,790
times minus lambda
squared plus four,
725
00:43:52,790 --> 00:44:03,610
so the eigenvalues are
five, thanks, thanks,
726
00:44:03,610 --> 00:44:08,470
so the eigenvalues are
zero, square root of five,
727
00:44:08,470 --> 00:44:11,410
and minus square root of five.
728
00:44:11,410 --> 00:44:13,970
And I would never write
down those three eigenvalues
729
00:44:13,970 --> 00:44:17,120
without checking the
trace to tell the truth.
730
00:44:17,120 --> 00:44:19,520
Because -- because we did a
bunch of calculations here
731
00:44:19,520 --> 00:44:22,990
but then I can quickly add up
the eigenvalues to get zero,
732
00:44:22,990 --> 00:44:27,270
add up the trace to get
zero, and feel that I'm --
733
00:44:27,270 --> 00:44:30,510
well, I guess that wouldn't have
caught my error if I'd made it
734
00:44:30,510 --> 00:44:34,280
-- if -- if that had been a
four I wouldn't have noticed,the
735
00:44:34,280 --> 00:44:37,660
determinant isn't anything
greatly useful here, right,
736
00:44:37,660 --> 00:44:41,310
because the determinant
is just zero.
737
00:44:41,310 --> 00:44:44,190
And so I never would
know whether that five
738
00:44:44,190 --> 00:44:48,990
was right or wrong, but
thanks for making it right.
739
00:44:48,990 --> 00:44:49,490
OK.
740
00:44:52,690 --> 00:44:54,380
Ha.
741
00:44:54,380 --> 00:45:00,760
Question two c, whoever
wrote this, probably me,
742
00:45:00,760 --> 00:45:03,270
said this is not difficult.
743
00:45:03,270 --> 00:45:06,570
I don't know why I put that in.
744
00:45:06,570 --> 00:45:10,640
just -- it asks for the
projection matrix onto
745
00:45:10,640 --> 00:45:12,050
the column space of A4.
746
00:45:15,402 --> 00:45:17,360
How could I have thought
that wasn't difficult?
747
00:45:17,360 --> 00:45:25,580
It looks extremely difficult.
what's the projection matrix
748
00:45:25,580 --> 00:45:27,505
onto the column space of A4?
749
00:45:32,250 --> 00:45:34,680
I don't know whether that --
this is not difficult is just
750
00:45:34,680 --> 00:45:37,650
like helpful or -- or insulting.
751
00:45:37,650 --> 00:45:41,380
Uh, what do you think?
752
00:45:41,380 --> 00:45:43,870
The -- what's the
column space of A4 here?
753
00:45:46,910 --> 00:45:52,800
Well, what's our first question
is is the matrix singular
754
00:45:52,800 --> 00:45:54,280
or invertible?
755
00:45:54,280 --> 00:45:58,210
If the answer is invertible,
then what's the column space?
756
00:46:00,770 --> 00:46:03,820
If -- if this matrix A4 is
invertible, so that's my guess,
757
00:46:03,820 --> 00:46:07,200
if this problem's easy it has
to be because this matrix is
758
00:46:07,200 --> 00:46:08,600
probably invertible.
759
00:46:08,600 --> 00:46:12,940
Then its column
space is R^4, good,
760
00:46:12,940 --> 00:46:15,260
the column space
is the whole space,
761
00:46:15,260 --> 00:46:18,320
and the answer to this easy
question is the projection
762
00:46:18,320 --> 00:46:23,180
matrix is the identity, it's the
four-by-four identity matrix.
763
00:46:23,180 --> 00:46:26,130
If this matrix is invertible.
764
00:46:26,130 --> 00:46:27,510
Shall we check invertibility?
765
00:46:27,510 --> 00:46:29,840
How would you find
its determinant?
766
00:46:29,840 --> 00:46:33,770
Can we just like take the
determinant of that matrix?
767
00:46:33,770 --> 00:46:37,010
I could ask you how -- so there
-- there are twenty-four terms,
768
00:46:37,010 --> 00:46:39,950
do we want to write all
twenty-four terms down?
769
00:46:39,950 --> 00:46:42,470
not in the remaining
ten seconds.
770
00:46:42,470 --> 00:46:43,930
Better to use cofactors.
771
00:46:43,930 --> 00:46:47,890
So I go along row
one, I see one --
772
00:46:47,890 --> 00:46:52,060
the only nonzero is this guy,
so I should take that one times
773
00:46:52,060 --> 00:46:53,380
the cofactor.
774
00:46:53,380 --> 00:46:55,680
Now so I'm down to
this determinant.
775
00:46:55,680 --> 00:46:56,640
OK.
776
00:46:56,640 --> 00:47:00,970
So now I'm -- look at this first
column, I see one times this,
777
00:47:00,970 --> 00:47:05,640
there's the cofactor of the
one, so I'm using up row one --
778
00:47:05,640 --> 00:47:09,530
row one and column one of
this three-by-three matrix,
779
00:47:09,530 --> 00:47:12,060
I'm down to this
cofactor, and by the way,
780
00:47:12,060 --> 00:47:14,370
those were both
plus signs, right?
781
00:47:14,370 --> 00:47:15,410
No, they weren't.
782
00:47:15,410 --> 00:47:17,240
That was a minus sign.
783
00:47:17,240 --> 00:47:18,490
That was a --
784
00:47:18,490 --> 00:47:22,540
that was a minus, and then that
was a plus, and then this, so
785
00:47:22,540 --> 00:47:24,670
what's the determinant?
786
00:47:24,670 --> 00:47:25,400
Nine.
787
00:47:25,400 --> 00:47:25,900
Nine.
788
00:47:25,900 --> 00:47:27,960
Determinant is nine.
789
00:47:27,960 --> 00:47:31,190
Determinant of A4 is nine.
790
00:47:31,190 --> 00:47:32,580
OK.
791
00:47:32,580 --> 00:47:38,640
Where A3, so my guess is I'll
put that on the final this
792
00:47:38,640 --> 00:47:43,320
year, the -- probably the odd-
numbered ones are singular
793
00:47:43,320 --> 00:47:47,830
and the even-numbered
ones are invertible.
794
00:47:47,830 --> 00:47:50,710
And I don't know
what the determinants
795
00:47:50,710 --> 00:47:55,750
are but I'm betting that
they have some nice formula.
796
00:47:55,750 --> 00:47:56,290
OK.
797
00:47:56,290 --> 00:48:02,770
So, recitations this week
will also be quiz review
798
00:48:02,770 --> 00:48:08,740
and then the quiz is
Wednesday at one o'clock.
799
00:48:08,740 --> 00:48:10,290
Thanks.