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GILBERT STRANG: OK.
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So this is a "prepare the way"
video about symmetric matrices
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and complex matrices.
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We'll see symmetric matrices
in second order systems
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of differential equations.
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Symmetric matrices are the best.
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They have special
properties, and we
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want to see what are the special
properties of the eigenvalues
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and the eigenvectors?
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And I guess the
title of this lecture
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tells you what those
properties are.
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So if a matrix is
symmetric-- and I'll
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use capital S for a
symmetric matrix--
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the first point is the
eigenvalues are real,
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which is not automatic.
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But it's always true if
the matrix is symmetric.
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And the second, even
more special point
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is that the eigenvectors are
perpendicular to each other.
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Different eigenvectors
for different eigenvalues
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come out perpendicular.
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Those are beautiful properties.
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They pay off.
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So that's the symmetric matrix,
and that's what I just said.
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Real lambda, orthogonal x.
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Also, we could look at
antisymmetric matrices.
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The transpose is
minus the matrix.
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In that case, we don't
have real eigenvalues.
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In fact, we are sure to have
pure, imaginary eigenvalues.
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I times something on
the imaginary axis.
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But again, the eigenvectors
will be orthogonal.
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However, they will
also be complex.
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When we have
antisymmetric matrices,
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we get into complex numbers.
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Can't help it, even
if the matrix is real.
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And then finally is the
family of orthogonal matrices.
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And those matrices have
eigenvalues of size 1,
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possibly complex.
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But the magnitude
of the number is 1.
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And again, the eigenvectors
are orthogonal.
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This is the great family of
real, imaginary, and unit
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circle for the eigenvalues.
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OK.
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I want to do examples.
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So I'll just have an
example of every one.
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So there's a symmetric matrix.
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There's a antisymmetric matrix.
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If I transpose it,
it changes sign.
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Here is a combination, not
symmetric, not antisymmetric,
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but still a good matrix.
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And there is an orthogonal
matrix, orthogonal columns.
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And those columns have length 1.
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That's why I've got the
square root of 2 in there.
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So these are the
special matrices here.
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Can I just draw a little
picture of the complex plane?
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There is the real axis.
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Here is the imaginary axis.
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And here's the unit circle,
not greatly circular but close.
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Now-- eigenvalues
are on the real axis
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when S transpose equals S.
They're on the imaginary axis
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when A transpose equals minus A.
And they're on the unit circle
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when Q transpose
Q is the identity.
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Q transpose is Q
inverse in this case.
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Q transpose is Q inverse.
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Here the transpose
is the matrix.
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Here the transpose
is minus the matrix.
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And you see the
beautiful picture
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of eigenvalues, where they are.
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And the eigenvectors for
all of those are orthogonal.
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Let me find them.
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Here that symmetric matrix
has lambda as 2 and 4.
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The trace is 6.
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The determinant is 8.
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That's the right answer.
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Lambda equal 2 and 4.
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And x would be 1
and minus 1 for 2.
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And for 4, it's 1 and 1.
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Orthogonal.
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Orthogonal eigenvectors--
take the dot product of those,
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you get 0 and real eigenvalues.
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What about A?
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Antisymmetric.
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The equation I-- when I do
determinant of lambda minus A,
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I get lambda squared plus
1 equals 0 for this one.
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That leads me to lambda
squared plus 1 equals 0.
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So that gives me lambda is
i and minus i, as promised,
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on the imaginary axis.
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And I guess that that matrix
is also an orthogonal matrix.
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And those eigenvalues,
i and minus i,
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are also on the circle.
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So that A is also a Q. OK.
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What are the
eigenvectors for that?
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I think that the
eigenvectors turn out
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to be 1 i and 1 minus i.
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Oh.
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Those are orthogonal.
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I'll have to tell you
about orthogonality
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for complex vectors.
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Let me complete these examples.
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What about the
eigenvalues of this one?
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Well, that's an easy one.
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Can you connect that to A?
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B is just A plus 3
times the identity--
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to put 3's on the diagonal.
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So I'm expecting
here the lambdas
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are-- if here they
were i and minus i.
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All I've done is add 3 times the
identity, so I'm just adding 3.
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I'm shifting by 3.
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I'll have 3 plus
i and 3 minus i.
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And the same eigenvectors.
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So that's a complex number.
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That matrix was not
perfectly antisymmetric.
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It's not perfectly symmetric.
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So that gave me a 3 plus i
somewhere not on the axis
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or that axis or the circle.
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Out there-- 3 plus
i and 3 minus i.
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And finally, this one,
the orthogonal matrix.
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What are the
eigenvalues of that?
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Let's see.
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I can see-- here I've added 1
times the identity, just added
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the identity to minus 1, 1.
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So again, I have this minus
1, 1 plus the identity.
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So I would have 1 plus i and
1 minus i from the matrix.
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And now I've got a division
by square root of 2,
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square root of 2.
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And those numbers
lambda-- you recognize
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that when you see that number,
that is on the unit circle.
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Where is it on the unit circle?
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1 plus i.
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1 plus i over square root of 2.
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Square root of 2
brings it down there.
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There's 1.
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There's i.
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Divide by square root of 2.
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That puts us on the circle.
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That's 1 plus i over
square root of 2.
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And here is 1 plus i, 1 minus
i over square root of two.
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Complex conjugates.
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When I say "complex
conjugate," that means
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I change every i to a minus i.
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I flip across the real axis.
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I'd want to do that in a minute.
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So are there more lessons
to see for these examples?
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Again, real eigenvalues and
real eigenvectors-- no problem.
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Here, imaginary eigenvalues.
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Here, complex eigenvalues.
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Here, complex eigenvalues
on the circle.
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On the circle.
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OK.
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And each of those
facts that I just
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said about the location
of the eigenvalues--
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it has a short proof, but maybe
I won't give the proof here.
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It's the fact that
you want to remember.
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Can I bring down again, just
for a moment, these main facts?
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Real, from
symmetric-- imaginary,
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from antisymmetric--
magnitude 1, from orthogonal.
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OK.
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Now I feel I've talking
about complex numbers,
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and I really should say-- I
should pay attention to that.
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Complex numbers.
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So I have lambda as a plus ib.
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What do I mean by the
"magnitude" of that number?
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What's the magnitude
of lambda is a plus ib?
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Again, I go along a, up b.
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Here is the lambda,
the complex number.
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And I want to know
the length of that.
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Well, everybody knows
the length of that.
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Thank goodness Pythagoras
lived, or his team lived.
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It's the square root of
a squared plus b squared.
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And notice what that-- how do I
get that number from this one?
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It's important.
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If I multiply a plus
ib times a minus
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ib-- so I have lambda-- that's
a plus ib-- times lambda
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conjugate-- that's a minus ib--
if I multiply those, that gives
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me a squared plus b squared.
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So I take the square
root, and this
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is what I would call the
"magnitude" of lambda.
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So the magnitude of a number
is that positive length.
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And it can be found-- you
take the complex number
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times its conjugate.
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That gives you a squared
plus b squared, and then
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take the square root.
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Basic facts about
complex numbers.
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OK.
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What about complex vectors?
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What is the dot product?
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What is the correct
x transpose x?
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Well, it's not x transpose x.
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Suppose x is the vector 1 i, as
we saw that as an eigenvector.
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What's the length
of that vector?
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The length of that vector is
not 1 squared plus i squared.
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1 squared plus i squared would
be 1 plus minus 1 would be 0.
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The length of that vector
is the size of this squared
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plus the size of this
squared, square root.
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Here we go.
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The length of x squared-- the
length of the vector squared--
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will be the vector.
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As always, I can find
it from a dot product.
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But I have to take
the conjugate of that.
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If I want the
length of x, I have
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to take-- I would usually
take x transpose x, right?
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If I have a real vector x,
then I find its dot product
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with itself, and
Pythagoras tells me
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I have the length squared.
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But if the things are complex--
I want minus i times i.
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I want to get lambda
times lambda bar.
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I want to get a positive number.
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Minus i times i is plus 1.
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Minus i times i is plus 1.
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So I must, must do that.
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So that's really what
"orthogonal" would mean.
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"Orthogonal complex
vectors" mean--
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"orthogonal vectors" mean that
x conjugate transpose y is 0.
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That's what I mean by
"orthogonal eigenvectors"
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when those eigenvectors
are complex.
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I must remember to take
the complex conjugate.
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And I also do it for matrices.
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So if I have a symmetric
matrix-- S transpose S.
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I know what that means.
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But suppose S is complex.
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Suppose S is complex.
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Then for a complex matrix, I
would look at S bar transpose
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equal S.
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Every time I transpose,
if I have complex numbers,
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I should take the
complex conjugate.
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MATLAB does that automatically.
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If you ask for x prime,
it will produce-- not just
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it'll change a column to
a row with that transpose,
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that prime.
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And it will take the
complex conjugate.
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So we must remember
always to do that.
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Yeah.
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And in fact, if S
was a complex matrix
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but it had that property--
let me give an example.
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So here's an S, an
example of that.
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1, 2, i, and minus i.
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So I have a complex matrix.
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And if I transpose it and
take complex conjugates,
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that brings me
back to S. And this
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is called a "Hermitian matrix"
among other possible names.
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Hermite was a important
mathematician.
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He studied this
complex case, and he
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understood to take the conjugate
as well as the transpose.
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And sometimes I would write
it as SH in his honor.
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So if I want one
symbol to do it-- SH.
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In engineering, sometimes
S with a star tells me,
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take the conjugate when
you transpose a matrix.
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So that's main facts about-- let
me bring those main facts down
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again-- orthogonal eigenvectors
and location of eigenvalues.
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Now I'm ready to solve
differential equations.
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Thank you.