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DAVID SHIROKOFF: Hi everyone.
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Welcome back.
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So today we're going to tackle
a problem in complex matrices.
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And specifically, we're going
to look at diagonalizing
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a complex matrix.
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So given this matrix
A, we're asked
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to find its eigenvalue matrix
lambda, and its eigenvector
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matrix S.
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And one thing to note
about this matrix A
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is that if we take its
conjugate transpose,
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it's actually equal to itself.
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So in Professor
Strang's book, he
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combines this notation
to be superscript
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H to mean conjugate transpose.
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So if you were to take the
transpose of this matrix
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and then conjugate
all the elements,
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you would find that A equals
its conjugate transpose,
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and we call this
property Hermitian.
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So I'll let you think
about this for a moment
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and I'll be back in a second.
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OK, welcome back.
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So what's the first step in
computing the eigenvectors
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and eigenvalues of the matrix?
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It's to take a look at the
characteristic equation.
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So specifically, we take
det of A minus lambda i.
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And quite possibly, the only
thing new with this problem
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is that the entries of
the matrix A are complex.
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Now, you may have already seen
that lambda's being complex,
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but we're going to work
this out explicitly.
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So if I take the
determinant, we get
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det of 2 minus lambda, 3 minus
lambda, we have 1 minus i,
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1 plus i.
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We want to set this to 0.
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This then gives us a
polynomial for lambda.
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1 plus i, 1 minus i,
set it equal to 0.
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We can expand out this term.
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You get 6 minus 5 lambda
plus lambda squared.
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These two terms you'll
note are complex conjugates
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of each other.
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This tends to make
things simple.
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So we have minus 1 minus i
squared is going to give us 2.
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Because they're
differences of squares,
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the cross terms
involving i cancel,
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and we get the
characteristic equation.
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Lambda squared minus 5
lambda plus 4 equals 0.
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And specifically, we can
factorize this equation.
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We see that there's
roots of minus 1--
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or factorizes into
factors of lambda minus 1
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and lambda minus 4, which then
give us roots of lambda is 1
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and lambda is 4.
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So when one curious
point to note
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is that the eigenvalues
are real in this case.
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1 and 4 are real, whereas the
matrix that we started with
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was complex.
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And this is a general property
of Hermitian matrices.
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So even though they might
be complex matrices,
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Hermitian matrices always
have real eigenvalues.
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So this is the first step when
asked to diagonalize a matrix.
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The second step is to
find the eigenvectors.
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And to do that
what we have to do
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is we have to look at
the cases for lambda
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equals 1 and lambda
is equal 4 separately.
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So let's first look at the
case of lambda is equal to 1.
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And specifically, we're going
to be looking for a vector such
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that A minus lambda*I
times the vector v is 0.
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And if we've done
things properly,
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this matrix A minus
lambda*I should be singular.
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So if we take A minus
lambda*I, we're going to get 1,
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1 minus i; 1 plus
i, 3 minus 1 is 2.
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And I'll write out components
of v, which are v_1 and v_2.
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And we want this to be 0.
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And you'll note that it's almost
always the case that when we
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work out A minus lambda*I,
the second row is going to be
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a constant multiple
of the first row.
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And this must be the case
because these two rows must be
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linearly dependent on each other
for the matrix A minus lambda*I
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to be singular.
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So if you look at
this you might think
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that these two rows
aren't necessarily
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linearly independent.
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But the point is that there's
complex numbers involved.
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And indeed, actually if we
were to multiply this first row
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by 1 plus lambda, we would
get 1 plus lambda and 2.
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And you note that that's
exactly the second row.
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So this second row is
actually 1 plus lambda times
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the first row.
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So these rows are
actually linearly
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dependent on each other.
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So what values of v_1
and v_2 can we take?
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Well, we just need to make
this top row multiplied
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by v_1 and v_2 equal to 0.
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And then because the second
row is a constant multiple
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of the first row, we're
automatically guaranteed
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that the second equation holds.
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So just by looking
at it, I'm going
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to take v_1 is equal to 1
minus i, and v_2 is negative 1.
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So we see that 1 times 1 minus
i minus 1 times 1 minus i
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is going to give us 0.
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So this is one solution.
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And of course, we can take
any constant multiple times
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this eigenvector,
and that's also
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going to be an eigenvector.
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So I'll just write this out.
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1 minus i, minus 1 is the
eigenvector for lambda
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is equal to 1.
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For lambda is equal to 4,
again, A minus lambda*I is going
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to give us negative 2,
1 minus i; 1 plus i,
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3 minus lambda's
going to be minus 1.
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And I'll call this
vector u_1 and u_2.
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And again, we want u_1 and
u_2 equal to 0-- or sorry,
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the matrix multiplied by
[u 1, u 2] is equal to 0.
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And just by looking
at this again,
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we see that the second row is
actually a constant multiple
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of the first row.
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For example, if we were
to multiply this row
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by negative 2, and
this row by 1 plus i,
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we would see that they're
constant multiples
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of each other.
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So I can take u to be, for
example, 1, and 1 plus i.
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How did I get this?
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Well I just looked at
the second equation
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because it's a little
simpler, and I said, well,
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if I have 1 plus I here, I
can just say multiply it by 1.
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And then minus 1 times 1
plus i, when I add them up,
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is going to vanish.
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So this is how I
get the second one.
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Now there's something
curious going on,
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and this is going to be another
property of Hermitian matrices.
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But if you actually take a
look at this eigenvector,
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it will be orthogonal
to this eigenvector
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when we conjugate the
elements and dot the two
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vectors together.
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So this is another
very special property
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of complex Hermitian matrices.
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OK, so the last step now is to
construct these matrices lambda
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and S. Now we already
know what lambda
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is because it's the diagonal
matrix with the eigenvalues 1
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and 4.
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So we have 1, 0; 0 and 4.
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Now I'm going to do
something special for S.
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I've noted that these
two vectors u and v
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are orthogonal to each other.
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So what do I mean by orthogonal?
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Specifically, if I were to
take v conjugate transpose
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and multiply it by
u, we would end up
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getting 1 plus i minus 1.
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This would be v
conjugate transpose.
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1, 1 plus i, and we see that
when we multiply these out
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we get 0.
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So when we have
orthogonal eigenvectors,
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there's a trick that we can do
to build up this matrix S and S
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inverse.
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What we can do is we
can normalize u and v.
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So specifically, we can take any
constant multiple of u and v,
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and it's still going
to be an eigenvector.
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So what I'm going to do is
I'm going to take u and v
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and multiply them
by their length.
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So for example u, the amplitude
of its top component is 1.
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The amplitude of its
bottom component is 2.
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So notice that the modulus of
the complex number 1 plus I
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is 2.
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So we have-- sorry, it's root 2,
the complex modulus is root 2.
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So the amplitude of the
entire vector is root 3.
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It's 1 plus 2 squared quantity
square rooted, so it's root 3.
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So what we can do
is we can build up
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this matrix S using a
normalization factor of 1
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over root 3.
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And I'm going to take
the-- the first column is
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the first eigenvector that
corresponds to eigenvalue 1.
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And then the second column is
the second eigenvector which
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corresponds to eigenvalue 4.
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And the reason I
put in this root 3
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here is to make this column unit
length 1, and this column unit
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length 1.
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And the reason I do this is
because now this matrix S,
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it's possible to check that this
matrix S is actually unitary,
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which means that its inverse
is actually just equal to it's
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conjugate transpose.
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So this is a very special
property of the eigenvectors
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of a Hermitian matrix.
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And then lastly, I'm
just going to write down
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the diagonalization
of A. So if I have A,
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because I have its
eigenvector matrix S,
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and its eigenvalue
matrix lambda,
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it's possible to decompose A
into a product of S lambda S
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inverse.
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And because S is
unitary, its inverse
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is actually its
conjugate transpose.
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So just to put the
pieces together,
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we have A is equal to S-- which
is 1 over root 3 1 minus i,
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minus 1; 1, 1 plus i-- times
the diagonal matrix [1, 0; 0, 4]
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times S inverse, which is going
to be its conjugate transpose.
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So what I do is I conjugate
each element, so 1 minus i
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becomes 1 plus i and vice versa.
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And then I take the transpose.
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So I get 1 plus i.
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Transposing swaps
the minus 1 and 1.
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And at the end of the day, I get
S inverse is just this matrix
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here.
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And if you were to multiply
these matrices out,
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you would see it you
actually do recover A.
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So just to summarize
quickly, even though we
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were given a complex matrix A,
the process to diagonalize A
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is the same as what
we've seen before.
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The first step is to find
the characteristic equation
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and the eigenvalues.
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And then the second step is
to find the eigenvectors,
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and you do this in
the same procedure.
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But in general, the
eigenvectors can be complex.
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And for this very special
case, when A is Hermitian,
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the eigenvalues are real,
and the eigenvectors
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are orthogonal to each other.
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So I think I'll conclude here,
and I'll see you next time.