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