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.