The solution of some matrix
equations can show a very
useful phenomenon, which
is that they may have
eigenvectors and eigenvalues The
essence of an eigenvector
is that when we multiply this
eigenvector by our matrix,
we're going to get back the
same vector, possibly
multiplied by a number.
And that number is called
the eigenvalue.
If we think by analogy with
ordinary geometrical vectors,
when a matrix multiplies one of
its own eigenvectors, the
result is a vector that we can
think of as pointing in the
same direction, but with a
length multiplied by the
eigenvalue.
The word eigen comes from
German, where, loosely
speaking, it means
own or self.
In fact, it is etymologically
related to the English word
own, eigen and own.
For an eigenvector of a given
matrix, when we multiply it by
that matrix, we get back the
self same vector, though
possibly of a different
length.
So this is where the idea
of self comes in with
eigenvectors.
The behavior of eigenvectors and
eigenvalues of matrices is
found in a generalized form in
quantum mechanics, where we
may have eigenfunctions instead
of eigenvectors.
Though, in a deeper sense,
the ideas of what we call
eigenfunctions and what we call
eigenvectors are really
the same thing.
And we will, in the end, be able
to use the matrix vector
form, even when we are dealing
with eigenfunctions instead of
what we might think of
as eigenvectors.
Hence, the essence of the ideas
of eigenvectors and
eigenvalues with matrices is
particularly important in the
mathematics we need for
quantum mechanics.
So let's look at a matrix
eigenequation.
An equation of this form, a
matrix A times a vector d
equal to a number lambda times
the same vector, for a square
matrix, is called an
eigenequation.
So this lambda is called
the eigenvalue.
This d is called
an eigenvector.
And if there are solutions of
this equation, and that's not
necessarily the case, they
may only exist for
specific values of lambda.
So if this equation has a
solution, which it might not,
if there are solutions, they
may exist only for specific
values of lamda, for specific
eigenvalues.
So let's look explicitly at how
we're going to solve an
eigenequation for matrices.
So we start out with
our eigenequation.
And then we rewrite it, rather
trivially, by putting in the
identity operator here.
We could always do that because
the identity operator
does nothing.
I times d, this vector,
is just d.
So we can always put an I
in there if we want to.
And so we can rewrite
this equation.
We can take this lambda
I over to the left.
So we've got this now just a
matrix A minus lambda I times
d is equal to zero.
So all we're doing here is
rewriting our matrix.
We have to be a little careful
with the notation here.
This zero here is not
the number zero.
It's actually a vector with
elements that are all zero.
And this notation is
somewhat loose.
We should have been clearer that
this is actually a vector
with elements zero and
not just a number.
But once we've sorted that
out, we realize that what
we've done is we've rewritten
this equation up
here in this form.
It's exactly the same equation,
but now it's got a
different character to it.
It's saying, some matrix B,
multiplying this vector, is
going to give us a zero
vector as a result.
Now why did we do that?
Well, we know something
about this.
If some matrix B times a vector
is to give us a zero
vector as a result--
and the vector d itself is not
going to be zero vector.
That would be trivial--
that means this matrix B
cannot have an inverse.
If it did have an inverse, which
we could write as B to
the minus 1, then we can
multiply both sides of this
equation here.
B to the minus 1 times B is,
of course, just I, on the
presumption that this
inverse exists.
So that means B to the minus
1 times B times the
vector d is just d.
And that must be, of course,
looking at the other side of
the equation, B to the minus 1
operating on this zero vector,
the vector with elements
all zero.
But any finite matrix
multiplying a zero vector must
give a zero vector.
So there is no non-zero
solution d.
There's no vector d that isn't
already just all zeros that
can solve this equation
here if this
matrix B has an inverse.
So by what we call reductio ad
aburdum, B has no inverse.
If it did have an inverse, we'd
get to a silly situation
here which can't exist.
So therefore, B cannot
have an inverse.
That's a reductio ad
absurdum proof.
So if we have this equation
here, which was just a rewrite
of our original eigenequation,
for this equation to have any
solutions, it must be
the case that the
matrix B has no inverse.
Now if the matrix B, which,
remember, is just our original
matrix minus the eigenvalue
times the identity matrix, if
that matrix B has no inverse, it
means, from the properties
of our determinant, that the
determinant of that matrix has
to be zero.
Remember, we said that before.
If the determinant of a
matrix is zero, the
matrix has no inverse.
And if the matrix has
no inverse, the
determinant is zero.
That's one of the magic
properties of determinants.
This equation here will allow
us to construct what's
sometimes called the
secular equation.
And the solutions of that
secular equation will give us
the values lambda for which our
eigenequation is going to
have a solution.
And once we figured out those
eigenvalues lambda, the values
of lambda for which our
eigenequation has a solution,
it's relatively straightforward
to deduce what
these eigenvectors are.
So suppose we want to find the
eigenvalues and eigenvectors,
if they exist, of a two by
two matrix like this.
It's a square matrix.
It happens to be
a complex one.
Imagine any elements on
the off diagonal.
So we write the determinant
condition for finding
eigenvalues.
The determinant of this
matrix has to be zero.
That means the determinant of
all of this matrix added up in
here has to be zero.
Now, the determinant of all of
this stuff here, we can add
everything together.
We take our lambda and actually
multiply the identity
matrix by it.
So that's this matrix
lambda lambda here.
We then add these two matrices
with the minus sign in here.
And we get this matrix.
So our equation, now, is we want
the determinant of all of
this matrix to equal zero.
So our secular equation becomes,
the determinant
equals zero here, gives
us all of this here.
That is, the determinant was
this term times this term
minus this term times
this term.
So here we are.
That's this term times
this term.
And minus this term times
this term is that there.
We've got to watch our minus
signs carefully here because
we get an I squared
in the middle.
But what that turns into is 1.5
minus lambda all squared
minus a quarter.
That's what we get from
multiplying these two minus
0.5's with the additional
I squared in here.
It has to be zero.
Or, equivalently, we can
multiply this out.
And what we've got is a
quadratic lambda squared minus
3 lambda plus 2 has
to equal zero.
Well, of course, we can solve
that quadratic equation.
That's straightforward enough.
And if we do so, we'll
get these roots here.
So lambda 1 equals 1.
Or another root is lambda 2,
we'll call it, equals 2.
These are the eigenvalues.
These are the values of that
parameter lambda for which
there will be solutions of our
original eigenvalue equation.
So we substitute these values
back into our original
eigenvalue equation and deduce
the corresponding
eigenvectors.
Let's just see that working.
So our original eigenvalue
equation was this one.
And we've now figured out what
values of lambda will allow
this equation to have
a solution.
So for a given eigenvalue that
we choose here, we're trying
to find out, then, what will be
the eigenvector, that is,
what will be the values
of d1 and d2 here.
Rewriting this equation here,
we can simply take this over
to the left hand side.
So 1.5 minus lambda, that's
having taken that over on the
diagonal here, minus 0.5i 0.5i,
as before, and then,
again, we've got minus
lambda d2 to put on
the diagonal here.
So we've taken all the stuff
on the right and pushed it
into the matrix on the left.
We're left with a zero
vector on the right.
We can evaluate this now for
a specific eigenvalue.
So let's take, say, the first
eigenvalue, lambda 1 equals 1,
that we just figured out.
So putting lambda 1 equal to 1
into this matrix here gives us
this matrix equation.
And now, conveniently, we can
actually go back to the
writing out the matrix
equation as a
couple of linear equations.
So this matrix equation here is
the same thing as these two
linear equations here.
And from either one of these
equations, it doesn't matter
which one--
this was the first of them.
This is the second of them--
we can now deduce
the eigenvector.
Because either one of these
equations simply gives us that
d2 is equal to i d1.
So we could take this
equation here.
We'd have 0.5 d1 is equal
to 0.5i d2 over here.
In other words, d1
is equal to i d2.
And you get the same answer if
you took the second equation.
Now, with eigenvectors, we are
free to choose any one of the
elements here.
It doesn't really matter what
we choose for one of them.
All we have is a relation
between the two of them.
So let's choose, say, d1 is
equal to 1, which gives us the
eigenvector v1 here,
v1 equal to 1i.
And similarly, if we went
through the same process using
lambda 2 equal to 2, we would
get the other eigenvector,
which would be 1 minus i.
For larger matrices with
eigensolutions, that is, for
some n by n matrices here, we
would have correspondingly
higher order polynomial
secular equations.
And those might get
hard to solve.
But they, in principle,
might have solutions.
And they would then give us n
eigenvalues and eigenvectors
if those eigenvalues and
eigenvectors exist, if the
solutions exist.
So a three by three matrix can
have three eigenvalues and
three corresponding
eigenvectors.
Note, incidentally, as I said
before, when we had the
freedom to fix one of our
elements in our two by two
matrix eigensolution,
eigenvectors can always be
multiplied by any constant and
still be eigenvectors.
It doesn't matter.
