Let be a two-dimensional Euclidean space with orthonormal basis
Let
be the operator defined by
so that
Geometrically, the operator acts on vectors in
by rotating them counterclockwise through an angle of
and then scaling it by
It is geometrically clear that
has no eigenvalues: rotating any vector by
results in a new vector which is not a scalar multiple of the original vector. Right? Maybe not, now that we know about complex numbers.
By definition, a nonzero vector is an eigenvector of
if and only if we have
for some scalar
This is the same thing as saying that the vector
belongs to the kernel of the operator
where
is the identity operator. This is in turn equivalent to saying that the kernel of
contains the nonzero vector
, which means that
is not invertible, which in turn means that
This chain of reasoning is true in general, i.e. we have the following general proposition.
Proposition 1: The eigenvalues of an operator are exactly the scalars
such that
So, according to Proposition 1, to find the eigenvalues of an operator we can try to solve the characteristic equation
The “unknown” in this equation is
Let us write down the characteristic equation of the rotation operator defined above. We have
so that and the characteristic equation of
is
There is no number which solves the characteristic equation, since for any such number the LHS is the sum of a nonnegative number and a positive number, which is a nonzero quantity. However, if we widen our scope of the number concept, this equation does have solutions, corresponding to the fact that
where is the imaginary unit, as introduced in Lecture 26. That is, while the characteristic equation has no real solutions, it has the two distinct complex solutions
In fact, this is always the case: the main advantage of complex vector spaces is that operators on such spaces always have eigenvalues.
Theorem 1: If is a linear operator on an
-dimensional complex vector space, then
has
(not necessarily distinct) complex eigenvalues
.
Proof: The eigenvalues of are the solutions of the characteristic equation,
Now, since
where is any basis of
the determinant
is a polynomial function of
and the highest degree term of which is
But the fundamental theorem of algebra says that every polynomial of degree
has
(not necessarily distinct) roots in
.
— Q.E.D.
The above is saying that if we consider the rotation operator of our example as on operator on a complex vector space, then it does have eigenvalues, even though it did not when considered as an operator on a real vector space. Now comes the question of what the eigenvectors corresponding to these eigenvalues are. In order for the solutions of the characteristic equation to actually correspond to eigenvalues of the operator
there must be nonzero vectors
such that
Let us see if we can actually calculate and
. We have that
Thus, satisfies
if and only if
are complex numbers, not both zero, such that
or equivalently
By inspection, and
are solutions to the above equations, whence
is an eigenvector of Similarly,
satisfies
if and only if
are complex numbers, not both zero, such that
or equivalently
By inspection, and
are solutions of these equations, whence
is an eigenvector of Now, what is the whole point of calculating eigenvalues and eigenvectors? Well, if
is a basis of
then we will have found that the matrix of the operator
in this basis is diagonal,
which is a more convenient matrix representation of then that given by
since we can use it to easily do computations with
So, we wish to show that
is a linearly independent set in
This follows immediately if
are orthogonal, so let’s see if we get lucky:
We didn’t get lucky, and worse than that this scalar product calculation suggests that something unpleasant happens when we start computing scalar products with complex numbers. Indeed, if we modify the above calculation by computing the scalar product of with itself, we find that
This is disturbing, since it says that the nonzero vector is orthogonal to itself, or equivalently that it has zero length,
The original of this problem is that, unlike squares of real numbers, squares of complex numbers can be negative. We have to modify the scalar product for complex vector spaces to accommodate this, we insist that a complex scalar product
is an antilinear function of its first argument:
where if then
is the complex conjugate of
Complex vector spaces with which come with a complex scalar product are the complex version of Euclidean spaces, and they have a special name.
Definition 1: A Hilbert space is a pair consisting of a complex vector space
together with a complex scalar product.
Continuing with our running example, let us re-compute the inner product of the eigenvectors of the rotation operator that we found above. Interpreting
as a complex inner product, we now find that
so that actually are orthogonal with respect to the complex scalar product on
in which the basis
is orthonormal. Thus
is a basis of the complex vector space
consisting of eigenvectors of the operator
meaning that while
has no eigenvalues or eigenvectors when considered as an operator on a Euclidean space, it is in fact semisimple when considered as an operator on a Hilbert space.
Even though they may initially seem more complicated, Hilbert spaces are actually easier to work with than Euclidean spaces — linear algebra runs more smoothly over the complex numbers than over the real numbers. For example, it is possible to give a succinct necessary and sufficient criterion for an operator on a (finite-dimensional) Hilbert space
to be semisimple. As in the case of a Euclidean space, define the adjoint of
to be the unique operator
such that
Theorem 2: is a semisimple operator if and only if it commutes with its adjoint, meaning that
I regret that we will not have time to prove this Theorem.