Let be a vector space, and let us consider the algebra
as a kind of ecosystem consisting of various life forms of varying complexity. We now move on to the portion of the course which is concerned with the taxonomy of linear operators — their classification and division into various particular classes.
The simplest organisms in the ecosystem are operators which act by scaling every vector
by a fixed number
; these are the single-celled organisms of the operator ecosystem.
Definition 1: An operator is said to be simple if there exists a scalar
such that
— Q.E.D.
Simple operators really are very simple, in the sense that they are no more complicated than numbers. Indeed, Definition 1 is equivalent to saying that where
is the identity operator, which plays the role of the number
in the algebra
meaning that it is the multiplicative identity in this algebra. Simple operators are extremely easy to manipulate algebraically: if
then we have
for any nonnegative integer and more generally if
is any polynomial in a single variable then we have
Exercise 1: Prove the above formula.
The formula even works in the case that
is a negative integer, provided that
equivalently, the simple operator
is invertible if and only if
its inverse being
If
and
are simple operators, then they commute,
just like ordinary numbers, and more generally
for any polynomial in two variables.
Exercise 2: Prove the above formula.
Another way to appreciate how truly simple simple operators are is to look at their matrices. In order to do this, we have to restrict to the case that the vector space is finite-dimensional. If
is
-dimensional, and
is any basis of
then the matrix of
relative to
is simply
where the off-diagonal matrix elements are all equal to zero. For this reason, simple operators are often called diagonal operators.
Most operators in are not simple operators — they are complicated multicellular organisms. So, to understand them we have to dissect them and look at their organs one at a time. Mathematically, this means that, given an operator
we look for special vectors in
on which
acts as if it was simple.
Definition 2: A nonzero vector is said to be an eigenvector of an operator
if
for some The scalar
is said to be an eigenvalue of
The best case scenario is that we can find a basis of entirely made up of eigenvectors of
Defintion 3: An operator is said to be semisimple if there exists a basis
of
consisting of eigenvectors of
Such a basis is called an eigenbasis for
As the name suggests, semisimple operators are pretty simple, but not quite as simple as simple operators. In particular, every simple operator is semisimple, because if is simple then every nonzero vector in
is an eigenvector of
and hence any basis in
is an eigenbasis for
The converse, however, is not true.
Let be an
-dimensional vector space, and let
be a semisimple operator. By definition, this means that there exists a basis
in
consisting of eigenvectors of
This in turn means that there exist numbers
such that
If then
is simple, but if these numbers are not all the same then it is not. However, even if all these numbers are different, the matrix of
relative to
will still be a diagonal matrix, i.e. it will have the form
For this reason, semisimple operators are often called diagonalizable operators. Note the shift in terminology from “diagonal,” for simple, to “diagonalizable,” for semisimple. The former term suggest an immutable characteristic, independent of basis, whereas the latter indicates that some action must be taken, in that a special basis must be found to reveal diagonal form. More precisely, the matrix of a semisimple operator is not diagonal with respect to an arbitrary basis; the definition only says that the matrix of
is diagonal relative to some basis.
Most linear operators are not semisimple — indeed, there are plenty of operators that have no eigenvectors at all. Consider the operator
which rotates a vector counterclockwise through the angle
The matrix of this operator relative to the standard basis
of is
If then
, so that
is a simple operator:
are eigenvectors, with eigenvalues
If
then
and again
is simple, with the same eigenvectors and eigenvalues
However, taking any other value of
for example
rotation through a right angle, it is geometrically clear that
is never a scalar multiple of
so that
has no eigenvectors at all. In particular, it is not semisimple.
Let us now formulate necessary and sufficient conditions for an operator to be semisimple. In this endeavor it is psychologically helpful to reorganize the eigenvector/eigenvalue definition by thinking of eigenvalues as the primary objects, and eigenvectors as secondary objects associated to them.
Defintion 4: The spectrum of an operator is the set
defined by
For each the set
defined by
is called the –eigenspace of
The dimension of
is called the geometric multiplicity of
In these terms, saying that is a simple operator means that the spectrum of
consists of a single number,
and that the corresponding eigenspace exhausts
At the other extreme, the rotation operator considered above has empty spectrum,
and thus does not have any eigenspaces.
Proposition 1: For any , for each
the eigenspace
is a subspace of
Proof: First, observe that because
Second, is closed under scalar multiplication: if
then
Third, is closed under vector addition: if
then
— Q.E.D.
So, the eigenspaces of an operator constitute a collection of subspaces of
of
indexed by the numbers
A key feature of these subspaces is that they are independent of one another.
Theorem 1: Suppose that are distinct eigenvalues of an operator
Let
be nonzero vectors such that
for each
Then
is a linearly independent set.
Proof: We prove this by induction on The base case is
and in this case the assertion is simply that the set
consisting of a single eigenvector of
is linearly independent. This is true, since eigenvectors are nonzero by definition.
For the induction step, suppose that is a linearly dependent set. Then, there exist numbers
not all equal to zero, such that
Let us suppose that Applying the operator
to both sides of the above vector equation, we get
On the other hand, we can multiply the original vector equation by any scalar and it remains true; in particular, we have
Now, subtracting this third equation from the second equation, we obtain
By the induction hypothesis, is a linearly independent set, and hence all the coefficients in this vector equation are zero. In particular, we have
But this is impossible, since and
Hence, the set
cannot be linearly dependent — it must be linearly independent.
— Q.E.D.
Restricting to the case that is finite-dimensional,
Theorem 1 has the following crucial consequences.
Corollary 1: is semisimple if and only if
Proof: Suppose first that is semisimple. By definition, this means that the span of the eigenspaces of
is all of
Thus
By Theorem 1, we have
and hence
Conversely, suppose that the sum of the dimensions of the eigenspaces of is equal to the dimension of
For each
let
be a basis of the eigenspace
Then, by Theorem 1, the set
is a linearly independent set, and hence a basis of the subspace of
Thus
Since by hypothesis we have
this implies that
which in turn implies that
Thus is a basis of
consisting of eigenvectors of
whence
is semisimple.
— Q.E.D.
Corollay 3: If then
is semisimple.
Proof: To say that is equivalent to saying that the spectrum of
consists of
distinct numbers,
Sampling a collection of nonzero vectors from each corresponding eigenspace,
we get a set of eigenvectors of
By Theorem 1,
is a linearly independent set, hence it is a basis of
— Q.E.D.
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