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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