Over , the rotation example shows that an operator need not have any eigenvector. Over an algebraically closed field (Chapter 4), such as , this cannot happen.
Theorem 7.5.1 (Existence of eigenvectors). Let be algebraically closed and let be a finite-dimensional vector space over . Then every has an eigenvector: there exist , , and with .
Proof. Let and take any nonzero vector . The vectors , must be linearly dependent, so there exist , not all zero, with Let , a nonzero polynomial with . It is not a nonzero constant, since for ; so . Since is algebraically closed, factors as with and (Proposition 4.3.17). Because evaluation at is a ring homomorphism (Proposition 7.2.2), This operator is not injective, since it kills . A composition of injective maps is injective, so some factor is not injective. Hence is an eigenvalue of (Proposition 7.4.2). ∎
Note. Equivalently, by Proposition 7.4.8(d), has a root because it is a nonconstant polynomial over an algebraically closed field. The proof above avoids determinants. Over , it breaks down exactly where fails to factor into linear factors; we return to real operators in Section 7.10.
A square matrix is upper triangular if for . If is a basis of , the matrix is upper triangular if and only if for every , that is, if and only if each subspace is -invariant.
Theorem 7.5.2 (Upper-triangular form). Let be algebraically closed, a finite-dimensional vector space over , and . Then there is a basis of in which the matrix of is upper triangular:
Proof. Induction on . If , every matrix of is upper triangular.
Let and assume the result in dimension . By Theorem 7.5.1, has an eigenvector , say . The line is -invariant, so induces with , where is the quotient map (Proposition 7.3.5). Intuitively, is “ modulo .” Since , the induction hypothesis gives a basis of with for each . Choose with .
The list spans : for , write ; then . Having elements, it is a basis. Finally, for , so . Together with , this says the matrix of in the basis is upper triangular. (Its first column is , and its lower right block is the matrix of in the basis , as in Proposition 7.3.5.) ∎
The diagonal of a triangular matrix carries the information about invertibility and eigenvalues. The next two results hold over any field.
Lemma 7.5.3 (Invertibility from the diagonal). Let have an upper-triangular matrix in the basis . Then is invertible if and only if all the diagonal entries are nonzero.
Proof. Write , with .
Suppose all . We show by induction on that . Since , and the right side lies in because do. So contains a basis, is surjective, and hence invertible (Corollary 6.1.16).
Suppose for some . Then , and also , so maps the -dimensional space into the -dimensional space . By Corollary 6.1.15, is not injective, so is not injective and not invertible. ∎
Corollary 7.5.4 (Eigenvalues from a triangular matrix). Let have an upper-triangular matrix in some basis. Then:
- is an eigenvalue of is not invertible for some .
- , and the number of times an eigenvalue appears on the diagonal of is its algebraic multiplicity.
In particular, every upper-triangular matrix representing has the same diagonal entries, up to order.
Proof. (a) The first equivalence is Proposition 7.4.2. For the second, apply Lemma 7.5.3 to , whose matrix is upper triangular with diagonal entries .
(b) The matrix is upper triangular with diagonal entries , and the determinant of a triangular matrix, over as over , is the product of its diagonal entries (Proposition 5.4.10). The statement about multiplicities is Lemma 7.4.11. The last sentence follows because does not depend on the basis. ∎
Note. Together, Theorem 7.5.2 and Corollary 7.5.4 show that over an algebraically closed field splits into linear factors, its roots being the diagonal entries of any triangular matrix of . We will soon see a more conceptual description of the algebraic multiplicity: it is the dimension of the generalized eigenspace (Theorem 7.6.14).