There are advantages to working with complex numbers.
Let be an arbitrary matrix, and an eigenvalue of . The geometric multiplicity of is defined as while its algebraic multiplicity is the multiplicity of viewed as a root of (as defined in the previous section).
For all square matrices and eigenvalues , . Moreover, this holds over both and (in
other words, both for real matrices with real eigenvalues, or more generally complex
matrices with complex eigenvalues)
This theorem has an important consequence. Let denote the (distinct) eigenvalues of the matrix . Then (working over if needed) we can write the characteristic polynomial of as where is the algebraic multiplicity of . From this factorization, we see that the sum of the algebraic multiplicities must equal the degree of , which equals : By the above theorem we have
Over the matrix is diagonalizable iff for each eigenvalue one has . If is a real
matrix and factors over into a product of linear terms, then the same holds over
.
- Proof
- We start with the complex case, as working over guaranteees complete
factorization of . By Theorem thm:geoalg, ; moreover, since for each , we have that iff
for each . But (over ), and is diagonalizable iff . This proves the result over .
If is a real matrix and the factorization of over yields only real eigenvectors, then factors completely into linear terms over , and the above argument can be repeated in this case to arrive at the same conclusion over .
An matrix is called defective if the sum of the geometric multiplicities over is strictly less than . Our last theorem shows that this happens iff there is some eigenvalue which is defective; that is, for which of for which . The defectiveness of a particular eigenvalue is then represented by the difference . This can be arbitrarily large, as the next exercise illustrates.
Let be the matrix with . Show that the eigenvalue has algebraic multiplicity , but
geometric multiplicity .
On the other hand, for almost all matrices (from a statistical point of view), factorization of into linear terms leads to distinct eigenvalues, which therefore must each have multiplicity equal to 1. Since the geometric multiplicity of a given eigenvalue must be at least 1 (as the corresponding eigenspace must be non-zero), we see that any eigenvalue with algebraic multiplicity 1 cannot be defective. Hence in this case, hence