TABLE OF CONTENTS Every eigenvalue leads to a solution of eq (7.1). If the eigenvalues are all separate and distinct [4], then
where A is n n and where the ? k's are the n eigenvalues of A. Because each X is n 1,eq (7.9) represents a system of n linear, simultanelous algebraic equations (see Section 5.6), and because the rank of [ A ? ? k I is equal to the rank of the augemented matrix[ A ? ? k I: O], a nontrivial solution exists if and only if det[ A ? ? k I] = 0.
The solution vector X k for each eigenvalue is called a characteristic vector or eigenvector and the set of n eigenvectors that are all n 1 constitute a linearly independent set if all of the eigenvectors are distinct. This can be proved by mathematical induction.
Suppose, for example, that there are two eigenvectors that derive from two unequal eigenvalues and that some linear combination, of them yields the n 1 null vector
If this is premultiplied by A, then
and by noting that AX = ?X this may be written as
If eq (a) is multiplied by ? 2 and the result subtracted from eq (c), the result is
or
However, it was postulated at the start of this exercise...
