ordinary eigenvector. Every × matrix has linearly independent generalized eigenvectors associated with it and can be shown to be similar to an "almost diagonal" matrix in Jordan normal form. That is, there exists an invertible matrix such that . The matrix in this case is called a generalized modal matrix for . If is an eigenvalue of algebraic multiplicity , then will have linearly independent generalized eigenvectors corresponding to . These results, in turn, provide a straightforward method for computing certain matrix functions of . Note: For an matrix over a field to be expressed in Jordan normal form, all eigenvalues of must be in . That is, the characteristic polynomial must factor completely into linear factors. For example, if has real-valued elements, then it may be necessary for the eigenvalues and the components of the eigenvectors to have complex values. The set spanned by all generalized eigenvectors for a given , forms the generalized eigenspace for . Examples Here are some examples to illustrate the concept of generalized eigenvectors. Some of the details will be described later. Example 1 This example is simple but clearly illustrates the point. This type of matrix is used frequently in textbooks. Suppose Then there is only one eigenvalue, , and its algebraic multiplicity is m = 2. Notice that this matrix is in Jordan normal form but is not diagonal. Hence, this matrix is not diagonalizable. Since there is one superdiagonal entry, there will be one generalized eigenvector of rank greater than 1 (or one could note that the vector space is of dimension 2, so there can be at most one generalized eigenvector of rank greater than 1). Alternatively, one could compute the dimension of the nullspace of to be p = 1, and thus there are m – p = 1 generalized eigenvectors of rank greater than 1. The ordinary eigenvector is computed as usual (see the eigenvector page for examples). Using this eigenvector, we compute the generalized eigenvector by solving Writing out the values: This simplifies to The element has no restrictions. The generalized eigenvector of rank 2 is then , where a can have any scalar value. The choice of a = 0 is usually the simplest. Note that so that is a generalized eigenvector, so that is an ordinary eigenvector, and that and are linearly independent and hence constitute a basis for the vector space . Example 2