Page "Generalized eigenvector" Paragraph 13

There are several ways to see that there will be one generalized eigenvector necessary.

Easiest is to notice that this matrix is in Jordan normal form, but is not diagonal, meaning that this is not a diagonalizable matrix.

Since there is 1 superdiagonal entry, there will be one generalized eigenvector ( or you could note that the vector space is of dimension 2, so there can be only one generalized eigenvector ).

Alternatively, you could compute the dimension of the nullspace of to be p = 1, and thus there are m-p = 1 generalized eigenvectors.