In the preceding example (17) we have found four left-hand Perron eigenvectors and three right-hand Perron eigenvectors. This means that there is at least one right-hand Perron eigenvector missing since the dimension of the right-hand eigenspace is equal to dimension of the left-hand eigenspace. It turns out that the missing right-hand Perron eigenvector must have negative and positive parts. Moreover the nonzero parts of this Perron eigenvector has zeros in places where the semipositive Perron eigenvectors have positive parts. Therefore the missing eigenvector can not be part of the nonnegative eigenspace. In general eigenvectors like these exist when a basic class points to two or more other basic classes, either directly or indirectly. In our example this occurs with the right-hand eigenvector at class 6. This class is pointing directly at classes 2 and 5. Both of which are basic.

(25)

Now, take w₂=b₂y₂, w₅=b₅y₅ and let q₆' satisfy q₆'A₆₆=λ^*q₆'»0 then

(26)

Therefore

(27)

Both the numerator and the denominator are positive. So, b₂ and b₅ must opposite in sign. To obtain w₆ we may proceed as follows. Rewrite equation (25) in the form

(28)

with b₂ and b₅ satisfying equation (27). Define

(29)

where a_6s is scalar, a_6r' a row vector and a_6c a colum vector. The matrix A_66(-1) is obtained by deleting e.g. the first row and column from the matrix A₆₆. Partition w₆ accordingly by writing w₆= (w_6s ; w_6c')'. Observe that since A₆₆ is irreducible, the dominant eigenvalue of A_66(-1) is smaller than λ. Hence (λI-A_66(-1))^-1>0. This inverse is not necessarily positive since the matrix might be reducible.

Let G=(0;I_66(-1)), then

(30)

thus

(31)

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