First we shall discuss a few special cases in which the subinvariance theorem plays a prominent role. These cases are simple and shed some light on how the method that we are developing works and give us insight into the structure of semipositive eigenvectors.

Consider the following matrix

(1)

with all A_ij<>0. From the Frobenius/Perron theorem there is a semipositive eigenvector w (the Perron eigenvector) corresponding to λ^*. If we look a the eigenvectorequation, we have

(2)

This means that for i=1,..n-1: w_i=θy_i>>0 or w_i=0. We shall prove that w_n can not be zero. Suppose to the contrary that w_n=0. Then we would have

(3)

The Frobenius/Perron Theorem states that w>0, implying that there is a k=1,...n-1 with w_k>>0. Therefore

(4)

Which is a contradiction. Therefore w_n>>0.

This is not a highly remarkable conclusion. Suppose that the matrix A represents an the coefficients matrix in the closed input/output model. The last sector provides inputs to all the other sectors. If the production in the last sector becomes zero, all intermediate deliveries become zero as well. This would imply that the required inputs are lacking so that the production in any sector is reduced to zero. Therefore the entire economy will can not produce anything at all. This means that the n^th sector is a key sector in this economy where the production is concerned. The dominance of this key sector is dealt with later. First a theorem is presented that will simplify our proofs and allow us to concentrate on the main issue.

Signtheorem I

Given a matrix A in the normal form II. Then for any semipositive left-hand and right-hand eigenvector, v and w respectively

1. v_i is either zero or positive.
2. w_j is either zero or positive.

(5)

(6)

When all the off-diagonal blocks in a certain row, say t, are zero, equation (6) becomes A_ttw_t=λ^*w_t. If λ_t=λ^* then either w_t=0 or w_t >>0 because of the irreducibility of A_tt. If λ_t<λ^* then w_t=0.

Now the case that at least one of the off-diagonal blocks is nonzero.

If λ_t=λ^* for a certain row t then the inequality A_ttw_t <=λ^*w_t is valid, implying on the grounds of the sub invariance theorem that w_t is either zero or positive.

If λ_t<λ^*, then w_t can be expressed in terms of some other w_j's with j<t. That is

(7)

With at least one A_tj semipositive. The inverse matrix is always positive (due to the inverse theorem). A_tjw_j is nonnegative. If A_tjw_j is zero then no contribution is made. When this is true for all i=1,...t-1 then w_t=0. Now consider the case that A_tjw_j is semipositive. This implies that (λ^*I-A_tt)^-1A_tjw_j is positive and thus w_t is positive. This completes the proof for the w_j's. The proof for v_t is analogous (for an alternative proof see Cooper, 1973).

Now we continue our example. Let us suppose that λ_n=λ^* (sector n attains the dominant eigenvalue). With the help of the second equation in (2) we may write

(8)

Applying the subinvariance theorem yields w_n=b_ny_n, where b_n is a nonzero scalar (implicitly we have assumed that b_n>0, but it possible to relax on this assumption later on). Substituting this solution in equation (2) gives equation (3). So all the other w_j's are equal to zero now. The entire right-hand Perron eigenvector can be written as w'=b_n(0,...0,y_n)', which is semipositive. Thus the dominant behaviour of the n^th sector shuts all the other sectors down and all the output in the economy will come from this sector.

Let us suppose that λ_n<λ^*. Now we can write

(9)

The inverse-matrix is positive (The inverse theorem), every A_n,j<>0, so w_n<>0 is only valid when at least one other w_i<> 0. This is certainly true because the dominant eigenvalue has to be attained somewhere. For at least one block the equation w_i=b_iy_i<>0 is valid. This means that even when the n-th sector does not show dominant behaviour it still will produce output together with all the other sectors which do show dominant behaviour (λ_i=λ^*). The left-hand eigenvector does not have such a simple solution so we drop this example for a while.

A matrix which has a simple solution for both the left and the right eigenvector is

(10)

The results on w_n for the matrix in (1) hold here too, but in a more roundabout way. Using signtheorem I it is very easy to show that

(11)

So w_n>>0. In the same way we can show that v₁'>>0, because

(12)

If λ_n=λ^* then w_n=b_ny_n and all the other w_i are equal to zero. We can prove this again by the same procedure we used before but now with the eigenvector equation A_n,n-1w_n-1+A_n.nw_n=λw_n. Similarily if λ₁=λ^* then v₁'=a₁q₁'and all other v_i=0.

Next, suppose that λ_n<λ^*. This implies w_n=(λ^*I-A_n,n)^-1A_n,n-1w_n-1. So, w_n-1<> 0 since w_n<>0. If λ_n-1=λ^* then w_n-1=b_n-1y_n-1 and the other w_i are equal to zero. If λ_n-1<λ^* then w_n-1=(λ^*I-A_n-1,n-1)^-1A_n-1,n-2w_n-2 etc., until we find the first equation in which the dominant eigenvalue is attained. If this is the case for the k^th equation the eigenvector w will look like
 

(13)

The nonzero components of this vector can be computed recursively, taking w_k=y_k as a starting vector and computing w_k+1,...w_n in increasing order. Analogous results hold for the left eigenvector by working from the other direction.

Note that all components of the eigenvector which are not equal to zero are positive. In an economy structured like in (10), sectors which are "high" in the chain will produce a positive output (this holds for all sectors) down to the highest member which shows dominant behaviour. This sector will cause the sectors which are lower in the chain to produce nothing. On the other hand the lower sectors in the chain will determine the prices since the left-hand eigenvector has positive parts up to the lowest sector which attains the dominant eigenvalue. The higher sectors will receive zero-prices. This fits remarkably well with the intuition. Key sectors high in the chain like industry and agriculture determine the extend of output. Sectors low in the chain like distributors and retailers determine prices. Note that the the dot product v'w, which is in this interpretation equal to the total revenue, is normally zero. Only when the dominant eigenvalue is attained in exactly one sector then v'w>0.

The next theorem asserts that w_n>>0 if every column of A contains a semipositive submatrix below its main diagonal. Similarly v₁>>0 if every row of A contains a semipositive submatrix left of its main diagonal.

Signtheorem II

Given a matrix A in the normal form II, consider the have following conditions

Then under

1. w_n>>0, and λ_n=λ^* impliesw=b_n(0,...0,y_n')'
2. v₁>>0, and λ₁=λ* implies v=a₁(q₁,0,...0)'

Suppose w_n=0. We look first at the (n-1)^th column. If A_n,n-1<>0 then w_n-1=0. Now look at the (n-2)^th column. If either A_n,n-2 or A_n-1,n-2 is not equal to zero then w_n-2=0 etc. Until we reach the first column. Every w_i is now equal to zero, but this a contradiction in the view of the Perron/Frobenius theorem. So a nonzero A_i,j in every column is a sufficient condition to ensure that w_n>>0. The proof for v₁ is analogous to this argument.

If λ_n=λ^* then all the w_i in n^th equation are set to zero. Surely w_n-1 is between them because condition (a) is valid and this means that A_n,n-1<> 0. So now look at equation n-1 too. There is now a A_i,n-2<> 0 in either equation (i=) n or n-1. This means that w_n-2=0. This story repeats itself until the first equation, making every w_i, except of course w_n, equal to zero. The proof for the left eigenvector is again analogous.

Signtheorem III

Let A be a matrix in normal form II and let its subdiagonal be composed of semipositive matrices (that is, A_i,i-1>0 for all i=2,...n). Then the other blocks in the matrix don't matter in determining the sign-pattern of a semipositive left and the right eigenvector. Moreover these semipositive eigenvectors satisfy

Where

Proof:

If these other blocks (A_ij with j i-2) are zero then we have (10). This matrix has all the properties of signtheorem II. If we add a nonzero A_ij-block to this matrix then this does not form an additional restriction on the eigenvectors. The blocks on the subdiagonal form are a sufficient condition to ensure that v₁ and w_n are positive.

As for the other w_j, they are certainly zero if λ_n=λ^*. When this is not the case then w_n can surely be expressed in terms of w_n-1. That is

(14)

If λ_n-1<λ^* then w_n-1 can surely be expressed in terms of w_n-2, etc. Until we find a block A_kk for which λ_k=λ^* (it has to be somewhere). Now k=M by definition. As before all the other w_j (j<M) are equal to zero and w_M=b_My_M. This means that

(15)

The expression for w_M+2 becomes

(16)

Etc. Because all the A_i+1,i are semipositive, A_i+1,iy_i are semipositive. The inverses are always positive and thus w_M,...w_n are positive. The terms that contain the other blocks have nothing to add to that. This completes the proof for the right eigenvector. The proof for the left eigenvector is of course analogous.

MAIN