Abar = <w,v>, scalar product of right and left eigenvectors corresponding to la1 when I choose w1=1 and v1=1

[Henry F. Mollet]

What is the Abar below that I’m defining in 4 different ways below, in matrix algebra or in the theory of the discrete z-transform  (analogue of the Laplace transform for continuous functions). Perhaps 3) is best to tell me if there is a mathematical special name for this. 4) might be good in terms matrix algebra. Could be that the Leslie matrix is so special, more like a sparse matrix with mostly zeros, that no name was coined for this special case.

1) Abar = sum [x la^(-x) l(x) m(x)] = sum [x la^(-x) Fj Product (Pj)  sum over all x =ages, product from j=1 to x-1; 
Gives the mean age of the reproducing females.
2) Abar = <w,v>, scalar product of right and left eigenvector when I choose w1=1 and v1=1;
3) Abar = - la [d(phi)/d(la)] at la = la1; where phi = phi(la) is the characteristic function;
 if I put r =ln(la) (instantaneous population growth rate), then Abar = - [d(phi)/d(r)] at r = r1.
4) AbarMatrix = Fstar*Nstar^2; Abar=AbarMatrix(1,1); (see complete script below);
I suspect that this has been derived before by a matrix algebra mathematician. I derived it almost by trial and error.
Henry

N.B. More explanation if needed
Note that the Leslie matrix (A = F + T) is a special matrix with non-zero element only on the first row (no other restrictions) and the first sub-diagonal (have to be <1). It can be used for species that have only one way to produce offspring which leads to an  age-stuctured life-history table and applies to most animals. Everything works also for a stage-based model where we also have diagonal elements but T(1,1) has to be zero (ok to have F(1,1)) because we need a “starter” age-class for the “w1=1 an v1=1” bit to be defined clearly. While Abar can be calculated for species with different types of offspring (many plants) it can no longer be interpreted as the mean age of the reproducing females and is therefore not a useful quantity.

For a Leslie matrix, the “special “ scalar product <w,v> when w1=1 and v1 = 1  has an easily understood meaning and is called Abar. Abar is the mean age of the reproducing parents at the stable age distribution. Abar can also be calculated as Abar = - la1 d(phi)/d(la), where phi = phi (la) is the characteristic function. If we set the characteristic function to one we get the characteristic equation (Euler sum =1), whose solution  will produce  all the solution including the largest and real one, namely lambda1 (la1)

Given a Leslie matrix A = F + T, where 
F is the discounted fertility matrix with non-zero elements on the first row only and no restrictions on their value, and
T is a transmission matrix with non-zero elements on the first sub-diagonal only which have to be smaller than one are the survival rates, I can show (and others have done this before) that
Abar = <W,V> when I choose w1=1 and v1 = 1 to calculate the scalar product.

The Leslie matrix can be represented in a life cycle graph with a node for each age class. Arcs back to first node, from age at first to last reproduction represent discounted fertilities on the first row  and  arrows between notes represent survival rates on the first subdiagonal. If the life cycle graph is z-transformed by dividing everything by la1, then the age-structure (= right eigenvector) corresponding to la1 can be read off the graph starting out with w1 = 1, w2 = P1/la1 and so forth. If the arrows are reversed (corresponds to transposition of the A-matrix) then the reproductive value (= left eigenvectors) can be read off the graph. It’s a bit more complicated but we start again with v1 = 1 but then v2 has the most terms and it’s easier to start with the last one which is v(last) = F(1, last)/la1. The characteristic function can also be read off this graph by summing over all the loops. This works also for stage-based models with matrix elements on the diagonal but the resulting self-loops have to be reduced first based on the signal-flow graphs methods introduced by Mason.

% Octave script for the calculation of  Abar      (A.14)
% 1. Calculation of dominant eigenvalue la1 of projection matrix A = F + T
lambda1 = max(eig(A));
% 2. Calculation of Generation Time (GT) and mu1 from the fundamental matrix N
% Note that here ‘T’ is used for Transition Matrix T and Generation Time is abbreviated with GT
N=inv(eye(length(T))-T);                                              % N = (I - T)-1 (Caswell 2001, p. 118)
R=F*N;                                                                         % Caswell (2001, p. 126, eq. 5.64)
Ro=max(eig(R))                                                            % Ro is dominant eigenvalue of R-matrix
GenerationTime=log(Ro)/log(lambda1)                         % GT = ln(Ro)/ln(la1) = ln(Ro)/r
Mu1Matrix=F*N^2./Ro;                                                 % After (Cochran and Ellner 1992, Table 2)
Mu1=Mu1Matrix(1,1)                                                     % m1 is the (1,1) element of the m1-matrix 
% 3. Calculation of  Abar from new matrices F* and N*
Fstar=lambda1*F;                                                         % F* = l1 * F
Nstar= inv(max(eig(A))*eye(length(T))-T);                    % Replacing identiy matrix I with la1*I
AbarMatrix = Fstar*Nstar^2;                                         % This study
Abar=AbarMatrix(1,1)                                                    % Abar is the (1,1) element of the Abar-matrix