Symbolically raise square matrices to a symbolic power
The matrix A^^t returns a symbolic matrix power. I think it requires A to be a square matrix, otherwise it doesn't make much sense.
I'm guess there are multiple ways to do this, but a Hungarian mathematician taught me this method (since its nearly identical for calculating the matrix exponential).
(i) First, reexpress the general solution in terms of unknown matrices B1, B2, ... and eigenvalues: A^^t = B1 * lambda1^t + B2 * lambda2^t +...
When there are repeated eigenvalues you use the same trick as in solving differential equations. For example, one eigenvalue with an algebraic multiplicity of four would have the general form: A^^t = B1_1 * lambda1^t + B1_2 * t * lambda1^t + B1_3 * t^2 * lambda1^t + B1_4 * t^3 * lambda1^t + B2 * lambda2^t + ...
(ii) Then we can generate additional equations by incrementing t. A^^t = B1_1 * lambda1^t + B1_2 * t * lambda1^t + B1_3 * t^2 * lambda1^t + B1_4 * t^3 * lambda1^t + B2 * lambda2^t + ... A^^(t+1) = B1_1 * lambda1^(t+1) + B1_2 * (t+1) * lambda1^(t+1) + B1_3 * (t+1)^2 * lambda1^(t+1) + ... A^^(t+2) = B1_1 * lambda1^(t+2) + B1_2 * (t+2) * lambda1^(t+2) + B1_3 * (t+2)^2 * lambda1^(t+2) + ... ... Do this to get enough linearly independent equations (where the variables are actually the matrices B1_1, B1_2, B1_3, B1_4, B2, B3...)
(iii) If there are no zero eigenvalues: Since these equations are true regardless the value of t, you can set t=0 and then solve the multilinear equation for each of the B matrices in terms of Identity, A^^1, A^^2, etc.
If zero is an eigenvalue: Same as above, but you may have to set t=1 instead of zero to avoid degenerate solutions, such as A^^t = Identity.
(iv) Then solve using linear algebra. You're augmented matrix will look something like this (assuming we set t=1)
(Header on the top row)
B1_1, B1_2, B1_3 , B1_4 | A, A^^2, A^^3, A^^4 ---------------------------------------------------- 1 , 1 , 1 , 1 | 1, 0 , 0 , 0 L1 , 2*L1 , 4*L1 , 8*L1 | 0, 1 , 0 , 0 L1^2, 3*L1^2, 9*L1^2 , 27*L1^2| 0, 0 , 1 , 0 L1^3, 4*L1^3, 16*L1^3, 64*L1^3| 0, 0 , 0 , 1
Thus using linear algebra techniques, you can solve for each B matrix in terms of the A matrix and its powers.
A^^t properties when calculated using the above method
When A is invertible, A^^t, t=-1 is the inverse. When A is not invertible, A^^t, t=-1 is the Drazin inverse. When A is not invertible, A^^t, t=0 need not be the identity matrix. A^^t1 . A^^t2 = A^^(t1 + t2)
How the above method is related to matrixexp
This same method can calculate matrix exponentials with two changes: (1) instead of multiplying each B matrix by lambda^t, multiply by exp(lambda*t) (2) instead of incrementing t, to get more equations, take the derivative with respect to t. The rest of the procedure is the same if I recall correctly at the moment.
---Conclusions I think I could throw together a hodge-podge of methods to go through this process, but I'm not familiar enough with Maxima to: make it robust enough to behave nicely when the matrix is not square; allow it to solve a simpler set of linear equations when the A matrix does not have a zero eigenvalue; make an example that shows how to use it; make documentation for Maxima so people know that A^^-1 is the Drazin inverse and not the Moore-Penrose inverse.