I was recently given a question asking me to find the sequences to a system of linear difference equations, in three sequences, xt, yt and zt. The matrix is below. \begin{pmatrix}1&1&4\\1&1&4\\4&4&-2\end{pmatrix}
The initial condition is given as (0,1,1).
I am afraid that it viewing the image would be necessary to understand the question here, unfortunately I am not experienced enough with this website enough to type out all the math.
(The eigenvectors and eigenvalues were all calculated in a previous subsection, and the calculations didn't seem relevant so I didn't include that in)
The solution uses a slightly different technique from what I am used to, which would be to diagonalize, solve the sequences with an arbitrary constant and multiply with the transition matrix to obtain an equation with two variables (In this case, 2 because one of the eigenvalues is 0), and solve for the constants. But here, that system would be inconsistent and unsolvable.
The answer uses a different equation. However, the answer provided does not recreate the correct values of (0,1,1) for the initial condition (t=0). (In the answer, x and y expression is the same, but clearly for the initial condition they are not the same).
I do understand the equation, but I am wondering why that technique in particular is able to obtain a meaningful solution (at least for t > 0, which it does seem to work for)? How is this equation any different from the "standard" way to solve the the system of difference equations?
The method itself is sound: by induction, $(x_t,y_t,z_t)^T = A^t(x_0,y_0,z_0)^T$. The initial values and $A^t$ let you compute any element of the three sequences directly, so there’s no need to introduce unknown constants as you might when solving a linear difference equation for a single sequence. However, the given solution has a subtle error in its handling of the eigenvalue $0$. In the diagonalization of $A^t$, the lower-right element of the diagonal matrix is identically $0$, but in order to obtain $A^0=I$, it needs to equal $1$ when $t=0$. Normally, the expression $0^0$ is indeterminate, but if you declare $0^0=1$ for this problem, then the correct solution can be obtained by using $0^t$ instead of $0$ for that matrix element. The last two lines of the solution then become $$= \frac16 \pmatrix{1&1&1\\1&1&-1\\1&-2&0} \pmatrix{4(6)^t\\-1(-6)^t\\-3(0)^t} \\ = \frac16 \pmatrix{4(6)^t-(-6)^t-3(0)^t \\ 4(6)^t-(-6)^t+3(0)^t \\ 4(6)^t+2(-6)^t},$$ which does produce the correct value at $t=0$ and obviously agrees with the given solution for $t\gt0$. If you don’t like defining $0^0$ specially for this context, an alternative is to use the Iverson bracket $[t=0]$ instead of $0^t$.