A discrete-time system is described by the state equation
$$ V(k+1) = A V(k) + B u(k) $$
with $V(k) = [x(k), y(k)]^T$ being the state vector, and
$$ A = \begin{bmatrix} 2 && - 3 \\ 0.5 && -0.5 \end{bmatrix} $$
$$ B = \begin{bmatrix} 0 \\ 1 \end{bmatrix} $$
$$ u(k) = 1, \ k \ge 0 $$
If the system is initially relaxed, i.e. $V(0) = [0,0]^T $, then what is the value of the following limit
$$ \displaystyle \lim_{k \to \infty } \dfrac{x(k)}{y(k)} $$
My approach: is detailed in my answer below.
I appreciate your comments, and alternative solutions.
My solution was based initially on fixed-point analysis.
If $V^* = \displaystyle \lim_{k \to \infty} V(k) $, then
$$ V^* = A V^* + B $$
So that
$ (I - A) V^* = B $
Now, $(I - A) = \begin{bmatrix} -1 && 3 \\ -0.5 && 1.5 \end{bmatrix} $
$\det(I - A) = (-1) (1.5) + 0.5 (3 ) = 0 $
So we cannot solve for $V^*$ by matrix inversion. However, if
$ V^* = [ x^*, y^*]^T $, then
$ \begin{bmatrix} -1 && 3 \\ -0.5 && 1.5 \end{bmatrix} \begin{bmatrix} x^* \\ y^* \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} $
This system of equations has no solution. This indicates that there is no fixed point, and that $V(k)$ is unbounded.
So I took a different approach, and used eigenstructure analysis.
Matrix $A$ can be diagonalized as follows
$ A = P L P^{-1} $
where
$ L = \begin{bmatrix} 1 && 0 \\ 0 && 0.5 \end{bmatrix} $
$ P = \begin{bmatrix} 3 && 2 \\ 1 && 1 \end{bmatrix} $
So that
$ V(k+1) = P L P^{-1} V(k) + B u(k) $
From which
$ P^{-1} V(k+1) = L P^{-1} V(k) + P^{-1} B u(k) $
Let $W(k) = P^{-1} V(k)$, and let $C = P^{-1}B $ then
$ W(k+1) = L W(k) + C $
Applying the $z$-Transform, and using the fact that $W(0) = 0$, we get
$ W(z) = (zI - L)^{-1} C \left( \dfrac{z}{z-1} \right) $
And we have
$ zI - L = \begin{bmatrix} z - 1 && 0 \\ 0 && z - 0.5 \end{bmatrix} $
$ P^{-1} = \begin{bmatrix} 1 && - 2\\ -1 && 3 \end{bmatrix} $
So,
$W(z) = \begin{bmatrix} \dfrac{z}{(z-1)^2} && 0 \\ 0 && \dfrac{z}{(z-1)(z-0.5)} \end{bmatrix} \begin{bmatrix} -2 \\ 3 \end{bmatrix} = \begin{bmatrix} - \dfrac{2z}{(z - 1)^2} \\ \dfrac{3z}{(z-1)(z - 0.5)} \end{bmatrix} $
Applying the inverse Z-Transform, we get
$ W(k) = \begin{bmatrix} -2 k \\ 6 - 6 (0.5)^k \end{bmatrix}$
Finally
$ V(k) = P W(k) = \begin{bmatrix} 3 && 2 \\ 1 && 1 \end{bmatrix} \begin{bmatrix} -2 k \\ 6 - 6 (0.5)^k \end{bmatrix} = \begin{bmatrix} -6 k + 12 - 12 (0.5)^k \\ - 2 k + 6 - 6 (0.5)^k \end{bmatrix} $
Now, it is evident that
$ \displaystyle \lim_{k \to \infty } \dfrac{x(k)}{y(k)} = \lim_{k \to \infty } \dfrac{-6 k + 12 - 12 (0.5)^k }{- 2 k + 6 - 6 (0.5)^k} = 3 $