The 'ratio' of a 2x2 matrix

Theorem

There are $x_n$ such that $$ A_n := \begin{pmatrix} a_n & b x_n \\ c x_n & d_n \end{pmatrix} = A^n $$

Proof by induction

For $n=1$ it is trivial: $x_1 := 1$.

Let the theorem be true for a $n$.

Then on the one hand $$ A^{n+1} = A^n \cdot \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a_n & b x_n \\ c x_n & d_n \end{pmatrix} \cdot \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} aa_n + cbx_n & b(a_n + dx_n) \\ c (d_n + ax_n) & dd_n + bcx_n \end{pmatrix}. $$

On the other hand $$ A^{n+1} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot A^n = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot \begin{pmatrix} a_n & b x_n \\ c x_n & d_n \end{pmatrix} = \begin{pmatrix} aa_n + cbx_n & b(d_n + ax_n) \\ c (a_n + dx_n) & dd_n + bcx_n \end{pmatrix}. $$

Thus $a_n + dx_n = d_n + ax_n =: x_{n+1}$. $\square$

In the sense of the above definition the matrix

$$A= \begin{pmatrix} 0 & 0\\ 1 & 0 \end{pmatrix}$$

has ratio $=0$. Since $A^2=0$, the ratio for $A^2$ is not defined.

It seems, that "ratio" is not a good definition.