The 'ratio' of a 2x2 matrix Define the 'ratio' of a 2x2 matrix $$A= \begin{pmatrix} a & b\\ c & d \end{pmatrix} $$
to be $\frac{b}{c}$ when $c\neq 0$. Show that the ratio of $A^n$ is equal to the ratio of $A$, when the ratio of $A^n$ is well-defined.
My instinct is to go with a proof by induction, but I really can't see a way to prove this.
 A: 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.
A: 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 $a_n + dx_n = d_n + ax_n =: x_{n+1}$. $\square$
A: Diagonalize the matrix.
If it can be diagonalized, you get $A =PDP^{-1}$.  The ratio depends on $P$, not on the diagonal matrix $D$.  All powers of $A$ use the same $P$.
If it can't be diagonalized, then $D$ becomes an upper triangular matrix with equal numbers on the diagonal, and again the ratio is independent of $D$
