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Suppose a matrix $A=\begin{pmatrix} a & b \\ a & 0 \end{pmatrix}$. I need to calculate $\lim_{n\to\infty}A^n$, so I started doing literally power by power to see a pattern, and this pattern emerges:

$$A^{n+1}=\begin{pmatrix} (A^n_{1,1}+A^n_{1,2})a & A^n_{1,1}b \\ (A^n_{2,1}+A^n_{2,2})a & A^n_{2,1}b \end{pmatrix}$$

At $n=\infty$ supposedly $n=n+1$ so I define $\begin{pmatrix} x & y \\ z & t \end{pmatrix}$ as $A^n=A^{n+1}$

so we have that

$$ \begin{pmatrix} x & y \\ z & t \end{pmatrix}= \begin{pmatrix} (x+y)a & xb \\ (z+t)a & zb \end{pmatrix} $$

or

$$ \left\{ \begin{aligned} (x+y)a &= x\\ xb&=y \\ (z+t)a &= z \\ zb&=t \end{aligned} \right. $$

simplifying we get to

$$ \left\{ \begin{aligned} b+1&=\frac 1a\\ ab+a&=1 \end{aligned} \right. $$

Isolating $b$ in the first we have that $b=\frac 1a-1$; substituting on the second:

\begin{align} a(1/a-1)+a&=1 \\ 1-a+a=1 \\ 1=1 \end{align}

The problem is I don't understand the result. Does this mean that $\lim_{n\to\infty}A^n$ is defined no matter what values $a$ and $b$ we have? What does it mean?

Garmekain
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Let $B:=\lim_{n \to \infty}A^n$ if this limit exists. You noted that such a limit must satisfy $BA=B$, so after some calculations you found $b+1=1/a$. Thus the limit can only exist if $a$ and $b$ satisfy that equation.

You said $a$ and $b$ are integers, and there are two solutions to $b+1=1/a$. In the one case you can calculate $A^2$ and recognize what will happen as you take higher powers of $A$. In the second case, you can look at the determinant of $A$ and note that $\text{det}(A^n)=\text{det}(A)^n$, which will tell you something about whether or not $B$ can exist.

kccu
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