induction on matrices with powers + addition and limit

Question

$A= \begin{bmatrix} 1-q && p \\ q && 1-q \end{bmatrix}, 0<p<1, 0<q<1,$

Using mathematical induction show that

$A^n$ = $\frac{1}{p+q}\begin{bmatrix} q && p \\ q && p \end{bmatrix}$ + $\frac{(1-p-q)^n}{p+q}\begin{bmatrix} p && -p \\ -q && q \end{bmatrix}$

Assuming $|1-p-q|<1$ find $\lim_{n\to \infty} P^n$.

I know $A^{n+1} = A^n *A$, so assumption is $A^k = \frac{1}{p+q}\begin{bmatrix} q && p \\ q && p \end{bmatrix}$ + $\frac{(1-p-q)^k}{p+q}\begin{bmatrix} p && -p \\ -q && q \end{bmatrix}$

And we need to show that

$A^{k+1} =\frac{1}{p+q}\begin{bmatrix} q && p \\ q && p \end{bmatrix}$ + $\frac{(1-p-q)^{k+1}}{p+q}\begin{bmatrix} p && -p \\ -q && q \end{bmatrix}$

So $A^{k+1} = A^k * A = \frac{1}{p+q}\begin{bmatrix} q && p \\ q && p \end{bmatrix}\begin{bmatrix} 1-q && p \\ q && 1-q \end{bmatrix}$ +$\frac{(1-p-q)^{k+1}}{p+q}\begin{bmatrix} p && -p \\ -q && q \end{bmatrix}\begin{bmatrix} 1-q && p \\ q && 1-q \end{bmatrix}$

and I'm stuck from here on! I can't seem to get past this point. When I do multiplication on them it looks nothing like it's supposed to.

Answer 1

If this is true for all $n=1,2,3\ldots$ then $A=A^1$ must be eqial to

$$A= \begin{bmatrix} 1-q && p \\ q && 1-q \end{bmatrix}=\frac{1}{p+q}\begin{bmatrix} q && p \\ q && p \end{bmatrix} + \frac{(1-p-q)^1}{p+q}\begin{bmatrix} p && -p \\ -q && q \end{bmatrix}$$

For the ellement in the upper left corner we have $$1-q=\frac{q}{p+q}+p\left(\frac{1}{p+q}-1\right)=\frac{q}{p+q}+\frac{p}{p+q}-p=1-p$$

which is evidently correct only if $p=q$. May me there is a misprint in the conditions of the question?

Answer 2

It's probably a typo. The correct statement is perhaps $$ A=\pmatrix{1-q&p\\ q&1-\color{red}{p}} \ \Rightarrow\ A^n= \frac1{p+q}\pmatrix{p&p\\ q&q} + \frac{(1-p-q)^n}{p+q}\pmatrix{q&-p\\ -q&p} $$ or its transposed version, so that you are asked to determine the power of a stochastic matrix.