Sorry if the question is phrased in a clumsy way. I would definitely appreciate any help to make it clearer.
This is a question from A-Level Further Math (Cambridge Internation Edition)
Searchable Version:
Consider the matrix $A=\begin{pmatrix}1&0\\1&1\end{pmatrix}$. It is given that $A^n=\begin{pmatrix}1&0\\n&1\end{pmatrix}$.
Proof:
Let $P_k$ be the statement that, for some value $n=k$, $A^k=\begin{pmatrix}1&0\\k&1\end{pmatrix}$. To show this result is true, we must show the first case is true. This is achieved by considering $A^1=\begin{pmatrix}1&0\\1&1\end{pmatrix}$, which is true. Hence $P_1$ is true. Then consider $A^kA$, which is $$\begin{pmatrix}1&0\\k&1\end{pmatrix}\begin{pmatrix}1&0\\1&1\end{pmatrix}=\begin{pmatrix}1&0\\k+1&1\end{pmatrix}.$$ So it is true that $P_k\Rightarrow P_{k+1}$. Since $P_1$ is true and $P_{k} \Rightarrow P_{k+1}$, by the principle of mathematical induction, $P_n$ is true for all for all $n\geqq1$.
The problem with this proof is why are we trying to prove $A^{k+1}=A^k\cdot A$ instead of trying to prove $A\cdot A^k$, I realize the result is the same, but I am can't help but feeling a little confused as to what $A^{k+1}$ means, especially in light of the non-conversability of matrix multiplication.
In a set with a non-associative operation, powering becomes problematic from $A^3$ onwards; is $A^3=A^2A=(AA)A$ or is it $AA^2=A(AA)$? If we don't know that $(AA)A$ and $A(AA)$ are the same then we're stuck.
But matrix multiplication is associative and this problem doesn't arise. If multiplication is associative then there are two ways to potentially define powering recursively.
(i) $A^1=A$ and $A^{n+1}=A^nA$, and
(ii) $A^1=A$ and $A^{n+1}=AA^n$.
Whichever way you choose you can prove $A^mA^n=A^{m+n}$ by induction, and so that your definition is equivalent to the other one.