Need help with induction proof for a matrix raised to the 2^n

Question

I need some help proving this using induction. I can't seem to manipulate the algebra well enough.

Show that $$ \begin{bmatrix} 2 & 2 \\ 0 & 2 \end{bmatrix}^{2^n} = \begin{bmatrix} 2^{\left(2^n\right)} & 2^{\left(2^n +n\right)} \\ 0 & 2^{\left(2^n\right)} \end{bmatrix} $$ for n $\geq0$

Answer 1

If we define $m = 2^n$, then it is enough to show that$$ \begin{bmatrix} 2 & 2 \\ 0 & 2 \end{bmatrix}^{m} = \begin{bmatrix} 2^{m} & m\,2^{m} \\ 0 & 2^{m} \end{bmatrix} $$ for $m\geq1$. It is clear that the statement is true for $m=1$. Supose that the statement is true for $m=k$ where $k\ge 1$ is an integer. We want to show that it is also true for $m=k+1$. We have$$ \begin{bmatrix} 2 & 2 \\ 0 & 2 \end{bmatrix}^{k+1} =\begin{bmatrix} 2 & 2 \\ 0 & 2 \end{bmatrix} \,\begin{bmatrix} 2 & 2 \\ 0 & 2 \end{bmatrix}^{k} \\ = \begin{bmatrix} 2 & 2 \\ 0 & 2 \end{bmatrix}\,\begin{bmatrix} 2^{k} & k\,2^{k} \\ 0 & 2^{k} \end{bmatrix} = \begin{bmatrix} 2^{k+1} & (k+1)\,2^{k+1} \\ 0 & 2^{k+1} \end{bmatrix}, $$ which completes the proof.

Answer 2

Base Step: $${\begin{bmatrix} 2 & 2 \\ 0 & 2\end{bmatrix}^{2^0} = \begin{bmatrix} 2^{2^0} & 2^{2^0+0} \\ 0 & 2^{2^0}\end{bmatrix}}$$

Iterative Step: $$\begin{align}\begin{bmatrix} 2 & 2 \\ 0 & 2\end{bmatrix}^{2^{n+1}} ~&={{\left(\begin{bmatrix} 2 & 2 \\ 0 & 2\end{bmatrix}^{2^{n}}\right)}^2}&\text{Because}~& A^{2^{n+1}}={(A^{2^n})}^2\\[1ex] &= &\text{Assuming}~& \begin{bmatrix} 2 & 2 \\ 0 & 2\end{bmatrix}^{2^{n}} = \begin{bmatrix} 2^{2^n} & 2^{2^n+n} \\ 0 & 2^{2^n}\end{bmatrix}\\[1ex] &= &\text{Because}~&\begin{bmatrix} a & b \\ 0 & a\end{bmatrix}^{2} = \begin{bmatrix} a^2 & 2ab \\ 0 & a^2\end{bmatrix}\\[1ex] &= {\begin{bmatrix} 2^{2^{n+1}} & 2^{2^{n+1}+n+1} \\ 0 & 2^{2^{n+1}}\end{bmatrix}}\end{align}$$