Problem source

The matrices $\mathbf{I}$ and $\mathbf{J}$ are 
\[
\mathbf{I}=\begin{pmatrix}1 & 0\\
0 & 1
\end{pmatrix}\quad\mbox{ and }\quad\mathbf{J}=\begin{pmatrix}1 & 1\\
1 & 1
\end{pmatrix}
\]
respectively and $\mathbf{A}=\mathbf{I}+a\mathbf{J},$ where $a$
is a non-zero real constant. Prove that 
\[
\mathbf{A}^{2}=\mathbf{I}+\tfrac{1}{2}[(1+2a)^{2}-1]\mathbf{J}\quad\mbox{ and }\quad\mathbf{A}^{3}=\mathbf{I}+\tfrac{1}{2}[(1+2a)^{3}-1]\mathbf{J}
\]
and obtain a similar form for $\mathbf{A}^{4}.$

If $\mathbf{A}^{k}=\mathbf{I}+p_{k}\mathbf{J},$ suggest a suitable
form for $p_{k}$ and prove that it is correct by induction, or otherwise.

Solution source

If $\mathbf{J}=\begin{pmatrix}1 & 1\\
1 & 1
\end{pmatrix}$, them $\mathbf{J}^2=\begin{pmatrix}2 & 2\\
2 & 2
\end{pmatrix} = 2\mathbf{J}$. Therefore $\mathbf{J}^n = 2\mathbf{J}^{n-1} = 2^{n-1}\mathbf{J}$

Let $\mathbf{A}=\mathbf{I}+a\mathbf{J}$ then 
\begin{align*}
\mathbf{A}^2 &=\l \mathbf{I}+a\mathbf{J}\r^2 \\
&= \mathbf{I}+2a\mathbf{J} + a^2\mathbf{J}^2 \\
&= \mathbf{I}+2a\mathbf{J} + 2a^2\mathbf{J} \\
&= \mathbf{I}+(2a+ 2a^2)\mathbf{J} \\
&= \mathbf{I}+\frac12(1+4a+ 4a^2-1)\mathbf{J} \\
&= \mathbf{I}+\frac12((1+2a)^2-1)\mathbf{J} \\
\end{align*}

\begin{align*}
\mathbf{A}^3 &=\l \mathbf{I}+a\mathbf{J}\r^3 \\
&= \mathbf{I}+3a\mathbf{J} + a^2\mathbf{J} + a^3\mathbf{J}^3 \\
&= \mathbf{I}+3a\mathbf{J} + 6a^2\mathbf{J} + 4a^3\mathbf{J} \\
&= \mathbf{I}+(3a+ 6a^3+4a^3)\mathbf{J} \\
&= \mathbf{I}+\frac12(1+3\cdot2a+3\dot4a^2+ 8a^3-1)\mathbf{J} \\
&= \mathbf{I}+\frac12((1+2a)^3-1)\mathbf{J} \\
\end{align*}

\begin{align*}
\mathbf{A}^4 &=\l \mathbf{I}+a\mathbf{J}\r^4 \\
&= \mathbf{I}+4a\mathbf{J} + 6a^2\mathbf{J}^2 + 4a^3\mathbf{J}^3+a^4\mathbf{J}^4 \\
&= \mathbf{I}+4a\mathbf{J} + 12a^2\mathbf{J} + 16a^3\mathbf{J}+8a^4\mathbf{J}\\
&= \mathbf{I}+(4a+ 12a^3+16a^3+8a^4)\mathbf{J} \\
&= \mathbf{I}+\frac12(1+4\cdot2a+6\cdot4a^2+ 4\cdot8a^3+16a^4-1)\mathbf{J} \\
&= \mathbf{I}+\frac12((1+2a)^4-1)\mathbf{J} \\
\end{align*}

Claim: $\mathbf{A}^k = \mathbf{I} + \frac12 ((1+2a)^{k}-1)\mathbf{J}$

Proof: Firstly, note that $\mathbf{I}$ commutes with everything, so we can just apply the binomial theorem as if we were using real numbers:

\begin{align*}
\mathbf{A}^k &=\l \mathbf{I}+a\mathbf{J}\r^k \\
&= \sum_{i=0}^k \binom{k}{i}a^i\mathbf{J}^i \\
&= \mathbf{I} + \sum_{i=1}^k \binom{k}{i}a^i2^{i-1}\mathbf{J} \\
&= \mathbf{I} + \frac12\l\sum_{i=1}^k \binom{k}{i}a^i2^{i}\r\mathbf{J} \\
&= \mathbf{I} + \frac12\l\sum_{i=0}^k \binom{k}{i}a^i2^{i} - 1\r\mathbf{J} \\
&= \mathbf{I} + \frac12\l(1+2a)^k - 1\r\mathbf{J}
\end{align*}
as required