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\begin{document}
\title{Raising a Matrix to Power n}
\author{Amit Yaron}
\date{Aug 1, 2021}

\maketitle
Hello, another one from YouTube. The reason I want to solve it is
that the solver is Michael Penn, a serious math problem solver.

So, the matrix to be raised to power $n\in\mathbb{N}$ is:

\begin{multline*}
A=\begin{pmatrix}4 & -\sqrt{5}\\
2\sqrt{5} & -3
\end{pmatrix}
\end{multline*}

Raisning matrix A to power n is easy if matrix A is diagonalizabe,
i.e. there exists a matrix M. such that:

\begin{multline*}
M^{-1}AM
\end{multline*}

is diagonal. To find if it is, let us calculate A's characteristic
polynomial $P_{A}(t)$:

\begin{multline*}
P_{A}(t)=det(tI-A)=det\begin{pmatrix}t-4 & \sqrt{5}\\
-2\sqrt{5} & t+3
\end{pmatrix}=(t-4)(t+3)-\sqrt{5}(-2\sqrt{5})=\\
=t^{2}-t-12+10=t^{2}-t-2=(t-2)(t+1)
\end{multline*}

So, we have a distinct quadratic polynomial with two distinct zeros,
thus the matrix A is diagonizable.

The 2 eigen-values of matrix A are:

\begin{multline*}
\lambda_{1}=2,\ \lambda_{2}=-1
\end{multline*}

So, there exist a diagonal matrix D and a square matrix M, such that:

\begin{align*}
D=M^{-1}AM & =\begin{pmatrix}\begin{array}{rr}
2 & 0\\
0 & -1
\end{array}\end{pmatrix}\Rightarrow\\
\Rightarrow & MD=AM
\end{align*}

Now, let us find the elements of M, by solving systems of linear equations.
Let

\begin{multline*}
M=\begin{pmatrix}\begin{array}{rr}
m_{11} & m_{12}\\
m_{21} & m_{22}
\end{array}\end{pmatrix}
\end{multline*}

So,

\begin{multline*}
AM=MD=\begin{pmatrix}\begin{array}{rr}
2m_{11} & -m_{12}\\
2m_{21} & -m22
\end{array}\end{pmatrix}
\end{multline*}

System 1:

\begin{multline*}
\begin{pmatrix}\begin{array}{rr}
4 & -\sqrt{5}\\
2\sqrt{5} & -3
\end{array}\end{pmatrix}\begin{pmatrix}m_{11}\\
m_{21}
\end{pmatrix}=\begin{pmatrix}2m_{11}\\
2m_{21}
\end{pmatrix}\Rightarrow\\
\Rightarrow\begin{cases}
4m_{11}-\sqrt{5}m_{21} & =2m_{11}\\
2\sqrt{5}m_{11}-3m_{21} & =2m_{21}
\end{cases}\Rightarrow\\
\Rightarrow\begin{cases}
2m_{11}-\sqrt{5}m_{21} & =0\\
2\sqrt{5}m_{11}-5m_{21} & =0
\end{cases}
\end{multline*}

From both equations, we can get that:

\begin{multline*}
2m_{11}-\sqrt{5}m_{21}=0\Rightarrow2m_{11}=\sqrt{5}m_{21}\Rightarrow m_{11}=\frac{\sqrt{5}}{2}m_{21}
\end{multline*}

So, let us take:

\begin{multline*}
\begin{cases}
m_{11} & =\frac{\sqrt{5}}{2}\\
m_{21} & =1
\end{cases}
\end{multline*}

System 2:

\begin{multline*}
\begin{pmatrix}\begin{array}{rr}
4 & -\sqrt{5}\\
2\sqrt{5} & -3
\end{array}\end{pmatrix}\begin{pmatrix}\begin{array}{l}
m_{12}\\
m_{22}
\end{array}\end{pmatrix}=\begin{pmatrix}\begin{array}{l}
-m_{12}\\
-m_{22}
\end{array}\end{pmatrix}\Rightarrow\\
\Rightarrow\begin{cases}
4m_{12}-\sqrt{5}m_{22} & =-m_{12}\\
2\sqrt{5}m_{12}-3m_{22} & =-m_{22}
\end{cases}\Rightarrow\\
\Rightarrow\begin{cases}
5m_{12}-\sqrt{5}m_{22} & =0\\
2\sqrt{5}m_{12}-2m_{22} & =0
\end{cases}
\end{multline*}

From both equation, we get that:

\begin{multline*}
m_{22}=\sqrt{5}m_{12}
\end{multline*}

So, let us take:

\begin{multline*}
\begin{cases}
m_{12} & =1\\
m_{22} & =\sqrt{5}
\end{cases}
\end{multline*}

Thus, our matrix is:

\begin{multline*}
M=\begin{pmatrix}\begin{array}{rr}
m_{11} & m_{12}\\
m_{21} & m_{22}
\end{array}\end{pmatrix}=\begin{pmatrix}\begin{array}{rr}
\frac{\sqrt{5}}{2} & 1\\
1 & \sqrt{5}
\end{array}\end{pmatrix}\Rightarrow\\
\Rightarrow M^{-1}=\frac{adj(M)}{det(M)}=\frac{\begin{pmatrix}\begin{array}{rr}
\sqrt{5} & -1\\
-1 & \frac{\sqrt{5}}{2}
\end{array}\end{pmatrix}}{\frac{3}{2}}=\begin{pmatrix}\begin{array}{rr}
\frac{2\sqrt{5}}{3} & -\frac{2}{3}\\
-\frac{2}{3} & \frac{\sqrt{5}}{3}
\end{array}\end{pmatrix}
\end{multline*}

Finally,

\begin{multline*}
\begin{pmatrix}\begin{array}{rr}
4 & -\sqrt{5}\\
2\sqrt{5} & -3
\end{array}\end{pmatrix}^{n}=A^{n}=(MDM^{-1})^{n}=MD^{n}M^{-1}=\\
\begin{pmatrix}\begin{array}{rr}
\frac{\sqrt{5}}{2} & 1\\
1 & \sqrt{5}
\end{array}\end{pmatrix}\begin{pmatrix}\begin{array}{rr}
2 & 0\\
0 & -1
\end{array}\end{pmatrix}^{n}\begin{pmatrix}\begin{array}{rr}
\frac{2\sqrt{5}}{3} & -\frac{2}{3}\\
-\frac{2}{3} & \frac{\sqrt{5}}{3}
\end{array}\end{pmatrix}=\begin{pmatrix}\begin{array}{rr}
\frac{\sqrt{5}}{2} & 1\\
1 & \sqrt{5}
\end{array}\end{pmatrix}\begin{pmatrix}\begin{array}{rr}
2^{n} & 0\\
0 & (-1)^{n}
\end{array}\end{pmatrix}\begin{pmatrix}\begin{array}{rr}
\frac{2\sqrt{5}}{3} & -\frac{2}{3}\\
-\frac{2}{3} & \frac{\sqrt{5}}{3}
\end{array}\end{pmatrix}=\\
=\begin{pmatrix}\begin{array}{rr}
2^{n-1}\sqrt{5} & (-1)^{n}\\
2^{n} & (-1)^{n}(\sqrt{5)}
\end{array}\end{pmatrix}\begin{pmatrix}\begin{array}{rr}
\frac{2\sqrt{5}}{3} & -\frac{2}{3}\\
-\frac{2}{3} & \frac{\sqrt{5}}{3}
\end{array}\end{pmatrix}=\begin{pmatrix}\begin{array}{rr}
\frac{5\cdot2^{n}}{3}+\frac{2}{3}(-1)^{n+1} & -\frac{2^{n}\sqrt{5}}{3}+\frac{\sqrt{5}(-1)^{n}}{3}\\
\frac{2^{n+1}\sqrt{5}}{3}+\frac{2\sqrt{5}(-1)^{b+1}}{3} & -\frac{2^{n+1}}{3}+\frac{5(-1)^{n}}{3}
\end{array}\end{pmatrix}
\end{multline*}

and that for sure is a good place to stop,
\end{document}

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