Expression for arbitrary powers of a particular $2\times2$ matrix Given$$\mathbf{M}=
        \begin{pmatrix}
        7 & 5  \\
        -5 & 7  \\
        \end{pmatrix}
$$, what's the formula matrix for $\mathbf{M}^n$?  The eigenvalues and eigenvectors are complex and need to generate a real number formulas for each component of the resulting matrix.
 A: From what my professor states, the matrix (a,b;-b,a) is an isomorphic representation of a+bi.  This complex number is also isomorphically represented in polar coordinates as r*e^(i*theta), where r=√(a²+b²) and theta=tan^-1(b/a).  This form is easier to find M^n=(r^n)*e^(i*n*theta).  Essentially, r, theta change into r'=r^n, theta'=n*theta.  This new polar coordinate isomorphs back into standard form with a'+b'i=r'*cos(theta')+i*r'*sin(theta')=((√(a²+b²))^n)*cos(n*tan^-1(b/a))+i*((√(a²+b²))^n)*sin(n*tan^-1(b/a)).  This new complex number isomorphs back into matrix form as ( ((√(a²+b²))^n)*cos(n*tan^-1(b/a)) , ((√(a²+b²))^n)*sin(n*tan^-1(b/a)) ; -((√(a²+b²))^n)*sin(n*tan^-1(b/a)) , ((√(a²+b²))^n)*cos(n*tan^-1(b/a)) ).  This formula works with positive a values.  However, negative a values requires a phase shift of pi to put theta in the second or third quadrant.  That phase shift is suppose to work, but doesn't on the computer question.  Positive a is accepted.
A: $\textbf M=\begin{bmatrix}7&5\\-5&7\end{bmatrix}$
The eigenvalues will be the roots of the characteristic polynomial, $\lambda^2-(\mathrm{tr}\ \textbf M)\lambda+\det\textbf M=0$.  $\lambda^2-14\lambda+74=0$, so $\lambda=7\pm5i$.
Thus, the diagonalization of $\textbf M=\textbf A^{-1}\cdot\begin{bmatrix}7+5i&0\\0&7-5i\end{bmatrix}\cdot\textbf A$.
We know that $\textbf A$ must be the corresponding eigenvectors in column form, so we must solve $\textbf M \begin{bmatrix}v_1\\v_2\end{bmatrix}=\lambda\begin{bmatrix}v_1\\v_2\end{bmatrix}$. We obtain 
$$\begin{align} 7v_1+5v_2&=\lambda v_1 \\ -5v_1+7v_2&=\lambda v_2 \end{align}$$
Or
$$\begin{align} \mp 5v_1i+5v_2&=0 \\ -5v_1\mp 5v_2i&=0 \end{align}$$
You should be able to solve it from here.
