$n$-th power of a complex non-diagonalizable $2 \times 2$ matrix

Question

I have a complex matrix in the form $$ \begin{pmatrix} a-ib & ic \\ -ic & a+ib \end{pmatrix} $$ which is not diagonalizable according to my understanding. How to calculate the $n$-th power of this matrix? Can we use eigenvalue decomposition or Cayley-Hamilton theorem on this? Or is there a way to approximate the polynomials?

Answer 1

The characteristic polynomial is $$X^2-2aX+a^2+b^2-c^2$$ To be nondiagonalizable, this would have to equal $(X-r)^2$ for some $r$. Examining the linear term, $r$ would have to be $a$. Examining the constant term, $b^2$ would have to equal $c^2$. So as noted in the comments, $c=\pm b$ is required for this to be not diagonalizable. And in that situation, $a$ is the one eigenvalue.

So now consider $$M-aI=i\begin{pmatrix} -b & c \\ -c & b \end{pmatrix}$$ Its kernel is spanned by $\begin{pmatrix} c \\ b \end{pmatrix}$. And $\frac{-i}{2b}\begin{pmatrix} -c \\ b \end{pmatrix}\mapsto\begin{pmatrix} c \\ b \end{pmatrix}$.

It follows from the usual diagonalization/Jordan-normalization process that $$M=\begin{pmatrix} c & \frac{ic}{2b} \\ b & -\frac{i}{2} \end{pmatrix}\begin{pmatrix} a & 1 \\ 0 & a \end{pmatrix}\begin{pmatrix} c & \frac{ic}{2b} \\ b & -\frac{i}{2} \end{pmatrix}^{-1}$$

and then

$$M^n=\begin{pmatrix} c & \frac{ic}{2b} \\ b & -\frac{i}{2} \end{pmatrix}\begin{pmatrix} a & 1 \\ 0 & a \end{pmatrix}^n\begin{pmatrix} c & \frac{ic}{2b} \\ b & -\frac{i}{2} \end{pmatrix}^{-1}$$ $$M^n=\begin{pmatrix} c & \frac{ic}{2b} \\ b & -\frac{i}{2} \end{pmatrix}\begin{pmatrix} a^n & na^{n-1} \\ 0 & a^n \end{pmatrix}\begin{pmatrix} c & \frac{ic}{2b} \\ b & -\frac{i}{2} \end{pmatrix}^{-1}$$