For a given $2 \times 2$ matrix $A$ ($\in \mathbb{R}^{2 \times 2})$ I want to compute $A^n$ using singular value decomposition, where the intermediate terms are "nice".
Assuming we have a singular value decomposition $A = U \Sigma V^T$ we can state $A^n = (U \Sigma V^T)^n$. Assuming n is big enough, this expands into
$$A^n = (U \Sigma V^T)(U \Sigma V^T)(U \Sigma V^T)(U \Sigma V^T)^{n-3}$$
Reassociating the terms gives
$$A^n = (U \Sigma) (V^T U) \Sigma (V^T U) (\Sigma V^T)(U \Sigma V^T)^{n-3}$$
If $V^T U = I$ ($I$ being the identity matrix) this greatly simplifies to
$$A^n = U \Sigma \Sigma \dots \Sigma \Sigma V^T = U \Sigma^n V^T$$
where $\Sigma^n$ is straightforward to compute as $\Sigma$ is a diagonal matrix.
If more generally $V^T U = mI$ where $m \in \mathbb{R}$ is some scalar, the equation reduces to
$$A^n = U (m \Sigma)^n V^T$$
which is also fairly straightforward to compute.
What restrictions or conditions do I need to apply to A such that there exists a decomposition $A = U \Sigma V^T$ where $V^T U = mI$ to make computing powers of A a little nicer?