Matrix power ($A^k$)

For a square matrix $A$ and a nonnegative integer $k$, the matrix power is the repeated productwith $A^0=I$.

A cube root of a matrix $A$ is a square matrix $B$ of the same size over the specified field satisfying $B^3=A$. Every complex invertible matrix has one: put it in Jordan normal form, choose a scalar cube root of each nonzero eigenvalue, and apply the truncated binomial cube root of the identity plus a nilpotent matrix on each block. The Nilpotent Jordan block $J_3(0)$ has none, even though its square is nonzero. If $B^3=J_3(0)$, then $B^9=0$, making $B$ a nilpotent matrix of size three; the Cayley-Hamilton theorem then gives $B^3=0$, a contradiction.

Compute a nonnegative integer power $A^n$ by repeated squaring. If $n=2k$, use $A^n=(A^k{)}^2$; if $n=2k+1$, use $A^n=A(A^k{)}^2$. This needs $O(\log n)$ matrix multiplications and permits fast evaluation of fixed-order linear recurrences via a companion matrix. It is distinct from computing the matrix exponential $e^A$.

If a diagonalizable matrix has distinct eigenvalues $r,s$, its spectral projections of a diagonalizable matrix are $(A-sI)/(r-s)$ and $(rI-A)/(r-s)$. Raising the diagonal form to a nonnegative integer power gives the displayed formula. Equivalently, interpolate the scalar function $t^n$ by a degree-one polynomial at $r,s$, then apply it to $A$. The diagonalizability assumption matters: knowledge of two eigenvalues alone does not exclude nontrivial Jordan blocks.