# k-fold matrix product

For $k \in \mathbb{N}$, $B,C \in \mathbb{R^{n,n}}$, given the matrices $B,C$ , calculate all powers $B^k$ and $C^k$

I'm a bit puzzled by this task. I assume it's supposed to practice handling expotentiated matrices, but since the number of natural numbers is infinite I must be missing something about this task.

What am I missing?

Edit: Here are the matrices $$B= \begin{pmatrix} -10 & -2 & -14 \\ -5 & 1 & -3 \\ 7 & 1 & 9 \\ \end{pmatrix}$$

$$C= \begin{pmatrix} 11 & -7 & -6 \\ -10 & 8 & 6 \\ 30 & -21 & -17 \\ \end{pmatrix}$$

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Are $B,C$ arbitrary or do they have some special structure or property? – vadim123 Jun 6 '13 at 13:31
Please give us the specific matrices. There might be some eigenvalue trick that allows for a simple closed form for the powers. – Andreas Caranti Jun 6 '13 at 13:31
added the matrices – Rickyfox Jun 6 '13 at 13:38

The first matrix has minimal polynomial $x^3$. This means that $B^3 = 0$, so powers are not a problem beyond that.
The second one has minimal polynomial $x^2 - x$, so it's diagonalizable, and one can see (although it is not needed in the following, see below) that it can be conjugated to $$D = \begin{pmatrix} 0 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}.$$ Note that $D^k = D$ for all $k$ (for that you just need the minimal polynomial, you do not even need to know the multiplicity of the eigenvalues), and that there exists an invertible matrix $P$ such that $$C = P^{-1} D P.$$ So powers are easy here too.
But note that all of this could have been obtained without any reference to eigenvalues (although they help to understand what's going on). In fact, you could have computed $B^2$, and then $B^3 = 0$, so you know $B^k = 0$ for $k \ge 3$. And you could have noted that $C^2 = C$, so that $C^k = C$ for all $k$.