GRE subject question regarding 100th degree of $3 \times 3$ matrix.

Question

Let

$$M = \begin{bmatrix} 0 &1 & 0 \\ 0 & 0 & 1\\ 1 & 0 & 0 \\ \end{bmatrix}$$

How can I quickly find $M^{100}$? Do I need to find a pattern of how many it takes to get to the identity? That seems tedious with a $3 \times 3$ matrix. This question comes from the GRE subject exam.

Answer 1

You could recognize this is a permutation matrix. It moves $e_1$ to $e_3$, moves $e_2$ to $e_1$, and $e_3$ to $e_2$. If you are familiar with cycle notation for permutation groups, this is the cycle $(132)$.

What is $(132)^{100}$? What is that written back again as a $3\times3$ matrix in this manner?

Answer 2

Systematically, recall the Hamilton--Cayley theorem which states that the characteristic polynomial of a matrix annihilates this matrix, so you could compute the characteristic polynomial to start: quickly, $$ \det (c\boldsymbol I - \boldsymbol M) = c^3 - 1, $$ thus $\boldsymbol M^3 = \boldsymbol I$, and $\boldsymbol M^{100} = \boldsymbol M$.

Of course this method would also become entangled if you are dealing with matrices of bigger size. Then a possible alternative might be finding the Jordan canonical form, which has a relatively simple form for matrix powers.