This is a problem from a on-line source which yet comes with a solution (self-studier; not h.w.).
Let $E = \mathbb Q(\theta)$, where $\theta$ is a root of the irreducible polynomial \[ X^3 -3X + 1. \] I was wondering how one gets the matrix of the endomorphism $E \to E$ induced by multiplication by $\theta^2$.
I can see that $\theta^2 \cdot 1 = \theta^2$, $\theta^2 \cdot \theta = \theta^3 = 3\theta - 1$, and $\theta^2 \cdot \theta^2 = \theta \cdot \theta^3 = 3\theta^2 - \theta$.
This is where I fall apart. The solution gives the matrix \begin{bmatrix} 0&- 1&0 \\ 0&3&-1 \\ 1&0&3 \end{bmatrix}
I can that see if you multiply this by a $3 \times 1$ column vector of $1$'s, $4\theta^2$ + $2\theta$ $- 1$ emerges.
But I don't see how to derive the matrix; and, further, if it is unique. That is to say, if the $-1$ in the top row was in a different position in the top row, it seems you would still get $-1$ in the top position in the product.
Only a little of this is mine — except for the misstatements and confusion.
Thanks.