Suppose we have a field $\mathbb{Q}(\alpha)$ where the minimum polynomial of $\alpha$ is $P = x^3-2x^2+3x-5$
Prove that the mulitplication-by-$\alpha$ map $\phi$ : $\mathbb{Q}(\alpha) \rightarrow \mathbb{Q}(\alpha)$ given by $\phi(\beta) = \alpha\beta$ is a $\mathbb{Q}-$linear transformation
My reasoning:
Suppose $\beta, \gamma \in \mathbb{Q}(\alpha)$ and suppose $c$ is any scalar
$\phi(\beta + \gamma) = \alpha(\beta + \gamma) = \alpha\beta + \alpha\gamma = \phi(\beta)+\phi(\gamma)$
$\phi(c\beta) = \alpha c\beta = c\alpha\beta = c\phi(\beta)$
Compute $\phi(1), \phi(\alpha), \phi(\alpha^{2})$
My reasoning:
$\phi(1) = \alpha1 = \alpha$
$\phi(\alpha) = \alpha\alpha = \alpha^{2}$
$\phi(\alpha^{2}) = \alpha\alpha^{2} = \alpha^{3}$
Compute the matrix $A$ that represents $\phi$ with respect to the basis {$1, \alpha, \alpha^{2}$}
My reasoning:
I know that a matrix $A$ with respect to the basis {$1, \alpha, \alpha^{2}$} has columns with vectors $\phi(1), \phi(\alpha), \phi(\alpha^{2})$. Thus, I get the following matrix $A$:
$$ \begin{bmatrix} \alpha & 0 & 0 \\ 0 & \alpha^{2} & 0 \\ 0 & 0 & \alpha^{3} \end{bmatrix} $$
I was wondering if my reasoning was correct or if I have made mistakes (which I believe that I have). Thank you in advance for any help given.
Here's the matrix representation: $$\begin{bmatrix} 0 & 0 & 5\\ 1 & 0 & -3\\ 0 & 1 & 2 \end{bmatrix}.$$
$\phi(1)=\alpha=0\cdot 1+1\cdot \alpha+0\cdot\alpha^2$.
$\phi(\alpha)=\alpha^2=0\cdot 1+0\cdot\alpha +1\cdot \alpha^2$.
$\phi(\alpha^2)=\alpha^3= 5\cdot 1 -3\cdot\alpha+ 2\cdot\alpha^2$.
Notice that on the right of the $\cdot$'s I am writing basis elements.