I am studying canonical forms of matrices, and I'm a little stuck on something:
Consider a finite dimensional $\mathbb{F}-$vector space $V$ and an endomorphism $\alpha:\;V\to V$, and let $V_{\alpha}$ be the corresponding $\mathbb{F}[X]-$module i.e. $p\cdot \mathbf{v}=p(\alpha)[\mathbf{v}]$. Suppose: $$V_{\alpha}\cong \frac{\mathbb{F}[X]}{(X^k)}$$ as $\mathbb{F}[X]-$modules. Then the action of $X$ in $\mathbb{F}[X]/(X^k)$ viewed as an $\mathbb{F}-$vector space is: $$X\cdot (\mathbf{v}_1,\cdots,\mathbf{v}_k)=(0,\mathbf{v}_1,\cdots,\mathbf{v}_{k-1})$$
I can't immediately see why this is the case? I realise that $V_{\alpha}\cong \mathbb{F}[X]/(X^k)$ also as $\mathbb{F}-$modules, and I know that $1,X,\cdots ,X^{k-1}$ forms a basis of $\mathbb{F}[X]/(X^k)$, but I'm having trouble putting all this together. Would appreciate it if someone could help see this in clearer light.
If you write $x = \bar{X}$, then, as you have observed, $V_\alpha$ is an $\mathbf{F}$-vector space with $\mathbf{F}$-basis $\{1, x, x^2, \dots, x^{k-1}\}$. Since $V_\alpha$ is the $\mathbf{F}[X]$-module corresponding to $\alpha$, by definition, $\alpha$ is recovered from the action of $X$ on $V_\alpha$. And, the action of $X$ on a residue class in $\mathbf{F}[X]/(X^k)$ is
$$X \cdot \bar{f} = \overline{Xf} = x \bar{f}.$$
Your question boils down to figuring out how $x$ acts on your basis vectors $\{1, x, x^2, \dots, x^{k-1}\}$.
Do you see the claim now?