Show that $\phi: \mathbb{R}_3[x]\rightarrow\mathbb{R}^3, \phi(p):=[p(-1), p(0), p(1)] $ is a linear transformation
Solution 1:
Apparently, the problem is essentially solved within the formulation and comments, but because there are a lot of upvotes and no answer, I propose to write one up.
As OP rightly observes, the map in question obeys the rule: $$\phi( \alpha p + \beta q ) = \alpha \phi(p) + \beta \phi(q)$$ (note that the expression $\alpha \begin{bmatrix} p(-1) \\ p(0) \\ p(1) \end{bmatrix} + \beta \begin{bmatrix} q(-1) \\ q(0) \\ q(1) \end{bmatrix} $ is $\alpha \phi(p) + \beta \phi(q)$ ).
It remains to see that this condition implies linearity of $\phi$, which is defined by the following contidions:
- For any $p,q \in \mathbb R_3[x]$, it holds that $\phi(p + q) = \phi(p) + \phi(q)$.
- For any $p \in \mathbb R_3[x]$, $\alpha \in \mathbb R$, it holds that $\phi(\alpha p) = \alpha \phi(p)$.
We know that $\phi$ satisfies the condition:
- For any $p,q \in \mathbb R_3[x]$, $\alpha,\beta \in \mathbb R$, it holds that $\phi(\alpha p + \beta q) = \alpha \phi(p) + \beta \phi(q)$.
Now, putting $\alpha = \beta = 1$ in 3., we recover condition 1. Putting $\beta = 0$ and $q$ arbitrary, we recover condition 2. So, both conditions hold, and $\phi$ is indeed linear.
As an aside, note that if $\phi$ is linear, i.e. 1. and 2. hold, then: $$ \phi(\alpha p + \beta q) = \phi( \alpha p) + \phi(\beta q)= \alpha \phi(p) + \beta \phi(q)$$ so condition 3. follows. This means that 3. is in fact a different formulation of linearity (in fact, it might be taken as an alternative definition).