Prove that T is a linear transformation Does it matter that in the first line it's written $T(\alpha p+ \beta g)$ and not $T(\alpha p(t)+ \beta g(t))$ but at the end it is written with $\alpha T(p(t)) + \beta T(g(t)))$ with the $t$'s.

Define $T : \mathbb{P}_3 \to \mathbb{R}^4$ by $$T(p) = \begin{bmatrix}
p(-3) \\
p(-1) \\
p(1) \\
p(3)
\end{bmatrix}$$ Show that $T$ is a linear transformation. 



There is a technical difference between $f$ and $f(x)$ that I want to elaborate on.
$f$ is a function. $f(x)$ is a function evaluated at an indeterminate (this might be the wrong word for it, but my brain is on the fritz) point, $x$. Thus the sentence "$f(x)$ is differentiable" is technically incorrect because $f(x)$ is the value of $f$ at $x$ and is therefore a number and not a function. In contrast, "$f$ is differentiable" is correct (of course, its only correct when it is in fact true). For another example, "$f$ is given by $x^2+1$" is also wrong. $f$ is given by $\{(x,y)\in S:x^2+1=y\}$, or however your underlying foundation specifies functions. It is $f(x)$ that is given by $x^2+1$.
In practice, this difference rarely is relevant and the vast majority of people are content to wave their hands and ignore it. Instead, they use $f(x)$ to refer to both the function and the function evaluated at the indeterminate point.
Ironically, the main place that people attend to this difference is done incorrectly. People write $f\in\mathbb{Z}[x]$ thinking that that's preferred to $f(x)$, but $f$ is not an element of $\mathbb{Z}[x]$ because that's not a set of functions - it's a ring of numbers. It happens to be that these numbers have a particular correspondence to functions from $\mathbb{Z}$ to $\mathbb{Z}$ and that $\mathbb{Z}[x]$ is isomorphic to an interesting subset of that set of functions, but they are distinct structures.
