3
$\begingroup$

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.

enter image description here

$\endgroup$
6
$\begingroup$

It doesn't really matter. But it doesn't look very good. I would've written $\alpha T(p) + \beta T(g)$ the last time. Also, why choose the letters $p$ and $g$? Why not $f$ and $g$, or $p$ and $q$? That would make the proof look nicer, at least to my eyes.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.