doubt about a linear function verification exercise good evening everyone, I need to determine whether the following function is linear or not: $f : V → R^3 $ given that $V = R[x]_≤4 $ and $f(p(x)) =
(p(0), p′(0), p(1))$.
This is what I did, I just need a confirmation about its correctness:
$p(x)=ax^4+bx^3+cx^2+dx+e, p(y)=fy^4+gy^3+hy^2+ig+l$
therefore, $$f(p(x) + λp(y))=f(e+λl,0,a+b+c+d+λ(f+g+h+i+l))$$
and $$f(p(x))+λp(y))=(e,0,a+b+c+d)+λ(l,0,f+g+h+i+l)=(e+λl,0,a+b+c+d+λ(f+g+h+i+l))$$ since the two expressions give the same result the function is linear. is this correct? thanks in advance
 A: Sort of. Almost. Not quite.
Firstly, you've made something of a classic error when dealing with vector spaces of polynomials. When you wanted two arbitrary elements of the vector space $\Bbb{R}[x]_{\le 4}$, you've produced $p(x)$ and $p(y)$, which is not correct. What you are denoting here is the same polynomial, but with two different dummy variables $x$ and $y$ substituted in. So, if you're saying that
$$p(x) = ax^4 + bx^3 + cx^2 + dx + e,$$
then $p(y)$ is not a polynomial function of $y$ with new coefficients. Instead, we would have
$$p(y) = ay^4 + by^3 + cy^2 + dy + e.$$
Think about it; this is just how evaluating a function works: if you replace $x$ with $y$ in $p(x)$, then you should replace all instances of $x$ with $y$ (and make no other changes).
If you want two polynomials, you'd want to change the polynomial $p$ to something else (e.g. $q$). Then you could say that $p$ and $q$ take the desired forms:
\begin{align*}
p(x) &= ax^4 + bx^3 + cx^2 + dx + e \\
q(x) &= fx^4 + gx^3 + hx^2 + ix + l.
\end{align*}
What we name the dummy variable is immaterial. I could have just as easily chosen to define them as such:
\begin{align*}
p(y) &= ay^4 + by^3 + cy^2 + dy + e \\
q(m) &= fm^4 + gm^3 + hm^2 + im + l,
\end{align*}
and I would have been defining exactly the same polynomials. That is to say, the choice of dummy variable doesn't matter, but conventionally, we tend just to use $x$.
Remarkably, this error (which tends to sink students early on, in my experience) seems not to have slowed you down. Most of what you've written from then on is all correct. The only problem is the second coordinate, which you have evaluated as $0$. What you should get is
$$p'(x) = 4ax^3 + 3bx^2 + 2cx + d$$
and so
$$p'(0) = 4a \cdot 0^3 + 3b \cdot 0^2 + 2c \cdot 0 + d = d,$$
and similarly $q'(0) = i$. You can also compute $(p + \lambda q)'(x)$ and $(p + \lambda q)'(0)$ in a similar way, and verify that the two sides remain equal.
