Problem: Let $n \in \mathbb{N} $ and $\mathcal{P} _n(\mathbb{R} )$ be the set of polynomials with degree less than or equal to $n$. Consider a map
\[
T: \mathcal{P} _n \to \mathcal{P} _{n-1}, \quad f\mapsto f',
\]
where $f'$ ss the derivative of $f$ with respect to $x$.
$T$ is a linear transformation if
\[
T(\alpha v_1 + v_2) = \alpha T(v_1) + T(v_2),
\]
for $v_1, v_2 \in V$ and $\alpha \in \mathbb{R} $.
The null-space of $T$ is defiend as
\[
\mathcal{N} (T) = \left\{ f(x) = \sum_{i=0}^{n} a_i x^i \in \mathcal{P} _n: T(f) = 0 \right\}
\]
Use rank-nullity theorem to find the range of $T$.
Solution: Given that a map $T$ between two vector spaces $V$ and $W$, we say $T$ is a linear transformation if
\[
T(\alpha v_1 + v_2) = \alpha T(v_1) + T(v_2),
\]
for $v_1, v_2 \in V$ and $\alpha \in \mathbb{R} $.
Let $f, g\in \mathcal{P} _n (\mathbb{R} )$ and $\alpha \in \mathbb{R} $. Then,
\begin{align*}
T(\alpha f + g) & = \left( \alpha f + g \right)' = (\alpha f)' + g'\\
& = \alpha f' + g' = \alpha T(f) + T(g).
\end{align*}
Thus, $T$ is a linear transformation.
We now will use the rank-nullity theorem to find the range of $T$. According to the rank-nullity theorem,
\begin{align*}
& \operatorname{rank}(T) + \operatorname{nullity}(T) = \dim \left( \mathcal{P} _n \right) \\
\implies & \operatorname{rank}(T) + 1 = n + 1 \\
\implies & \operatorname{rank}(T) = n = \dim \left( \mathcal{P} _{n-1} \right).
\end{align*}
Hence, the $\operatorname{rank}(T) = n$ and therefore, $T$ is surjective.