Problem: Let $\mathcal{P} _3$ be the space of all real polynomials with degree at most $3$. Consider the homomorphism
\begin{gather*}
T: \mathcal{P} _3 \rightarrow \mathcal{P} _3, \\
a_0 + a_1 x + a_2 x^2 + a_3 x^3 \mapsto a_0 + (a_0 + a_1)x + (a_2 + a_3)x^3.
\end{gather*}
Find the following.
-
The kernel of $T$ and hence determine the nullity of $T$.
-
$T^{-1} \{ 2- x^3 \} $ and
-
$T^{-1} \{ 1 + x^2 \} $.
Solution: Given that the map $T$ is a homomorphism.
-
Recall that the kernel of a homomorphism is the collection of all elements which maps to zero. That is,
\begin{align*}
\ker T & = \left\{ p(x) \in \mathcal{P} _3: T(p(x)) = 0 \right\}.
\end{align*}
Let $p(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3$. Then,
\begin{align*}
\ker T & = \left\{ p(x) \in \mathcal{P}_3 : a_0 + (a_0 + a_1) x + (a_2 + a_3) x^3 = 0 \right\} \\
& = \left\{ p(x) \in \mathcal{P} _3: a_0 = a_0 + a_1 = a_2 + a_3 = 0 \right\} .
\end{align*}
So, we have
\begin{align*}
& a_0 = 0 \\
& a_0 + a_1 = 0 \implies a_1 = 0 \\
& a_2 + a_3 = 0 \implies a_2 = -a_3.
\end{align*}
Thus,
\[
\ker T = \left\{ a_2 x^2 - a_2 x^3 : a_2 \in \mathbb{R} \right\}.
\]
Therefore, A basis for the null space is $\{ x^2 - x^3 \} $. Hence, the nullity of $T$ is $1$.
-
We need to determine the inverse image of $2 - x^3$. Let
\begin{align*}
& a_0 + a_1 x + a_2 x^2 + a_3 x^3 \in T^{-1} \{ 2 - x^3 \} \\
\implies & T\left( a_0 + a_1 x + a_2 x^2 + a_3 x^3 \right) = 2 - x^3 \\
\implies & a_0 + (a_0 + a_1)x + (a_2 + a_3) x^3 = 2 - x^3 \\
\implies & a_0 = 2, \ a_0 + a_1 = 0,\ a_2 + a_3 = -1 \\
\implies & a_0 = 2,\ a_1 = -2, \ a_3 = a_2 - 1.
\end{align*}
Thus,
\[
T^{-1} \{ 2 - x^3 \} = \left\{ 2 - 2x + a x^2 - (a - 1)x^3: a \in \mathbb{R} \right\} .
\]
-
We need to determine the inverse image of $1 + x^2$. Note that for any $p(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3\in \mathcal{P} _3$, the coefficient of $x^2$ in $T(p(x))$ is $0$. So $T(p(x)) = 1 + x^2$ is not possible for any $p(x)$. Therefore, $T^{-1} \left\{ 1 + x^2 \right\} = \emptyset $.