Hi Waheeb Here is a numerical illustration of the concept. The null space of \(T\), which consists of all polynomials mapped to the zero polynomial, includes all constant polynomials \(a\), since their derivatives are 0.
For example, consider the polynomial \(p(x) = 7\). Applying the transformation \(T\) gives us:
\[ T(7) = T(7 + 0 \cdot x + 0 \cdot x^2 + 0 \cdot x^3) = 0 + 2 \cdot 0 \cdot x + 3 \cdot 0 \cdot x^2 = 0 \]
Thus, \(p(x) = 7\) is in the pre-image of 0, demonstrating that any constant polynomial indeed maps to the zero polynomial under \(T\).
The pre-image of the zero polynomial under \(T\) consists of all constant polynomials in \(P_3\). The linear transformation \(T\) is akin to the differentiation operation in calculus, excluding the constant term’s derivative, which aligns with your findings.