Question

Let \( P_3(\mathbb{R}) \) be the vector space of all polynomials of degree at most three with real coefficients under usual polynomial addition and scalar multiplication. Let \( T: P_3(\mathbb{R}) \to \mathbb{R}^2 \) be the linear transformation defined as \[ T(p) = \left(p(1), p'(1)\right) \] for all \( p \in P_3(\mathbb{R}) \), where \( p' \) is the derivative of \( p \).

Then the nullity of \( T \) is equal to ______. (Answer in integer)

Hint:

Use rank-nullity: find the rank of T first by checking whether it is onto, then nullity equals dim(domain) minus rank.

Answer 1

Correct Answer: 2

Step 1: Note the dimension of the domain.
\( P_3(\mathbb{R}) \), the space of polynomials of degree at most \( 3 \), has basis \( \{1, x, x^2, x^3\} \), so \( \dim P_3(\mathbb{R}) = 4 \). The codomain is \( \mathbb{R}^2 \), which has dimension \( 2 \).

Step 2: Recall the rank-nullity theorem.
For a linear map \( T: V \to W \) with \( V \) finite dimensional,
\[ \dim V = \operatorname{rank}(T) + \operatorname{nullity}(T) \]
Here \( \dim V = 4 \), so nullity\( (T) = 4 - \operatorname{rank}(T) \). To use this, the rank of \( T \) is needed.

Step 3: Show that \( T \) is onto \( \mathbb{R}^2 \).
Take any target pair \( (a,b) \in \mathbb{R}^2 \). Choose the polynomial \( p(x) = a + b(x-1) \), which has degree at most \( 1 \), so it certainly lies in \( P_3(\mathbb{R}) \). Then
\[ p(1) = a + b(1-1) = a, \qquad p'(x) = b \implies p'(1) = b \]
So \( T(p) = (a,b) \), and since \( (a,b) \) was arbitrary, \( T \) is surjective. This means \( \operatorname{rank}(T) = \dim \mathbb{R}^2 = 2 \).

Step 4: Apply rank-nullity.
\[ \operatorname{nullity}(T) = \dim P_3(\mathbb{R}) - \operatorname{rank}(T) = 4 - 2 = 2 \]

Final Answer:
The nullity of \( T \) is \( 2 \). \[ \boxed{2} \]