Question

Let \( P_3 \) denote the real vector space of all polynomials with real coefficients of degree at most 3. Consider the map \[ T: P_3 \to P_3 \quad \text{given by} \quad T(p(x)) = p'(x) + p(x). \] Then

• A. \( T \) is neither one-one nor onto
• B. \( T \) is one-one but not onto
• C. \( T \) is both one-one and onto
• D. \( T \) is onto but not one-one

Hint:

For linear transformations, check for injectivity by solving \( T(p(x)) = 0 \), and check for surjectivity by testing whether every element of the codomain can be reached.

Answer 1

The Correct Option is C

Step 1: Analyzing injectivity (one-one).
To check if \( T \) is one-one, we need to determine if \( T(p(x)) = T(q(x)) \) implies \( p(x) = q(x) \). Since \( T(p(x)) = p'(x) + p(x) \), the map is injective because the solution to \( p'(x) + p(x) = 0 \) leads to only the zero polynomial as the solution.
Step 2: Analyzing surjectivity (onto).
To check if \( T \) is onto, we need to determine if for every polynomial \( r(x) \in P_3 \), there exists a polynomial \( p(x) \in P_3 \) such that \( T(p(x)) = r(x) \). It turns out that \( T \) is onto because we can always find a polynomial \( p(x) \) such that \( T(p(x)) = r(x) \) for any given polynomial \( r(x) \in P_3 \).
Step 3: Conclusion.
Thus, the correct answer is (C).