Question

Let \( V = \{ p : p(x) = a_0 + a_1 x + a_2 x^2, a_0, a_1, a_2 \in \mathbb{R} \} \) be the vector space of all polynomials of degree at most 2 over the real field \( \mathbb{R} \). Let \( T: V \to V \) be the linear operator given by
\[ T(p) = (p(0) - p(1)) + (p(0) + p(1)) x + p(0) x^2. \] Then the sum of the eigenvalues of \( T \) is \(\underline{\hspace{2cm}}\) .

Hint:

To find the sum of the eigenvalues of a linear operator, calculate the trace of its matrix representation relative to a basis.

Answer 1

Correct Answer: 1

The given linear operator is: \[ T(p) = (p(0) - p(1)) + (p(0) + p(1)) x + p(0) x^2. \] To find the sum of the eigenvalues, we express this operator in matrix form relative to the basis \( \{ 1, x, x^2 \} \). The matrix representation of \( T \) can be computed, and the trace of this matrix will give us the sum of the eigenvalues. The sum of the eigenvalues of \( T \) is equal to the trace of the operator, which can be computed as the sum of the diagonal elements of the matrix. Final Answer: \[ \boxed{2}. \]