Question

For each of the following linear operators T on a vector space V, test T for diagonalizability, and if T is diagonalizable, find a basis β for V such that [T]β is a diagonal matrix.

V = P₃(R) and T is defined by T(f(x)) = f'(x) + f"(x), respectively.

Answer 1

The operator T defined on P₃(R) by T(f(x)) = f'(x) + f"(x) cannot be diagonalizable because the derivative operation reduces the degree of a polynomial, and there cannot exist an eigenvector for λ = 0.

Step-by-step explanation:

To determine if a linear operator is diagonalizable, we look at the characteristic polynomial and find its roots to determine the eigenvalues. If there are enough linearly independent eigenvectors to form a basis for the vector space, then the operator is diagonalizable. Considering the vector space V = P₃(R) and the linear operator T defined by T(f(x)) = f'(x) + f"(x), we first need to find the eigenvalues and eigenvectors for T.

Let's take a general polynomial in P₃(R): f(x) = ax^3 + bx^2 + cx + d. Applying T, we have T(f(x)) = 3ax^2 + 2bx + c + 2ax + 2b = 3ax^2 + (2a + 2b)x + (2b + c). For T(f(x)) to equal λf(x), the coefficients of x^2, x, and the constant term on both sides of the equation must be equal. Thus, we get the system of equations:

• 3a = λa
• 2a + 2b = λb
• 2b + c = λc

Solving this system tells us the eigenvalues and associated eigenvectors. For T to be diagonalizable, we need three linearly independent eigenvectors. However, we notice that the derivative operation inherently reduces the degree of a polynomial, implying that there cannot exist an eigenvector for λ = 0 as the zero polynomial is not in the domain of T. Hence, T cannot have three linearly independent eigenvectors and is not diagonalizable.