1. **Problem Statement:** We have a linear transformation $T$ on the space of polynomials $P_2$ defined by $T(f(x)) = f'(x)$, where $f'(x)$ is the derivative of $f(x)$. We need to determine if $T$ has eigenvalues and if so, find the eigenvalues and eigenvectors. 2. **Recall the definition of eigenvalues and eigenvectors:** For a linear transformation $T$, a scalar $\lambda$ is an eigenvalue if there exists a nonzero vector $v$ such that $$T(v) = \lambda v.$$ Here, $v$ is the eigenvector corresponding to eigenvalue $\lambda$. 3. **Apply the definition to our problem:** Let $f(x) \in P_2$ be a polynomial of degree at most 2, so $$f(x) = a_0 + a_1 x + a_2 x^2,$$ where $a_0, a_1, a_2$ are constants. 4. **Compute $T(f(x))$:** The derivative is $$T(f(x)) = f'(x) = a_1 + 2 a_2 x.$$ This is a polynomial of degree at most 1. 5. **Set up the eigenvalue equation:** We want $$T(f(x)) = \lambda f(x),$$ or $$f'(x) = \lambda f(x).$$ Substituting, we get $$a_1 + 2 a_2 x = \lambda (a_0 + a_1 x + a_2 x^2).$$ 6. **Equate coefficients of powers of $x$ on both sides:** - Coefficient of $x^2$: Left side is 0, right side is $\lambda a_2$, so $$0 = \lambda a_2.$$ - Coefficient of $x$: Left side is $2 a_2$, right side is $\lambda a_1$, so $$2 a_2 = \lambda a_1.$$ - Constant term: Left side is $a_1$, right side is $\lambda a_0$, so $$a_1 = \lambda a_0.$$ 7. **Analyze the system:** From $0 = \lambda a_2$, either $\lambda = 0$ or $a_2 = 0$. - Case 1: $\lambda = 0$ - Then from $2 a_2 = 0$, we get $a_2 = 0$. - From $a_1 = 0$, since $\lambda a_0 = 0$, no restriction on $a_0$. - So $f(x) = a_0$ (a constant polynomial). - Check if $T(f) = 0 = 0 \cdot f$, which is true. - So $\lambda = 0$ is an eigenvalue with eigenvectors being all nonzero constant polynomials. - Case 2: $\lambda \neq 0$ and $a_2 = 0$ - Then $2 a_2 = 0 = \lambda a_1$ implies $a_1 = 0$. - From $a_1 = \lambda a_0$, we get $0 = \lambda a_0$ so $a_0 = 0$. - So $f(x) = 0$, which is not an eigenvector (must be nonzero). 8. **Conclusion:** The only eigenvalue is $\lambda = 0$ with eigenvectors being all nonzero constant polynomials. **Final answer:** - Eigenvalue: $0$ - Eigenvectors: All nonzero constant polynomials $f(x) = a_0$, $a_0 \neq 0$.