Is $T$ nilpotent?

linear-algebra

You are a bit confused:

Option (1)

This is false.No matter how many times you compose $T$ with itself. If you compose it $n$ times, it won't be $0$ on polynomials of degree higher than $n-1$.

Option (2)

This is true. As you noticed, the only way for $f(x)$ to be an eigenvector is to be a constant polynomial and clearly the derivative of a constant is $0$, so the only eigenvalue is $0$.

Option (3)

This is true because option 1 is false.

Option (4)

This is false because option 2 is true.

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