Prove that if V is a real vector space of even dimension 2k, then every operator T ∈ L(V ) such that $T^2 + T + I$ is nilpotent satisfies
$(T^2 + T + I)^k = 0$
Hint: For a,c, you may want to argue with the minimal polynomial.
I know that zero is an eigenvector so the minimal polynomial for the operator has dim at most 2k-1. The characteristic polynomial is p(z) = z^2k. How do I show the minimal poly has degree $k$? Thanks so much
You can do this quickly with two basic facts: (1) the minimal polynomial divides the characteristic polynomial. (2) the characteristic polynomial divides a power of the minimal polynomial.
If $T^2 + T + I$ is nilpotent, then the minimal polynomial of $T$ divides a power of $x^2 + x + 1$, so by (2) the characteristic polynomial of $T$ divides a power of $x^2 + x + 1$. Since $x^2 + x + 1$ is irreducible over the reals (and by degree considerations), we must have that the characteristic polynomial of $T$ is $(x^2 + x + 1)^k$. By (1), we have $(T^2 + T + I)^k = 0$.