$T$ is diagonalizable on finite dimensional v.s. $\implies$ $(T^2+T+I)(\vec v) \ne \vec 0 , \forall \vec v \ne \vec0$?

Question

Let $T$ be a diagonalizable (over $\mathbb R$) operator on a finite dimensional real vector space ; then is it true that there is no non-zero vector $\vec v$ such that $(T^2+T+I)(\vec v)=\vec 0$ ? (where $I$ is the identity map )

Answer 1

Yes true. Notice that if $\lambda$ is an eigenvalue for $T$ then for every polynomial $P$, $P(\lambda)$ is an eigenvalue of $P(T)$ hence if $v\ne0$ exists such that $(T^2+T+I)v=0$ then there's $\lambda$ eigenvalue of $T$ such that

$$\lambda^2+\lambda+1=0$$ which is a contradiction since in this case $\lambda\not\in\Bbb R$.