If $T^9=T^8$, prove that $T^2=T$. Suppose $V$ is a complex inner product space and $T $ is a normal operator on $V$ such that $T^9=T^8$. Prove that $T^2=T$. Give an example of a non-normal $T$ such that $T^9=T^8$ but $T^2 \ne T$.

The polynomial $x^9 - x^8 = x^8(x-1)$ annihilates $T$ so the minimal polynomial $m_T$ of $T$ divides $x^8(x-1)$.
On the other hand, $T$ is normal and hence diagonalizable so $m_T$ consists only of linear factors. Therefore $$m_T \in \{x, x-1, x(x-1)\}$$
and therefore $0 = T(T-I) = T^2 - T$.

As an example of a non-normal operator $T$ such that $T^9 = T^8$ but $T^2 \ne T$, we can take 
$$T = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{pmatrix}$$
with $m_T(x) = x^2(x-1)$.
