Proving an operator is self-adjoint I have a linear operator $T:V \to V$ (where $V$ is a finite-dimensional vector space) such that $T^9=T^8$ and $T$ is normal, I need to prove that $T$ is self-adjoint and also that $T^2=T$. 
Would appreciate any help given.
Thanks a million!
 A: Any normal matrix is diagonalizable and thus we can write $T = U\Lambda U^*$ for some unitary matrix $U$ - the matrix of eigenvectors - and diagonal matrix $\Lambda$ - the matrix of eigenvalues.
If $T^9 = T^8$, we have the following:
$$T^9v = T^8 v.$$
Suppose $v$ is an eigenvector of $T$, then $Tv = \lambda v$ and
$$T^9v = \lambda^9 v = \lambda^8 v = T^8 v$$
Thus we have that
$$\lambda^8(\lambda-1)v = 0.$$
Therefore $\lambda = 0,1$ and the eigenvalues are real.
We wish to show that $T^2 = T$. Using the fact that $T = U\Lambda U^*$, we have that
$$T^2 = TT = (U\Lambda U^*)(U\Lambda U^*) = U\Lambda^2 U^*.$$
This last equality follows from that fact that $U$ is unitary. However since $\lambda = 0,1$, we have that $\Lambda^2 = \Lambda$ and so $T^2 = T$.
To show $T = T^*$, we can again use the fact that the eigenvalues are real and that $T$ is normal:
$$T^* = (U\Lambda U^*)^* = (U^*)^*\Lambda^*U^* = U\Lambda^*U^*.$$
However since $\Lambda$ is a matrix of real numbers and is diagonal, $\Lambda^* = \Lambda$ and we are done.
A: Hint. Call the underlying field $\mathbb{F}$. As $T$ is normal and its characteristic polynomial can be split into linear factors $\underline{\text{over }\, \mathbb{F}}$ (why?), $T$ is unitarily diagonalisable over $\mathbb{F}$. Now the rest is obvious.
