$\dim E=n$ and the normal operator $A\colon E\to E$ has $n$ distinct eigenvalues If $\dim E = n$ and the normal operator $A\colon E \rightarrow E$ has $n$ distinct eigenvalues, how do I show that $A$ is self adjoint? 
 A: This is asking you to prove that a normal operator with $n$ distinct realeigenvalues over a vector space of dimension $n$ is self-adjoint. 
$A = EDE^*$ Therefore $A^* = (EDE^*)* = (E^*)^*(D^*)E^*= ED^*E^*$. $\bar{D} = D^*$ and as $D$ is real $\bar{D} = D$. So $A^* = EDE^* = A$. 
A: Since $A$ has all distinct eigenvalues, if some other operator $S$ commutes with it, then it's a polynomial of it. 
Choose an basis such that the matrix $M$ of $A$ is diagonal, and define an operator $U_M$ on the space of matrices where $U_M(B) = MB - BM$. Letting $E_{ij}$ be a unit matrix ($1$ in the $(i,j)$ slot and $0$ elsewhere), we see that $U_M(E_{ij}) = (A_{ii} - A_{jj})E_{ij}$, which you can verify directly. This is equal to $0$ iff $i=j$, since the eigenvalues of $A$ are distinct, so $\dim(\ker(U_M)) = M$.
Now, we know that the set of polynomials in $M$ clearly is a subspace of $U_M$. Thing is, since $M$ has all distinct eigenvalues, its minimal polynomial equals its characteristic polynomial, and thus has degree $n$. This means that the space of polynomials in $M$ also has dimension $n$, so that it is equal to $\ker(U_M)$.
By the above and because $A$ is normal, we know that $A^*$ is a polynomial in $A$. Thus, $\langle Ax,y\rangle = \langle x,p(A)y\rangle$ where $p$ is some polynomial. This implies that $\langle (A-I)x,(I-p(A))y\rangle = 0$ for any $x$ and $y$. In particular, take $x=y$. So, by this, we also know that $\langle (I-p(A))x,(A-I)x\rangle = 0$, implying that $\langle (A-p(A))x,(A-p(A))x\rangle = \|(A-p(A))x\|^2 = 0$ for all $x$, so that $A = p(A)$. So then $A$ is self-adjoint.
This is probably an odd way of doing it, I just kinda think like that. The other answer is more direct once you have the spectral theorem for normal operators.
