Prove that there is a real linear polynomial $f(x)$ Let $(V,<,>)$ be a real inner product space and $T \in \mathcal{L}(V,V)$ a normal  operator. Assume that the minimal polynomial of $T$ is a real irreducible quadratic. Prove that there is a real linear polynomial $f(x)$ such that $T^{*}=f(T)$
There is a Lemma in the book : Let $T$ be a normal operator on $V$. Then for all vectors $v \in V$, $|| T(v) ||=||T^{*}(v)||$. Also there is another defintion of being normal, $TT^*=T^*T$
$T^{*}$ denote to the adjoint of an linear transformation $T$ between inner product spaces. 
I believe a real irreducible quadratic something like, $X^2+1=0$
I do not know how would I get started.    
 A: Hint: The minimal polynomial has the form $m(x) = (x - \mu)(x - \overline{\mu})$ for some $\mu \in \Bbb C$.  It suffices to find a polynomial $p$ for which $p(\mu) = \overline{\mu}$ and $p(\overline{\mu}) = \mu$.
A: Here is a theorem from Axler's book "Linear Algebra Done Right" that should be helpful.

Let $V$ be a finite dimensional real inner product space, and $T$ a linear operator on $V$. Then $T$ is normal if and only if there is an orthonormal basis of $V$ with respect to which the matrix of $T$ has block diagonal form with each block either $1\times 1$ or $2\times 2$. The $2 \times 2$ blocks are of the form
  $$\bigl(\begin{smallmatrix}
a & -b\\ 
b & a
\end{smallmatrix}\bigr)$$
  with $b > 0$.

Since the minimal polynomial of $T$ is an irreducible quadratic, it has no real eigenvalues. Thus there are no $1\times 1$ blocks in the above decomposition. Using this, you can see exactly the form of $T$ and $T^*$.
The eigenvalues of the matrix $$\bigl(\begin{smallmatrix}
a & -b\\ 
b & a
\end{smallmatrix}\bigr)$$
are $a \pm bi$. So if there were more than one type of $2\times 2$ matrix showing up in the decomposition, then we would have more than two eigenvalues, and so our minimal polynomial would not be a quadratic. Therefore our decomposition is just a bunch of copies of
$$\bigl(\begin{smallmatrix}
a & -b\\ 
b & a
\end{smallmatrix}\bigr)$$
along the diagonal.
So lets find a polynomial that will take the transpose of this $2\times 2$. Since taking the transpose will make the upper-right entry $b$, I will first multiply the matrix by $-1$. This gives
$$\bigl(\begin{smallmatrix}
-a & b\\ 
-b & -a
\end{smallmatrix}\bigr).$$
But now the main diagonal is bad. We can fix it by adding $2a$ to the main diagonal. We can accomplish this by adding the matrix $2a I$.
So $$\bigl(\begin{smallmatrix}
a & -b\\ 
b & a
\end{smallmatrix}\bigr)^* = \bigl(\begin{smallmatrix}
a & -b\\ 
b & a
\end{smallmatrix}\bigr)^{tr} = -\bigl(\begin{smallmatrix}
a & -b\\ 
b & a
\end{smallmatrix}\bigr) + 2aI.$$
Therefore the polynomial we want is $f(x) = -x + 2a$.
You can now argue that this polynomial works for $T$.
