An operator $T:\mathbb{R}^4\to \mathbb{R}^4$ such that $T$ has no (real) eigenvalues. Give an example of an operator $T:\mathbb{R}^4\to \mathbb{R}^4$ such that $T$ has no (real) eigenvalues. 
How can I find this operator?
Thanks for your help.
 A: Consider a rotation about the origin. 
A: A simple example is the linear operator corresponding to $ \left( \begin{array}{ccc}
0 & -1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & -1\\
0 & 0 & 1 & 0\\
 \end{array} \right) $
A: Another angle to look at this problem is to take the differential equation $(D^2+1)^2[y]=0$
  (or $y''''+2y''+y=0$ if you prefer) and reduce it to a system of four first order ODEs by reduction of order $x_1=y,x_2=y',x_3=y'',x_4=y'''$. The matrix of this system of ODEs will have complex eigenvalues $i,-i$ corresponding to a pair of complex e-vectors and a pair of generalized complex e-vectors. This is the complementary matrix to the ODE.
$$ \left[ \begin{array}{cccc}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
-1 & 0 & -2 & 0 
\end{array} \right] $$
This matrix corresponds to $T: \mathbb{R}^4 \rightarrow \mathbb{R}^4$ with no real e-values.
A: First, find a polynomial of degree 4 with no real roots. Then form the companion matrix of this polynomial (look up "companion matrix" if you are not familiar with this term - you'll be glad you did). 
A: You can construct examples by doing the following:


*

*Choose any diagonal matrix with non-real values on it's diagonal:


$
 A =
 \begin{pmatrix}
  a_{1,1} & 0 & 0 & 0 \\
  0 & a_{2,2} & 0 & 0 \\
  0  & 0  & a_{3,3} & 0  \\
  0 & 0 & 0 & a_{4,4}
 \end{pmatrix}
$
By choosing correctly $M \in GL(4,\mathbb C)$, you can make the operator $T=M^{-1}AM$ real: $T : \mathbb R^4 \longrightarrow \mathbb R^4$.
