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.
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.
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).
Consider a rotation about the origin.
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 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) $
You can construct examples by doing the following:
$ 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: $T : \mathbb R^4 \longrightarrow \mathbb R^4$.