Question

Give an example of a linear operator on R^4 which has no eigenvectors

Answer 1

#5746 Build the linear operator on $\mathbb{R}^4$ that does not possess eigenvectors Solution It is obvious that in order a linear operator on $\mathbb{R}^4$ not to possess eigenvectors it is neccessary and sufficient for operator only to have complex eigenvalues with non-zero imaginary part.

$$A = \left( \begin{array}{cccc} 1 & -1 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 1 & 1 \end{array} \right)$$

$A$ possesses two eigenvalues of multiplicity 2, they are $\lambda_{1,2} = 1 \pm i$ (it is a consequence of that $A$ has block form), thus the equality $A\mathbf{v} = \lambda_{1,2}\mathbf{v}$, $\mathbf{v} \neq 0$ is impossible.