ThmDex – An index of mathematical definitions, results, and conjectures.
Result R4950 on D1851: Eigenvalue
Real matrix with no real eigenvalues
Formulation 0
Let $A \in \mathbb{R}^{2 \times 2}$ be a D4571: Real matrix such that
(i) \begin{equation} A = \begin{bmatrix} 0 & - 1 \\ 1 & 0 \end{bmatrix} \end{equation}
Then
(1) \begin{equation} \left\{ \lambda \in \mathbb{R} : \text{det}(A - \lambda I_2) = 0 \right\} = \emptyset \end{equation}
(2) \begin{equation} \left\{ \lambda \in \mathbb{C} : \text{det}(A - \lambda I_2) = 0 \right\} = \{ -i, i \} \end{equation}
Proofs
Proof 0
Let $A \in \mathbb{R}^{2 \times 2}$ be a D4571: Real matrix such that
(i) \begin{equation} A = \begin{bmatrix} 0 & - 1 \\ 1 & 0 \end{bmatrix} \end{equation}
We have \begin{equation} \text{det}(A - \lambda I_2) = \text{det} \begin{bmatrix} - \lambda & - 1 \\ 1 & - \lambda \end{bmatrix} = \lambda^2 + 1 \end{equation} The equation $\lambda^2 + 1 = 0$ has no real solutions, but we have \begin{equation} i^2 + 1 = -1 + 1 = 0 \end{equation} and \begin{equation} (-i)^2 + 1 = i^2 + 1 = 0 \end{equation} $\square$