Every n × n-matrix A with real entries has at least one real eigenvalue.

Question

I have a true/false question:

Every n × n-matrix A with real entries has at least one real eigenvalue.

I am thinking that this is true but I would like to hear other opinions.

Thanks

Answer 1

Nope. Try a rotation in $\Bbb R^2$.

Answer 2

Here's a counterexample:

\begin{bmatrix} 1 & 1 \\[0.3em] -4 & 1 \\[0.3em] \end{bmatrix}

Being clever, we can construct any number of $2 \times 2$ matrices such that its characteristic polynomial has negative discriminant, and thus has no real roots (eigenvalues).

Answer 3

Eigenvalues of a $2\times 2$ matrix $M$ are the roots of the characteristic polynomial $\lambda^2 - \text{tr}(M)\lambda + \det(M)$. All you need to do is find a $2\times 2$ matrix where $\text{tr}(M)^2-4\det(M)<0$ to find a counterexample.

Answer 4

Since the eigenvalues of a matrix $A\in\mathcal M_n(\Bbb R)$ are roots of its characteristic polynomial then the question is equivalent to say: "every polynomial with real coefficients has a real root" which is wrong.

Remark: The answer is yes if we more assume that $n$ is odd (why?).