(2n+1) Orthogonal Matrix with determinant 1 has at least one unit eigenvalue

Question

Suppose $A$ is any $2n + 1 \times 2n + 1$ matrix with determinant 1

Why would $A$ have a unit eigenvalue; and does this not hold for a $2n \times 2n$ matrix?

Answer 1

This follows from a more basic fact. Try to show the following:

Lemma: If $n$ is a positive odd integer and $A \in \mathbb{R}^{n\times n}$, then $A$ has a real eigenvalue.

Hint: Consider the charateristic polynomial $p_A(x)$ of $A$ and its limits when $x \to \pm \infty$.

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Now, for your second question, indeed, when $n$ is even $A$ may have no real eigenvalues. A simple example is given by:

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