Let $O\in \mathbb{R}^{n\times n}$ be an orthogonal matrix, i.e. $O^tO=I=OO^t$. Suppose its diagonal entries $\{O_{jj}\}_{j\in \{1,…,n\}}$ are (strictly) positive. Can $-1$ then be included in the spectrum of $O$?
Note that if the diagonal is required to be non-negative in stead of positive, then
$$\begin{pmatrix}
0&-1 & 0\\
0 & 0 & -1\\
-1 & 0& 0
\end{pmatrix}$$
provides a counterexample, since it has a non-negative diagonal yet includes -1 in its spectrum.
Apart from that, a straightforward calculation like (suppose, seeking a contradiction, that $v\in \mathbb{R}^n$ is a normalized eigenvector with eigenvalue -1)
$$-1=\langle v,-v\rangle=\langle v,Ov\rangle \geq \sum_{j=1}^n\left(|O_{jj}|v_j^2 – \left|\sum_{k\neq j}O_{jk}v_jv_k\right|\right)> -\left(\sum_{j=1}^n\sum_{k\neq j}O_{jk}^2\right)^{1/2}\geq-n^{1/2}$$
doesn't suffice to resolve the problem (because it didn't result in the desired contradiction). Likewise, writing out $\langle e_j,Ov\rangle = -\langle e_j,v\rangle$ doesn't seem to yield anything. I am aware that my statement, if true, would imply for odd $n$ that a positive diagonal would imply $\det O=1$. Also, note that there is a related 'hybrid' open question: what if the diagonal of $O$ is non-negative and non-zero?
EDIT: there are also counterexamples to the hybrid problem just mentioned: take
$$\begin{pmatrix}
\sin(\theta) & \cos(\theta) & 0\\
0 & 0 & -1\\
\cos(\theta) & -\sin(\theta)& 0
\end{pmatrix}$$
for an angle $\theta \in (0,\pi)$. (to quicker analyse examples in odd $n$, like $n=3$ here, it helps to prove the auxiliary lemma $\det O = -1 \Rightarrow -1 \in \sigma(O)$)
Best Answer
This is clearly impossible when $n=1$. It is also impossible when $n=2$, for, if one of the eigenvalues is $-1$, the other eigenvalue must be $\pm1$. Hence the trace is non-positive and the matrix cannot possibly possess a positive diagonal.
When $n\ge3$, let $e$ be the vector of ones. The Householder reflection matrix $Q=I-\frac{2}{n}ee^T$ will then satisfy your requirement.