Can orthogonal matrix with positive diagonal have -1 in its spectrum?

Question

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 $\det O = -1 \Rightarrow -1 \in \sigma(O)$)

Answer 1

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.

Answer 2

Let $$A=\begin{pmatrix} 0& 1&-1 \\ -1& 0& 1\\ 1&-1 &0 \end{pmatrix}$$ and let, $\forall \varepsilon \in \mathbb{R}$, $$U(\varepsilon) =\exp(\varepsilon A)=\sum_{n=0}^\infty \frac{1}{n!}(\varepsilon A)^n= \begin{pmatrix}1 & \varepsilon&-\varepsilon\\-\varepsilon & 1 & \varepsilon\\ \varepsilon &-\varepsilon& 1 \end{pmatrix}+{\cal O}(\varepsilon^2)$$ Note that the anti-symmetry of $A$ implies that $U(\varepsilon)$ is orthogonal (for all $\varepsilon$) and $\det U(\varepsilon)=\exp(\varepsilon \text{Tr}(A))=1$.

The product $$U(\varepsilon)\begin{pmatrix} 0 & -1 & 0 \\ 0 & 0 &-1 \\ -1 & 0& 0\end{pmatrix}=\begin{pmatrix} \varepsilon & ... & ... \\ ... & \varepsilon & ... \\ ... & ... & \varepsilon\end{pmatrix}+{\cal O}(\varepsilon^2)$$ is orthogonal by the group property of the orthogonal matrices, its diagonal is positive for sufficiently small $\varepsilon>0$. The determinant of this matrix is $-\det(U(\varepsilon))=-1$ which implies, since the dimension of the matrix is odd, that $-1$ is in its spectrum.