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

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

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.
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.