7
$\begingroup$

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

$\endgroup$
7
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.