Let $A \in M_{4 \times 4} (\Bbb R)$ be an orthogonal matrix with $\det(A)=-1$. Prove that $-1$ is an eigenvalue of $A$.
I'm a bit lost. I know about all the basic orthogonal matrices' properties (including the ones about scalar product). I also know that orthogonal matrices' eigenvalues are $\pm 1$. Any tips, please?
We have
$$A^T(A+I) = A^TA + A^T = I+A^T = (I+A)^T$$
so taking the determinant gives
$$- \det(A+I) = \det A^T \cdot \det (A+I) = \det(I+A)^T = \det(I+A)$$
Therefore $\det (A + I) = 0$ so $A + I$ is not invertible. Hence $-1$ is an eigenvalue of $A$.
All the eigenvalues of $A$ are complex numbers with absolute value equal to $1$. Besides, if $\lambda\in\mathbb{C}\setminus\mathbb{R}$ is an eigenvalue, then $\overline\lambda$ is also an eigenvalue and $\lambda\times\overline\lambda=|\lambda|^2=1$. Therefore, since the product of the eigenvalues is equal to $-1$, one of them must be $-1$.
The (complex) eigenvalues of an orthogonal matrix are $1,-1$ or pairs of complex numbers $\lambda, \bar\lambda$ with $|\lambda|=1$. That non-real eigenvalues appear in pairs follows from the fact that the characteristic polynomial has real coefficient.
Now, let us assume that the orthogonal matrix $O$ has the eigenvalues $1$ ($a$ times), $-1$ ($b$ times), and the pairs $\lambda_j, \bar\lambda_j$, $j=1, \dots , N$. We find that the determinant is given by $$\det O = 1^a (-1)^b \prod_j |\lambda_j|^{2} = (-1)^b\;.$$
So if $\det O =-1$, we have that the number $b$ of eigenvalues $-1$ is odd. So we have at least one eigenvector with eigenvalues $-1$.
The eigenvalues of any orthogonal matrix are unimodular complex numbers; this is easy to see: if
$O^TO = OO^T = I \tag 1$
and
$O \vec v = \mu \vec v, \tag 2$
we have
$\bar \mu \mu \langle \vec v, \vec v \rangle = \langle \mu \vec v, \mu \vec v \rangle = \langle O \vec v, O \vec v \rangle = \langle \vec v, O^TO \vec v \rangle = \langle \vec v, I \vec v \rangle = \langle \vec v, \vec v \rangle, \tag 3$
whence
$\vert \mu \vert^2 = \bar \mu \mu = 1 \Longrightarrow \vert \mu \vert = 1 \tag 4$
as claimed.
Now since $O$ is real, every complex eigenvalue must be matched by it's conjugtate; since
$\det O = \mu_1 \mu_2 \mu_3 \mu_4 = -1, \tag 5$
we may by a simple counting argument see that the numbers of purely real and genuinely complex (that is, not in $\Bbb R$) eigenvalues are constrained as follows: if there are no real eigenvalues, then we must have two complex conjugate pairs $\mu_2 = \bar \mu_1$, $\mu_4 = \bar \mu_3$, so $\det O = 1$, against our assumption; there cannot be precisely on real eigenvalue since then we would have an unmatched complex eigenvalue; if there are exactly two real eigenvalues, they must be $1$ and $-1$ in order to preserve (5); there cannot be three real eigenvalues since again we would have an unconjugated complex eigenvalue standing alone; if there are four real eigenvalues they must by (5) be either $1, 1, 1, -1$ or $1, -1, -1, -1$; thus in every admissible case it follows that $-1$ is an eigenvalue of $O$.
First, the eigenvalues of an orthogonal matrix have absolute values equal to $1$. Let $\lambda_1, \lambda_2, \lambda_3, \lambda_4$ be the eigenvalues of $A$. Then $$\det(A) = -1= \lambda_1 \lambda_2 \lambda_3 \lambda_4.$$ If one of the eigenvalues is complex, say $\lambda_1$, then its conjugate must also be one of the eigenvalues, say $\lambda_2$. Since the eigenvalues of an orthogonal matrix have unity absolute value, $\lambda_1 \lambda_2=1$, and $\lambda_3 \lambda_4=-1$. Since both of them cannot be complex conjugate of each other and have unity magnitude. One of them must be $1$ and and other one must be $-1$.
If all the eigenvalues are real then they must be either $1$ or $-1$. Since their product is $-1$, the number of eigenvalues having value $-1$ must be odd. Consequently, if $\det(A)=-1$ then there must be at least one eigenvalue (actually an odd number of eigenvalues) whose value is $-1$.
We know that for all eigenvalues of an orthogonal transformation, we have that $| \lambda |=1$. Moreover, if $\lambda$ is a complex eigenvalue, then its conjugate $\bar{\lambda}$ is also an eigenvalue. Since $\lambda \bar{\lambda}=1$, and $\det{A}=-1$ is the product of eigenvalues, one of the eigenvalues must be $-1$.
