If matrix $B$ is skew-symmetric with eigenvalue $r$, then $-r$ is also an eigenvalue of $B$

Question

I've considered the determinants of both $B$ and $-B^T$, and found that

$\det(B) = (-1)^n \det(B)$, where $B$ is a $n \times n$ matrix.

I've also tried the approach where $r$ is an eigenvalue of $B$ iff $\det(B-rI) = 0$, but as determinants are not preserved over addition/subtraction, I'm not really sure what to do from there.

Answer 1

Hint: Note that $r$ is an eigenvalue of $B$ if and only if $\det(B - rI) = 0$. Note also that $\det(A) = \det(A^T)$ for any matrix $A$.

Answer 2

Consider an arbitrary commutative ring $\mathbf{A}$, a natural number $n \in \mathbb{N}$ and an antisymmetric matrix $M \in \mathscr{M}_n(\mathbf{A})$, square of order $n$ over $\mathbf{A}$. I use the term antisymmetry as a synonym of skew-symmetry, meaning that the relation $M^{\mathrm{t}}=-M$ takes place.

Let $\chi_T \in \mathbf{A}[X]$ denote the characteristic polynomial of arbitrary matrix $T$. We then have by definition that: $$\begin{align*} \chi_{M^{\mathrm{t}}}&=\det\left(X\mathrm{I}_n-M^{\mathrm{t}}\right)\\ &=\det(X\mathrm{I}_n+M)\\ &=\det\left(-(-X\mathrm{I}_n-M)\right)\\ &=(-1)^n\det((-X)\mathrm{I}_n-M)\\ &=(-1)^n\chi_M(-X). \end{align*}$$ On the other hand, it is also generally the case that: $$\begin{align*} \chi_{M^{\mathrm{t}}}&=\det\left(X\mathrm{I}_n-M^{\mathrm{t}}\right)\\ &=\det(X\mathrm{I}_n-M)^{\mathrm{t}}\\ &=\det(X\mathrm{I}_n-M)\\ &=\chi_M. \end{align*}$$ We thus infer the relation $\chi_M=(-1)^n\chi_M(-X)$. If $\mathbf{A}$ is a field of characteristic different from $2$, we gather that the coefficients of $\chi_M$ at degrees $k$ of opposite parity to $n$ are null (and thus in particular there exists no invertible antisymmetric matrix square of odd order over a commutative field of characteristic distinct from $2$).

This means that given the decomposition $\chi_M=\displaystyle\prod_{k=1}^n(X-\lambda_k)$ for the characteristic polynomial of $M$ we also have the decomposition $\chi_M=\displaystyle\prod_{k=1}^n(X+\lambda_k)$. Therefore, if $\mathbf{A}$ is a field and $\lambda$ an eigenvalue of $M$ the opposite $-\lambda$ must also be an eigenvalue of $M$ of the same multiplicity.

In particular, when $\mathbf{A}$ is an algebraically closed field such that $\mathrm{char}\mathbf{A}\neq 2$, the characteristic polynomial of $M$ can be written as $\chi_M=X^h\displaystyle\prod_{\lambda \in R}\left(X-\lambda^2\right)^{k_{\lambda}}$, where $R \subseteq \mathbf{A}^{\times}$ is a finite subset such that $|R \cap \{\pm \mu\}| \leqslant 1$ for each scalar $\mu \in \mathbf{A}$ and $k \in \left(\mathbb{N}^{\times}\right)^R$ is the family of multiplicities, $k_{\lambda}$ being the common multiplicity that both $\lambda$ and $-\lambda$ have in $\chi_M$.