Eigenvalues of anti-symmetric real matrix and non-singularity

Question

I am trying to solve the following problem (book: Classical Mechanics by Goldstein, exercise 4.4):

By examining the eigenvalues of an anti-symmetric $3\times3$ real matrix $A$, show that $I\pm A$ is non-singular.

What I've tried to do:

$(I\pm A)\vec{v}=\lambda\vec{v}\Rightarrow (I\pm A - \lambda I)\vec{v}=0$
Let $B=(I\pm A - \lambda I)$, then, $\det(B)=(\pm 1-\lambda)^3-a_{23}a_{32}(\pm 1-\lambda)-a_{12}a_{21}(\pm 1-\lambda)-a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{13}a_{31}(\pm 1-\lambda)$

Well, that does not seem to be a good approach. Any tips?

Answer 1

There does not appear to be anything intrinsically three-dimensional about this problem. Indeed, in what follows we may allow $A$ to be any real skew-symmetric matrix.

The only real eigenvalue of a real skew-symmetric matrix $A = -A^T$ is $0$, for if

$A \vec x = \mu \vec x, \; \mu \in \Bbb R, \tag 1$

we can normalize $\vec x \ne 0$ so that

$\langle \vec x, \vec x \rangle = 1; \tag 2$

then from (1),

$\mu = \mu \langle \vec x, \vec x \rangle = \langle \vec x, \mu \vec x \rangle$ $= \langle \vec x, A \vec x \rangle = \langle A \vec x, \vec x \rangle = \langle \vec x, A^T \vec x \rangle = \langle \vec x, -A \vec x \rangle = \langle \vec x, -\mu \vec x \rangle = -\mu \langle \vec x, \vec x \rangle = -\mu, \tag 3$

whence

$\mu = 0; \tag 4$

thus the non-zero eigenvalues of $A$ must occur in complex conjugate pairs

$\alpha_k, \bar \alpha_k \in \Bbb C \setminus \Bbb R, \; 1 \le k \le \lfloor \dfrac{n}{2} \rfloor, \tag 5$

the eigenvalues of $I \pm A$ are thus respectively

$1 \pm \mu, 1 \pm \alpha_k, 1 \pm \bar \alpha_k = 1, 1 \pm \alpha_k, 1 \pm \bar \alpha_k, \tag 6$

none of which are $0$ since $\alpha_k \notin \Bbb R$; since no eigenvalue of $I \pm A$ is $0$,

$\det (I \pm A) \ne 0, \tag 7$

whence $I \pm A$ is non-singular.

Answer 2

Let $\lambda$ be an eigenvalue of $A$ and choose a unit eigenvector $\vec{v}\in E_\lambda$. Then \begin{align*} \lambda &= \lambda\cdot\lVert\vec{v}\rVert \\ &= \lambda\cdot\langle\vec{v}, \vec{v}\rangle \\ &= \langle\lambda\cdot\vec{v}, \vec{v}\rangle \\ &= \langle A\vec{v}, \vec{v}\rangle \\ &= \langle\vec{v}, A^\ast\vec{v}\rangle \\ &= \langle\vec{v}, -A\vec{v}\rangle \\ &= \langle\vec{v}, -\lambda\vec{v}\rangle \\ &= -\overline{\lambda}\langle\vec{v}, \vec{v}\rangle \\ &= -\overline{\lambda} \end{align*} This proves that $\lambda=-\overline{\lambda}$, which means that the real part of $\lambda$ is zero. Here, we have used the inner product on $\Bbb C^n$.

Now, note that the eigenvalues of $\alpha\cdot I+A$ are given by $\{\alpha+\lambda:\lambda\text{ is an eigenvalue of }A\}$.

How does this give our result?