Show $\mathbf{m} \times \mathbf{n}$ is an eigenvector of $I + \mathbf{m}\mathbf{n}^T$ for $\mathbf{m}$, $\mathbf{n}$ orthogonal real unit vectors

Suppose $\mathbf{m}$ and $\mathbf{n}$ are orthogonal, real unit vectors. I want to show that $\mathbf{m} \times \mathbf{n}$ is an eigenvector of the matrix $B$ defined by

$$B=I + \mathbf{m}\mathbf{n}^T,$$

where $I$ is the $3 \times 3$ identity matrix.

I can't see any nice, intuitive way to see why this should be the case, so I have resorted to looking at the $i$th component of the vector $B(\mathbf{m} \times \mathbf{n})$ and am hoping that I can show that it is some scalar multiple of the $i$th component of $\mathbf{m} \times \mathbf{n}$, but I have gotten stuck attempting this method.

What I have tried so far:

$\begin{aligned} B(\mathbf{m} \times \mathbf{n})_i&=(I+\mathbf{m}\mathbf{n}^T)(\mathbf{m} \times \mathbf{n})_i\\&=(\mathbf{m} \times \mathbf{n})_i+(\mathbf{m}\mathbf{n}^T)(\mathbf{m} \times \mathbf{n})_i \\&=\epsilon_{pqi}\mathbf{m}_p\mathbf{n}_q\,\,+(\mathbf{m}\mathbf{n}^T)_{ij}\,\epsilon_{pqj}\mathbf{m}_p\mathbf{n}_q \\&=\mathbf{m}_p\mathbf{n}_q(\epsilon_{pqi}+\mathbf{m}_i\mathbf{n}_j\,\epsilon_{pqj}) \end{aligned}$

Where $\epsilon_{ijk}$ is the Levi-Civita symbol, and the above working using the summation convention.

At this point, I'm unsure how this can be simplified further, and I'm not sure that this is necessarily the right course of action since it seems difficult to apply conditions such as the orthogonality of $\mathbf{m}$ and $\mathbf{n}$ anywhere in the expression. I know that this must mean $\mathbf{m} \cdot\mathbf{n}=\mathbf{m}_i\mathbf{n}_i=0$, but I don't see the relevance of that to anything above. Any guidance, particularly a more clear/intuitive way to show this result rather than symbolic manipulation, would be much appreciated.

Solution 1:

The approach I find most intuitive is geometric. $\mathbf m \times \mathbf n$ is orthogonal to $\mathbf n$, so it follows that $$ \mathbf m\mathbf n^T (\mathbf m \times \mathbf n) = \mathbf m [\mathbf n^T(\mathbf m \times \mathbf n)] = \mathbf m \cdot 0 = \mathbf 0. $$ Thus, we can conclude that $$ (I + \mathbf m \mathbf n^T)(\mathbf m \times \mathbf n) = \mathbf m \times \mathbf n + \mathbf m\mathbf n^T(\mathbf m\times \mathbf n) = \mathbf m \times \mathbf n + \mathbf 0 = \mathbf m \times \mathbf n, $$ so that $\mathbf m \times \mathbf n$ is an eigenvector associated with the eigenvalue $1$.

If you prefer symbolic manipulation, we could proceed with the following summation convention computation: $$ (\mathbf m \mathbf n^T)_{ij} \epsilon_{pqj}\mathbf m_p\mathbf n_q = \mathbf m_i\mathbf n_j \epsilon_{pqj}\mathbf m_p\mathbf n_q = \mathbf m_i \cdot [\epsilon_{pqj}\mathbf m_p\mathbf n_q\mathbf m_j] = \mathbf m_i \cdot 0 = 0. $$