 Solution

If $p=m\times n$, then $n^{T}p=0$, and hence $Bp=p$. Also $Bm=m$ because $m\perp n$. Thus $p$ and $m$ are linearly independent eigenvectors with eigenvalue one.

For every vector $v$,so $B-I$ is nonzero but has square zero:Consequently all three eigenvalues are one. The eigenspace iswhich has dimension two. It cannot supply three linearly independent eigenvectors, so