Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 1 / 6B / b

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

Multiplication givesThe vector $p=m\times n$ is an eigenvector with eigenvalue one. On the plane spanned by the orthonormal pair $m,n$, the matrix isIts eigenvalues are the roots of ${\mu }^2-3\mu +1$, namelyFor either root, an eigenvector is $m+({\mu }_{\pm }-1)n$. ThereforeThese positive eigenvalues are the squared singular values of $B$.