Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 1 / 6B (Vectors and Matrices)

A matrix and its matrix transpose have the same characteristic polynomial, becauseThus they have the same eigenvalues, including algebraic multiplicities.

The proposed sum identity is false. Forboth eigenvalues ${\lambda }_i$ are zero, whereashas eigenvalues $0,1$. HenceEquivalently, the two sides are generally $\mathrm{tr}(A^{T}A)$ and $\mathrm{tr}(A^2)$, which need not agree.

The product identity always holds, sinceTherefore

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 $B$.