Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2008 / Paper 2 / 16

The permutation matrix acts by $A(x,y,z{)}^T=(y,z,x{)}^T$. Applying this cyclic permutation three times returns every vector to itself, proving $A^3=I$. Explicitly $A^2=\left(\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right)$ and multiplication once more gives the identity matrix.

Using $A^3=I$, the matrix power reduces modulo three:

Since $A^{2}A=A^3=I=A{A}^2$, the inverse matrix is

Here $A^{3n+1}=A$, so its inverse matrix is

With $\nu =0$, the characteristic polynomial factors into $(1-\lambda )[{\lambda }^2-\lambda -({\mu }^2+4)]$, up to the overall sign convention. The eigenvalues are $1$ and $(1\pm \sqrt{17+4{\mu }^2})/2$. Integer values require an odd integer $m$ satisfying $(m-2\mu )(m+2\mu )=17$. The positive integer factorization $17=1\cdot 17$ forces $m=9$ andThe resulting eigenvalues and corresponding nonzero eigenvectors areAny nonzero scalar multiple is equally valid. The factorization also shows that no other positive integer $\mu$ works.

For $\nu =5$, $B$ is real symmetric. Sylvester's criterion says that all eigenvalues are positive exactly when its leading principal matrix minors are positive. These are $1$, $5-4=1$, and $\det B=1-{\mu }^2$. ThereforeAt either endpoint one eigenvalue is zero, so the strict positivity requirement excludes it.