Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 18

Let ${\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3$ be the Cartesian basis vectors, with the given vector $\mathbf{x}={\mathbf{e}}_1$. The permutation matrix cycles them forward:Thus three applications return the original vector.

Interchanging the rows and columns gives the matrix transposeIt cycles the three basis vectors in the opposite direction. Hence

The preceding cycles show the action on a full basis, not only on one direction: $A$ sends ${\mathbf{e}}_1\to {\mathbf{e}}_2\to {\mathbf{e}}_3\to {\mathbf{e}}_1$, and $A^{\mathsf{T}}$ reverses each arrow. Therefore either composition fixes every basis vector, and by linearity fixes every vector. ThusIn particular $A$ is an orthogonal matrix, $A^{\mathsf{T}}=A^{-1}=A^2$, and $A^3=I$.

For an eigenvector $\mathbf{v}=(v_1,v_2,v_3{)}^{\mathsf{T}}$, the equation $A\mathbf{v}=\lambda \mathbf{v}$ readsA nonzero solution requires ${\lambda }^3=1$. Write the three roots of unity in the requested polar form of a complex number:Indeed ${\mathbf{v}}_k=(1,{\lambda }_k^2,{\lambda }_k{)}^{\mathsf{T}}$ satisfies all three equations. These eigenvectors are unnormalized; any nonzero scalar multiples are equally valid.

Since $A^{\mathsf{T}}=A^{-1}$, each eigenvector ${\mathbf{v}}_k$ from part (d) remains an eigenvector, with reciprocal eigenvalue:The set of eigenvalues is again $\{1,e^{2\pi i/3},e^{4\pi i/3}\}$, with the two nonreal values exchanged relative to the same eigenvectors. Equivalently, for $\mu =e^{2\pi ik/3}$ one can choose the eigenvector $(1,\mu ,{\mu }^2{)}^{\mathsf{T}}$. Pairing each eigenvalue with its eigenvector matters in the next part.

The symmetric part of a matrix isIt satisfies $S^{\mathsf{T}}=S$. The complementary term $(A-A^{\mathsf{T}})/2$ is a skew-symmetric matrix.

Use the common eigenvectors of $A$ and $A^{\mathsf{T}}$. For each $k$,Thus, without constructing $S$ explicitly, its eigenvalues areThe eigenvalue $1$ has direction $(1,1,1{)}^{\mathsf{T}}$. The repeated eigenvalue $-1/2$ has a real eigenspace consisting of vectors whose coordinates sum to zero. This is the general rule for the symmetric part of an orthogonal matrix. Adding unrelated lists of eigenvalues would not suffice; here the shared eigenvectors justify the pairing.