Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2018 / Paper 1 / 6

A square matrix $A$ is a diagonalizable matrix over a specified field if some invertible matrix $P$ makes $P^{-1}AP$ diagonal. Equivalently, the underlying vector space has a basis of eigenvectors of $A$.

For example,The first is already diagonal. The second has only the eigenvalue one, of algebraic multiplicity two, but its eigenspace consists of $(x,0{)}^T$ and has dimension one. It cannot supply a two-vector eigenbasis.

Every row sums to six, so $\mathbf{e}=(1,1,1{)}^T$ is an eigenvector with eigenvalue six. Since $M$ is a real symmetric matrix, its other two eigenvalues, say ${\lambda }_2,{\lambda }_3$, are real. The trace givesAlso, for a real symmetric matrix, $\mathrm{tr}(M^2)$ is the sum of the squares of all its entries. HereThus ${\lambda }_2^2+{\lambda }_3^2=42-36=6$. Together with their zero sum, this implies

By the spectral theorem for real symmetric matrices, choose an orthonormal eigenbasis ${\mathbf{u}}_1,{\mathbf{u}}_2,{\mathbf{u}}_3$ corresponding to $6,\sqrt{3},-\sqrt{3}$, with ${\mathbf{u}}_1=(1,1,1{)}^T/\sqrt{3}$. Write $P=[{\mathbf{u}}_1{\mathbf{u}}_2{\mathbf{u}}_3]$ and $\mathbf{x}=P\mathbf{\xi }$. This orthogonal change of coordinates givesThe plane constraint is therefore ${\xi }_1=0$. On that plane the quadratic form vanishes exactly whenConsequently the entire set isThe two direction vectors are nonzero and not parallel, since ${\mathbf{u}}_2,{\mathbf{u}}_3$ are linearly independent. An invertible orthogonal transformation maps each full line to a full line. Thus the set consists of precisely two infinite lines through the origin.