Past exam of the mathematics course of the University of Cambridge / 2012 / ia / Paper 1 / 8B / a

The symmetry between the last two coordinates gives the eigenvector $(0,1,-1{)}^T$ with eigenvalue $2$. On vectors $(s,t,t{)}^T$, the remaining eigenvalue equation iswhose characteristic polynomial is ${\lambda }^2-5\lambda +4=(\lambda -1)(\lambda -4)$. Corresponding eigenvectors are $(1,-1,-1{)}^T$ and $(2,1,1{)}^T$. Thus all eigenvalues and eigenspaces areNonzero scalar multiples give all eigenvectors in each eigenspace. The three displayed vectors are mutually orthogonal and have squared norms $3,2,6$, respectively.

The quadratic expression is ${\mathbf{x}}^{T}A\mathbf{x}$. Normalize the preceding eigenvectors to obtain an orthonormal basisWith the orthogonal matrix $U=({\mathbf{u}}_1{\mathbf{u}}_2{\mathbf{u}}_3)$ and coordinates $\mathbf{y}=U^T\mathbf{x}$, the surface equation becomes $y_1^2+2y_2^2+4y_3^2=1$. All coefficients are positive, so it is an ellipsoid, with principal semiaxes $1,1/\sqrt{2},1/2$ along the respective eigenvectors. This is principal-axis reduction of a quadric.

A map to the unit sphere is $\mathbf{x}\mapsto \mathrm{diag}(1,\sqrt{2},2)U^T\mathbf{x}$. One explicit matrix isIndeed $B^{T}B=A$, so $|B\mathbf{x}{|}^2={\mathbf{x}}^{T}A\mathbf{x}$. Since $B$ is invertible, every point on the unit sphere has a preimage on the ellipsoid; the map is onto, not merely into. The requested $B$ is not unique, since an additional orthogonal matrix on the left preserves the unit sphere.