Past exam of the mathematics course of the University of Cambridge / 2012 / ia / Paper 1 / 8B (Vectors and Matrices)

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.

For a real orthogonal matrix, $P^{T}P=I$. ConsequentlyThus it preserves every real inner product, and in particular lengths and angles.

An eigenvalue can be complex even though $P$ is real, so use the complex Hermitian inner product. For a nonzero complex eigenvector $\mathbf{v}$, the real orthogonal matrix also satisfies $P^{\ast }P=P^{T}P=I$. Henceand $\boxed{|\lambda |=1}$. Taking complex conjugates in $P\mathbf{v}=\lambda \mathbf{v}$ gives $P\overline{\mathbf{v}}=\overline{\lambda }\overline{\mathbf{v}}$ because $P$ is real. The conjugate vector is nonzero, so $\boxed{{\lambda }^{\ast }=\overline{\lambda }\text{ is also an eigenvalue}}$.

For the unheaded geometrical conclusion, the third eigenvalue of $Q$ must be real: nonreal eigenvalues of a real matrix occur in conjugate pairs. Its modulus is one, so it is $1$ or $-1$. Exact algebraic multiplicity two for $1$ excludes a third $1$, leaving $-1$. By the supplied diagonalisability fact, the $1$-eigenspace is a two-dimensional plane and the $-1$-eigenspace is a line. They are orthogonal: preservation of the inner product between a fixed vector and a negated vector gives $\mathbf{u}\cdot \mathbf{v}=-\mathbf{u}\cdot \mathbf{v}$. Therefore $Q$ is the orthogonal reflection in its fixed plane. It fixes the two tangential directions and reverses the normal direction; with unit normal $\mathbf{n}$, $Q=I-2\mathbf{n}{\mathbf{n}}^T$.