Past exam of the mathematics course of the University of Cambridge / 2013 / ia / Paper 1 / 7B (Vectors and Matrices)

We prove independence of eigenvectors for distinct eigenvalues by induction on the number of vectors. A single eigenvector is nonzero and hence independent. Suppose the first $d-1$ are independent, and let ${\sum }_{j=1}^d{c}_j{v}_j=0$. Applying $A-{\lambda }_{d}I$ givesBy the induction hypothesis every coefficient in this shorter relation vanishes. The eigenvalues are distinct, so $c_j=0$ for $j<d$. The original relation becomes $c_d{v}_d=0$, and $v_d\ne 0$ gives $c_d=0$. Thus the entire family of eigenvectors is linearly independent. The argument works over either the real or complex field.

Represent the quadratic form by the symmetric matrixThe $xy$ block has characteristic polynomial $(2-\mu )(5-\mu )-4=(\mu -1)(\mu -6)$. Corresponding orthonormal eigenvectors, together with the $z$ direction, areThe eigenvalues are $1,6,-1$. To eliminate the linear term, solve $H\tilde{O}=-(0,3\sqrt{5},0{)}^T$, obtainingIndeed the first equation gives $x=y$, and the second then gives $3y=-3\sqrt{5}$. Define the new coordinates by $X=\tilde{O}+\tilde{x}{\tilde{e}}_1+\tilde{y}{\tilde{e}}_2+\tilde{z}{\tilde{e}}_3$. The original polynomial takes the value $-15$ at the center, so its translated and orthogonally diagonalized form isThusThe surface is an elliptic one-sheet hyperboloid, centered at $\tilde{O}$, with axis parallel to ${\tilde{e}}_3$. Its waist at $\tilde{z}=0$ is an ellipse with semiaxes $\sqrt{15}$ and $\sqrt{5/2}$. Every constant-$\tilde{z}$ section is a nonempty ellipse, so the two signs of $\tilde{z}$ belong to the same connected sheet. This is an application of principal-axis reduction of a quadric.