Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 1 / 13 / b

Multiply the quadratic form out directly:Thus $b=5$, $c=-6$ and $d=5$. Both eigenvalues are positive, so this is a positive-definite quadratic form; its positive unit level set is an ellipse. The next part exhibits its principal axes explicitly.

Let $Q=(x_1{x}_2)$ for the normalized eigenvectors already chosen. Then $Q^{T}Q=I$, $(x,y{)}^T=Q(x^{\prime },y^{\prime }{)}^T$, and orthogonal diagonalization givesThe ellipse has semi-axis lengths $1/\sqrt{2}$ and $1/\sqrt{8}$ along $x_1$ and $x_2$, respectively. Rescale $x^{\prime \prime }=\sqrt{2}{x}^{\prime }$ and $y^{\prime \prime }=\sqrt{8}{y}^{\prime }$ to obtain the unit circle $(x^{\prime \prime }{)}^2+(y^{\prime \prime }{)}^2=1$. Since $(x^{\prime },y^{\prime }{)}^T=Q^T(x,y{)}^T$,The identity $B^{T}B=A$ checks the rescaling directly. Swapping or reversing eigenvectors gives other valid coordinate matrices; the displayed one follows the explicit choices in part (a).