Past exam of the mathematical methods I course of the University of Oxford / 2013 / 9

For an ordinary real quadratic form, split the matrix into its symmetric part and skew-symmetric part:The scalar ${\mathbf{x}}^{T}K\mathbf{x}$ equals its transpose ${\mathbf{x}}^T{K}^T\mathbf{x}=-{\mathbf{x}}^{T}K\mathbf{x}$, so it is zero. ConsequentlyThis uses transpose because the source form is bilinear. A complex Hermitian form ${\mathbf{x}}^{†}A\mathbf{x}$ would instead require the conjugate-transpose decomposition.

The symmetric part of the displayed matrix isIt equals the matrix of 8 plus $3I$, so the same orthonormal eigenbasis gives eigenvalues $6,1,1$. In coordinates $u,v,w$ along ${\mathbf{v}}_3,{\mathbf{v}}_{-2},{\mathbf{e}}_2$, respectively, the level set isAll three eigenvalues are positive, so this is an ellipsoid with semi-axis lengths $1,\sqrt{6},\sqrt{6}$. The short axis is along $(2,0,1)$, and the plane of the longest axes is $2x_1+x_3=0$.

A natural longest-axis direction in the $x_1{x}_3$ plane is ${\mathbf{v}}_{-2}=(1,0,-2)/\sqrt{5}$. The angle $\eta$ of a unit vector to the plane $x_1=0$ has $\sin \eta =|v_1|$, givingfor that choice. Degeneracy qualification: the two longest semi-axes are equal, so “the longest axis” does not specify a unique direction. The equally valid longest axis ${\mathbf{e}}_2$ lies in $x_1=0$ and has angle zero. Generally $\mathbf{v}=\cos \psi {\mathbf{v}}_{-2}+\sin \psi {\mathbf{e}}_2$ is a longest-axis unit vector, withThis is the principal-axis freedom in a repeated ellipsoid eigenspace; a unique answer needs an extra axis-selection convention.

The coefficient matrix and right-hand side areFor an LU decomposition with unit diagonal in the lower factor, subtract row 1 from row 2, add twice row 1 to row 3, and then add the new row 2 to row 3. The elimination multipliers are $l_{21}=1$, $l_{31}=-2$, and $l_{32}=-1$. ThusMultiplication verifies $LU=A$. Forward substitution in $L\mathbf{y}=\mathbf{b}$ gives $y_1=4$, $y_2=-4$, and $y_3=-2$. Back substitution then gives $x_3=-1$, $x_2=5/3$, and $x_1=1/6$:Every pivot is nonzero, so no row permutation is needed and the linear system has a unique solution.