Past exam of the mathematical methods I course of the University of Oxford / 2016 resit / 9

Both expressions are the same scalar:This identity uses the ordinary transpose and remains true for complex entries. A Hermitian inner product uses a conjugate transpose instead and has conjugate symmetry, not this bilinear symmetry.

The $y$ coordinate is an independent eigen-direction, while the $xz$ block is $\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$. Thuswhich gives $\det (A-I)=0$. The eigenspaces areDivide the two diagonal $xz$ vectors by $\sqrt{2}$ to obtain an orthonormal eigenbasis. The eigenvalue one has multiplicity two; giving only the $y$ axis would omit an independent eigenvector.

Because $A$ is symmetric, translating $\mathbf{r}=\mathbf{x}+\mathbf{a}$ givesThe linear term vanishes when $\boxed{2A\mathbf{a}+\mathbf{b}=0}$. Solving gives $\boxed{\mathbf{a}=(-1,0,0{)}^{\mathsf{T}}}$. Since $\mathbf{b}=-2A\mathbf{a}$, the remaining constant isThis is completing the square for a quadratic form.

An orthogonal transformation to the eigenvectors preserves distance from the centre and writes the surface as $u^2+v^2+3w^2=3$. It is an ellipsoid with semi-axes $\sqrt{3},\sqrt{3},1$. ThereforeThe original $y$ axis is one of the eigen-directions with eigenvalue one, and the centre has $a_y=0$. Consequently $y^2\le 3$, attained when the other centred coordinates vanish, so $\boxed{y_{\max }=\sqrt{3}}$.

For $A^{\prime }$, the centre and constant remain $\mathbf{a}=(-1,0,0)$ and $C=3$. Its eigenvalues are $3,1,-1$, so rotated coordinates put the surface in the form $3u^2+w^2-y^2=3$. This is a one-sheet hyperboloid. Since $3\Vert \mathbf{x}{\Vert }^2\ge {\mathbf{x}}^{\mathsf{T}}{A}^{\prime }\mathbf{x}=3$, the nearest centre-to-surface distance is at least one, attained in the eigen-direction with eigenvalue three. The coordinate $y$ is unbounded, because arbitrarily large $|y|$ can be balanced by $u$ or $w$. Thus the nearest distance is one, and there is no maximum value of $y$.

For the printed nonsymmetric $A^{\prime \prime }$, use the symmetric part of a matrix:The skew-symmetric part contributes zero to this scalar. The same centre solves $2S\mathbf{a}+\mathbf{b}=0$, and again $C=3$. Sylvester's positive principal minors are $2,2,5/2$, so this surface is an ellipsoid.

Let ${\lambda }_{\ast }$ be the largest eigenvalue of $S$. Its characteristic equation is $4{\lambda }^3-20{\lambda }^2+27\lambda -10=0$, and the Rayleigh quotient givesFor the maximum $y$, the support function of a centred ellipsoid gives the bound below. Equivalently, apply the Cauchy-Schwarz inequality in the $S$-weighted inner product: $y^2\le 3{\mathbf{e}}_y^{\mathsf{T}}{S}^{-1}{\mathbf{e}}_y$. The inverse entry is ${\mathbf{e}}_y^{\mathsf{T}}{S}^{-1}{\mathbf{e}}_y=6/5$, and equality is attained for a centred vector proportional to $S^{-1}{\mathbf{e}}_y$. Thus

Using the eigenvalues of the raw nonsymmetric $A^{\prime \prime }$ would be incorrect: it is its symmetric part that determines the surface and its Euclidean geometry.