Distance extrema on a real quadratic surface ($x^{T}Ax=1,|x{|}^2=1/\lambda$)

For a real symmetric matrix $A$, every stationary squared distance $x^{T}x$ on $x^{T}Ax=1$ occurs in a positive-eigenvalue eigenspace. The Lagrange multiplier equation is $x=\Lambda Ax$; multiplying by $x^T$ gives $\Lambda =|x{|}^2>0$, hence $Ax=|x{|}^{-2}x$. If $A$ is a positive-definite matrix, the nearest and farthest squared distances are $1/{\lambda }_{\max }$ and $1/{\lambda }_{\min }$. With a repeated eigenvalue, the stationary set may contain an entire sphere within its eigenspace. For an indefinite or singular form the surface may be unbounded, and a stationary point need not be a global extremum.

Let $A$ be a real positive-definite matrix and let the columns of $U$ be an orthonormal basis of a two-dimensional linear subspace in three-dimensional Euclidean space. Points in that central plane are $x=Uq$, with $|x{|}^2=|q{|}^2$. Restricting the ellipsoid $x^{T}Ax=1$ gives $q^T(U^{T}AU)q=1$. The compressed matrix $U^{T}AU$ is a positive-definite matrix, and its extreme eigenvalues give the distance formulas in the title by the Rayleigh quotient. The two extremal axes are the images under $U$ of the corresponding eigenvectors. This restriction reduces a problem with two Lagrange multipliers to a two-dimensional quadratic form.