Past exam of the options II course of the University of Oxford / 2013 / F2 / b

The Lagrange multiplier theorem for one regular constraint says: if $f,k$ are $C^1$ near $x$, $f(x)=0$, $\nabla f(x)\ne 0$, and $x$ is a local extremum of $k$ restricted to $f^{-1}(0)$, thenFor several constraints with linearly independent gradients, the objective gradient lies in their linear span. The regularity assumption rules out singular points where the usual multiplier conclusion can fail.

On the constraint set, $\Vert x\Vert =1$, so $\nabla f(x)=2x\ne 0$. Part (a)(iv) and the Lagrange multiplier theorem give $2A^{T}Ax=2\lambda x$, henceSince $x$ is nonzero, it is an eigenvector. Moreover $\lambda =x^T{A}^{T}Ax=\Vert Ax{\Vert }^2\ge 0$, so the associated eigenvalue is real and nonnegative.

For $n\ge 1$, the unit sphere is nonempty and compact by the Heine-Borel theorem. The function $k(x)=\Vert Ax{\Vert }^2$ is continuous and therefore attains a minimum and a maximum on it by the extreme value theorem. Either point supplies a nonzero eigenvector in part (ii), with a real multiplier. Thus $A^{T}A$ always has a real eigenvalue, indeed a nonnegative one. This gives existence without assuming the spectral theorem first.