Past exam of the options II course of the University of Oxford / 2004 / F (Multivariable Calculus)

Write the variables as $(u,v)\in {\mathbb{R}}^n\times {\mathbb{R}}^k$, with $a=(u_0,v_0)$. The hypothesis makes $D_{u}f(a)$ invertible. The implicit function theorem supplies neighbourhoods $U,V$ of $u_0,v_0$ and a unique smooth map $\varphi :V\to U$ such thatIts derivative is $D\varphi (v_0)=-(D_{u}f(a){)}^{-1}{D}_{v}f(a)$. Thus the derivative of this parametrization has image exactly $\ker Df(a)$.

Since $h$ has a constrained maximum at $a$, differentiating $v\mapsto h(\varphi (v),v)$ gives zero in every direction. Therefore $Dh(a)$ vanishes on $\ker Df(a)$. In finite-dimensional linear algebra, $(\ker Df(a){)}^{\perp }$ is the row space of $Df(a)$. ConsequentlyThe full-rank hypothesis is what permits this Lagrange multiplier conclusion; it is not a consequence of constrained maximality alone at a singular constraint.

Positive definiteness gives a positive smallest eigenvalue $c$ with $x^{t}Ax\ge c|x{|}^2$. Hence the set $x^{t}Ax=y^{t}Ay=1$ is compact, and $h$ attains a maximum there. Its two constraint gradients are $(2Ax,0)$ and $(0,2Ay)$; they are linearly independent because $x,y$ are nonzero and $A$ is invertible.

Put $s=x^{t}Ay$. At any constrained extremum the multiplier equations areMultiplication by $x^t$ and $y^t$ gives $\lambda =\mu =s^2$. If $s=0$ the objective is zero. Otherwise $Ay=sAx$, so $y=sx$ and normalization gives $s^2=1$. Thus every maximum is at most one, and the value one is attained at $x=y$. This proves $\boxed{(x^{t}Ay{)}^2\le 1}$ on the normalized constraints.

For nonzero arbitrary vectors apply that result to $x/\sqrt{x^{t}Ax}$ and $y/\sqrt{y^{t}Ay}$. If either vector is zero the conclusion is immediate. ThereforeEquality for nonzero vectors holds precisely when they are linearly dependent. This is the Cauchy-Schwarz inequality for the inner product induced by $A$.