Let $ Q $ be an $ n \times n $ diagonal matrix with positive diagonal entries $\lambda_{1}<\cdots<\lambda_{n}$.
Find local minimizer(s) for the function $f : \mathbb{R}^{n} \times \mathbb{R}^{n} \rightarrow \mathbb{R} $ and show which 1st and 2nd order conditions satisfy:
I want to minimize: $ f(x, y)= {x}^{T} Q y $ subject to $ \left\| x \right\|^{2} + \left\| y \right\|^{2} = 1 $ for $ x, y \in \mathbb{R}^n $.
Here's what I have tried: I see that all feasible points are regular points, so I can find all minimizer candidates using Lagrange multipliers.
Letting $g(x,y) = |x|^2+|y|^2-1$, I set $$\nabla f + \lambda \nabla g = \left(\begin{array}{cc}Qy+\lambda 2x\\Qx+\lambda2y\end{array}\right):= {\bf 0}.$$
I'm not quite sure how to proceed here. Am I missing something in how to tackle this kind of problem?
A full answer, per the suggestion in the comments:
Your equation for the candidate minimizers (via Lagrange's method) can be written as $$ x = - \frac{1}{2 \lambda}Qy, \quad y = -\frac{1}{2\lambda} Qx $$ If we use the second equation to substitute into the first, we obtain $$ x = \frac{1}{4\lambda^2}Q^2x $$ So, $\lambda$ will correspond to a candidate minimizer if and only if $\frac{1}{4\lambda^2}$ is an eigenvalue of $Q$, and $x$ must be an associated eigenvector.
From there, we must scale $x$ appropriately so that it fits the constraint $|x|^2 + |y|^2 = 1$. In particular, $x$ must satisfy the condition $$ |x|^2 + \frac{1}{2|\lambda|}|Qx|^2 = 1 \implies\\ |x|^2 + \frac 1{2 |\lambda|}x^TQ^TQx = 1 $$ This simplifies significantly in the case that $Q$ is symmetric (which is true in this case since $Q$ is diagonal). In particular, since $Q = Q^T$, the equation above becomes $$ |x|^2 + \frac 1{2 |\lambda|}x^TQ^2x = 1 \implies\\ |x|^2 + \frac 1{2 |\lambda|}x^T(4 \lambda^2 x) = 1 \implies\\ (1 + 2 |\lambda|)|x|^2 = 1 \implies\\ |x|^2 = \frac{1}{1 + 2 |\lambda|} $$