Lagrange multipliers with $\lambda = 0$ problem

Question

We want to maximize the function $f(x_1,x_2,x_3,x_4) = \sum_{i=1}^{4}a_i^2x_i$ over the compact set $\Omega = \{x \in \mathbb R^4: |x| = 1, \langle x,a\rangle = 0\}$ where $a = (a_1,a_2,a_3,a_4)$ is some non zero vector

This is a continuous function over a compact set so it admits minimum and maximum. Thus we need to solve the system $\begin{cases}a_i^2 = 2\lambda_1 x_i +\lambda_2 a_i \\ |x|-1 = 0\\\langle x,a \rangle = 0\end{cases}$

I solved this for the case that $\lambda_1 \neq 0$. But when $\lambda_1 = 0$ I have too many variables and no way to isolate $x$ - how do I solve this?

Answer 1

If you have $\lambda_1=0$, then the first condition becomes

$$a_i^2=\lambda_2a_i, i=1,2,3,4$$

for some $\lambda_2$. That means

$$\sum_{i=1}^4a_i^2x_i=\sum_{i=1}^4\lambda_2a_ix_i=\lambda_2\sum_{i=1}^4a_ix_i=\lambda_2\langle a,x\rangle = 0.$$

In other words, your objective function is a multiple of one of of your conditions, so your objective function is the same ($0$) on the whole permissable set.

Answer 2

Hint If $\lambda_1 = 0$ then $a_i^2 = \lambda_2 a_i$ which implies $\lambda_2 = a_i \Rightarrow a = (\lambda_2, \lambda_2, \lambda_2, \lambda_2)$, or $\lambda_2 = 0$, and you should be able to compute a solution.

Answer 3

By your first equation, $\lambda_1 = 0$ can only occur if $a_1 = a_2 = a_3 = a_4 = \lambda_2$. By the third equation this means you are maximizing/minimizing the zero function.

EDIT: As Ingix pointed out, $a_i = 0$ is also possible and leads to the same conclusion.