Let f(x) = x $\bullet$ $Q x$, $g_{0}$ = $|x|^{2} - 1$.
Considering the minimization problem - minimize f subject to $g_{0}$ = 0 show that if v is any solution to the minimization problem then v is an eigenvector of Q and that the associated eigenvalue $\lambda_{0}$ is : $min_{ |x|^{2} = 1} $ x $\bullet$ $Q x$. I was able to show this part.
For the next part an extra constraint : v $\bullet$ x = 0 is added. And we are asked to show that if $v_{2}$ is any solution to the minimization problem then $v_{2}$ is an eigenvector of Q and the associated eigenvalue is $\lambda_{1}$ such that $\lambda_{1} \geq \lambda_{0}$. I got $\lambda_{1}$ = $min_{ |x|^{2} = 1 \text{ and } \text{v } \bullet \text{ x = } 0} $ x $\bullet$ $Q x$. However I am not able to show $\lambda_{1} \geq \lambda_{0}$. Any hint would be greatly appreciated.
Note : This is part of a school assignment so please do not give away the answer - just some useful hint.
Edit : I think I got the answer but still would like to be sure (dont think it will be right to post the answer here but its basically a simple contradiction that will arise if $\lambda_{1} \geq \lambda_{0}$ were not true)