I.6.6.1: Quadratic Form on a Sphere

We now discuss an actual example. Consider the problem of minimizing the function over , where we make no assumptions on the matrix , the vector , and the scalar . Instead of going the usual route of differentiating and solving the stationary equations, we use the decomposition approach.

Define or, letting , Then but also If then We distinguish three cases.

  • If the minimum is attained at and the minimum is equal to , where It follows that in this case is the smallest eigenvalue of , written as . If is the corresponding unit-length eigenvector, then the minimizer of is

  • If the minimum is not attained and for each . Thus as well.

  • If then we must distinguish two sub-cases. If then . If then again. Thus if we have , unless both and , in which sub-case we have equal to , the smallest eigenvalue of and the minimizer equal to any corresponding eigenvector.

Of course if we have . Thus

Now start with the alternative decomposition We want to show that although the intermediate calculations are different, the result is the same.

  • If and then , as before. But if If and the minimum is attained at , and . Because is less than or equal to , we still have equal to the smallest eigenvalue of .

  • If we stiil have .

  • If we distinguish three sub-cases. If then , as before. If then . And if the minimum is attained at and equal to . Again we have , unless both and , when is equal to .

We have solved the problem by using the decompositions~\eqref{E:g2} and~\eqref{E:g3}. But we can also interchange the order of the infimums and use or

Let's look at the problem . For a minimum we must have where the Lagrange multiplier is chosen such that . At the minimum