From an engineer point of view what I would do if I had a non-positive definite matrix is:
Obtain its eigen decomposition.
Changes the negative eigenvalues for zeroes.
I think you meant \alpha=\pi/4 (or \pi/4+\pi)
For M=2, the solution only allows these values for the lambdas. I am interested in a generic M which is less obvious.
I actually forgot to mention \lambda_1+\cdots+\lambda_M=M, i.e. the average lambda is one.
But you derived this conditions...
Hello buddies,
Here is my question. It seems simple but at the same time does not seem to have an obvious answer to me.
Given that you have two vectors \mathbf{u},\mathbf{v}.
They are orthogonal \mathbf{u}^T\mathbf{v}=0 by standard dot product definition.
They have norm one...