I am trying to prove the sentence "A short calculation shows the above leads to M u = λ u (symmetry of M is needed here). Therefore λ is the largest eigenvalue of M."
from this wikipedia site about Singular Value Decomposition. Shortly, show $\nabla \bar{x}^{T} M \bar{x} = \lambda ( \nabla \bar{x}^{T} \bar{x} )$
satisfying $M \bar{u} = \lambda \bar{u}$ where $\bar{u}$ is the maximum eigen vector, $\bar{M}$ is a symmetric matrix and $\lambda$ is eigen value, corresponding to the maximum eigen vector.
[Update]
Maximize $\bar{x}^{T} M \bar{x}$ over all $\bar{x}$ such that $||\bar{x}||=1$. Lagrange function is $f(x) = \bar{x}^{T} M \bar{x} + \mu (||\bar{x}||- 1)$
Now I need to calculate $\nabla f(x)$
calculating, and trying to show that the largest eigen value occur with the constraint...calculating and thinking.
Nabla operations here.