For the value $x^T\Lambda x$, where $\Lambda$ is positive semidefinite (and it is not positive definite), what determines the maximum value of the mutiplication when the value of $x$ is given. (If $\Lambda$ is PD, then it should be the maximum eigen value affects the value through decomposing $x$ to a linear combination of all orthogonal eigen vectors)
Let me explain more clearly. The reason I say $x$ is given is because the norm of $x$ is not the primary concern here. Say, we can treat $\Lambda$ as a system, and here the concerned thing is the property of the system, rather than how the input $x$ affect the result.
Yes, the largest eigenvalue will determine the maximum value of the quadratic form. One can see this by writing $\Lambda$ by a sum of rank one matrices. $$ \Lambda = \sum \lambda_i u_i u_i^T $$ where $u_i$ are the eigenvectors. Then, $$ x^T \Lambda x = \sum \lambda_i (x^T u_i)^2. $$ As the $u_i$ form an orthonormal basis, the maximum is achieved by letting $x = u_i$ for $i$ corresponding to the largest eigenvalue.