I would like to solve the following maximization problem:
$$ max_x (x+y)^TM(x+y) \\ s.t. \hspace{.25cm} x^Tx \leq 1 $$ Where M is symmetric and negative definite.
I can construct the lagrangian : $$ max_x (x+y)^TM(x+y) + \lambda[r^Tr -1] $$
I am able to derive the F.O.C.: $$ M(x+y) = -2\lambda x $$ When the constraint is not binding clearly x = - y, with unique max = 0. But I have no idea how to solve for the binding case.
If I assume invertibility of $[M + 2\lambda I]$ (which im not sure if I can do) I can arrive at: $$ x = -[M+2\lambda I]^{-1}My $$ Invoking the constraint (and symmetry): $$ 1 = y^T M [M+2\lambda I]^{-1}M^2[M+2\lambda I]^{-1}M y $$ Which is 1 equation in 1 unknown and should have a unique solution for $\lambda$ which would solve the problem, but I see no way to do so.
Thank You for your help.
$ \def\a{\alpha}\def\b{\beta} \def\l{\lambda}\def\s{\sigma} \def\o{{\tt1}}\def\p{\partial} \def\LR#1{\left(#1\right)} \def\trace#1{\operatorname{Tr}\LR{#1}} \def\qiq{\quad\implies\quad} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\c#1{\color{red}{#1}} \def\CLR#1{\c{\LR{#1}}} \def\fracLR#1#2{\LR{\frac{#1}{#2}}} $In the binding case the problem becomes $$\eqalign{ &\max_x\;&(x+y)^TM(x+y) \\ &{\rm s.t.}&x^Tx = \o \\ }$$ which can be solved by constructing the constrained variable $x$ from an unconstrained variable $u$ $$\eqalign{ x &= \frac{u}{\|u\|} \qiq \grad xu = \fracLR{\|u\|^2I-uu^T}{\|u\|^3} = \fracLR{I-xx^T}{\|u\|} \\ }$$ For typing convenience, introduce the vector $$\eqalign{ z &= x+y \qiq dz = dx \\ }$$ Write the objective function and calculate its unconstrained gradient $$\eqalign{ \phi &= z^TMz \\ d\phi &= 2Mz:dz = 2Mz:dx = 2\fracLR{I-xx^T}{\|u\|}Mz:du \\ \grad{\phi}{u} &= \fracLR{2}{\|u\|}\LR{I-xx^T}Mz \\ }$$ Setting the gradient to zero yields the eigenvalue equation of a rank-one matrix $\LR{xx^T}$ $$\eqalign{ \LR{xx^T}(Mz) &= (Mz) \\ \LR{xx^T}v &= \l v \\ }$$ whose only non-zero eigenpair is $(\l,v)=(\o,x).\;$
This leads to a soluble linear equation for the optimal $x$ vector in the binding case $$\eqalign{ &x = Mz = M(x+y) \\ &\LR{I-M}x = My \\ &x = \LR{I-M}^{-1}My \\ }$$