Tag: j$ and that $$ \begin{cases} a_{ij}=0 &\text{if $i+j\equiv 0\mod 2$

Maximizing quadratic forms

Post author By Math Dev
Post date March 26, 2020
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Consider the maximization problem $$\text{maximize} \quad Q(x)= \sum_{i<j} \Big(\sum_{k} a_{ik}a_{jk}\Big) x_i x_j \quad \text{subject to} \quad \sum_{i}x_i^2=1,$$ and let $M$ be maximum value obtained by $Q$ under such constraint. Suppose that $a_{ij}=a_{ji}$ for all $i,j$ and that $$ \begin{cases} a_{ij}=0 &\text{if $i+j\equiv 0\mod 2$,} \\ a_{ij}>0 & \text{if $i+j\equiv 1\mod 4$,} \\ a_{ij}<0 & \text{if […]

Tags } \\ a_{ij}>0 & \text{if $i+j\equiv 1\mod 4$, } \\ a_{ij}<0 & \text{if $i+j\equiv 3\mod 4$.} \end{cases} $$ I was wondering if one could get an interesting upper bound on $M$ by using th, $$ and let $M$ be maximum value obtained by $Q$ under such constraint. Suppose that $a_{ij}=a_{ji}$ for all $i, but still that does not require for every instance to use Lagrange multipliers (I am not interested in the precise value of $M$, Consider the maximization problem $$\text{maximize} \quad Q(x)= \sum_{i<j} \Big(\sum_{k} a_{ik}a_{jk}\Big) x_i x_j \quad \text{subject to} \, I am interested in a better upper bound than that provided by the Cauchy-Schwarz inequality, j$ and that $$ \begin{cases} a_{ij}=0 &\text{if $i+j\equiv 0\mod 2$, nor in where the maximum is attained).