Positivity of quadratic form minus linear form on the simplex

Question

Let us $a_{ij}$ be the elements of a $n$-dimensional covariance matrix. Can we prove the following?

$$ 1-\sum_{k=1}^n a_{ik} \lambda_k + \sum_{j=1}^n \sum_{k=1}^n \lambda_j a_{jk} \lambda_k > 0, \qquad i \in \{1, \dots, n\} $$

where the $\lambda_k$ are constrained by

$$ \sum_{k=1}^n \lambda_k = 1, \quad \lambda_i > 0 \text { for } i \in \{1, \dots, n\} $$

Or, in a more general way, what are the conditions that the elements of the covariance matrix should satisfy so that the above set of inequalities hold?

NOTE: in matrix form, if $A$ is a covariance matrix and $a_i$ is a row vector having the i-th row of $A$, the question is:

is $1-a_i \lambda + \lambda^T A \lambda >0 $ where $\lambda=[\lambda_1 \ldots \lambda_n]^T$ and $\lambda^T e_n $ =1 with $\lambda_k >0$ and $e_n$ a column vector of ones. The question should be formulated as: find the conditions for the elements of matrix $A$ so that the inequality holds for $i=1,\ldots n$.

Answer 1

Here is a sufficient condition. The function $f(\lambda) = 1-a_i \lambda + \lambda^T A \lambda >0$ restricted on the simplex can be written, with a change of coordinates $\lambda\rightarrow x$ given by $\lambda_i=x_i^2$, as: $$ f(\lambda) = f(x) = \left(\sum_k x_i^2\right)^2 - \left(\sum_k a_{ik}x^2_k\right)\left(\sum_i x_i^2\right) + \sum_{i,j}A_{i,j}x_i^2x_j^2. $$ Now, if $f(\lambda)\geq 0$ on the simplex, then this implies that $f(x)$ must be positive semidefinite on the unit sphere. Now, a homogeneous quartic PSD on the unit sphere is globally PSD. A sufficient condition for a quartic to be globally PSD is to choose a sum of squares representation, which must have a form $zPz^\top$ with $z=[x_1^2~\cdots~x_n^2]$ and $P\succeq 0$. This translates to a linear matrix inequality on elements of $A$, which is the final sufficient condition.