Let $A\in\mathbb{R}^{n\times n}$ be a symmetric matrix with eigenvalues $\lambda_i\in [0,1]$, and $e_i=(0, 0, .. , 1,\dots)$ be the standard vectors. I need for any vector $w \in \mathbb{R}^n$ that
$$ e_i^TA w \leq |e_i^Tw| $$
I tried to use a spectral decomposition but so far nothing worked. So maybe this doesn't hold true?
More precisely, I have the following problem:
Let $A= (\frac{1}{d}\sum_{j=1}^d v_i v_i^T)$ for some vectors $v_i \in\mathbb{R}^n$ such that each component is bounded by $|v_i^{(j)}|\leq \frac{1}{\sqrt{n}}$ so that $\|v_i\|\leq 1$ and therefore the eigenvalues of $A$ are contained in $[0,1]$. Consider the polynomial $p_m(x)=\sum_{i=0}^m(1-x)^ix$ defined on $[0,1]$. One can check that $p(x)\leq 1$ for all $x\in [0,1]$. Therefore we have
$$ \|p(A)w\|_2\leq\|w\|_2 $$ Now for any $k$ and any $w \in\mathbb{R}^n$ such that each component is bounded by $|w^{(j)}|\leq \frac{1}{\sqrt{n}}$, i need to show that $$ |e_k^T p(A)w|\leq \frac{C}{\sqrt{n}} $$ where $C>0$ is a constant independent of $n$ and $m$. Thank you verry much in advance!
If $w = \sum w_je_j$ and $A = (a_{ij}),$ then your inequality is: $$\sum _jw_je^T_iAe_j = \sum_jw_ja_{ij}\leq |w_i|, \forall\, i$$ so I don't think this holds since you can easily choose $w_i$'s to violate this. I think: $$A = \begin{bmatrix} \frac 12 & \frac 13 \\ \frac 13 & \frac 12 \end{bmatrix}$$ and $w = [1 \,\,\, 0]^T$ gives a contradiction as: $$w_1a_{11} + w_2a_{12} = \frac 12 < 1$$ but $$w_1a_{12} + w_2a_{22} = \frac 13\not <0.$$