Given a positive definite matrix $\mathbf{M}$ such that $\mathbf{w'Mw}=s$ where $s>=0$, and a constraint $\mathbf{w'1}=1$, where the bold faced $\mathbf{1}$ indicates a vector of $1$s, does it not follow that
$$\mathbf{Mw}=\frac{s}{\mathbf{w'w}}\mathbf{w}$$
That is, that $\mathbf{w}$ is an eigenvector of $\mathbf{M}$ and the scalar $\frac{s}{\mathbf{w'w}}$ is its associated eigenvalue?
Reasoning is as follows:
$$\mathbf{1'Mw}=k$$
(where $k$ is some unknown scalar)
Now, considering the constraint $\mathbf{w'1}=1$, the following must be true:
$$\mathbf{Mw}=k\mathbf{w}$$
That is, pre-multiplying this by $\mathbf{1'}$ will recover $\mathbf{1'Mw}=k$.
As for what the scalar $k$ is, pre-multiply the equation $\mathbf{Mw}=k\mathbf{w}$ by $\mathbf{w'}$:
$$\mathbf{w'Mw}=k\mathbf{w'w}$$
But $\mathbf{w'Mw}=s$, so $k=\frac{s}{\mathbf{w'w}}$ as in the proposition at the top of this post.
Counterexample: $\mathbf M=\pmatrix{2&1\\1&1}$, $\mathbf w=(1,0)^T$. $\mathbf M\mathbf w=(2,1)^T$, which cannot be a scalar multiple of $\mathbf w$.
What you’re claiming is that for every positive-definite matrix, every vector whose sum of components relative to the standard basis is $1$ is an eigenvector. Moreover, for this to hold, in an $n$-dimensional space the quantity ${\mathbf w'\mathbf M\mathbf w\over\mathbf w'\mathbf w}$ can take on at most $n$ values over all such vectors for a fixed $\mathbf M$. Using the matrix from above, this would mean that for all $\mathbf w=(t,1-t)^T$, $${\mathbf w'\mathbf M\mathbf w\over\mathbf w'\mathbf w} = {t^2+1 \over 2t^2-2t+1}$$ takes on at most two different values for all $t\in\mathbb R$.