I am having a look at the paper "Hodge structures of type $(n, 0, \ldots, 0, n)$" by Totaro. At the very beginning he says that a Hodge structure of type $(1, 0, 1)$ has always complex multiplication.
Here, type $(1, 0, 1)$ means that $V$ is a pure $\mathbb{Q}$-Hodge structure of weight 2 with Hodge decomposition $V_{\mathbb{C}}=V^{2, 0} \oplus V^{0, 2}$ and $V^{2, 0}$ one dimensional.
"Complex multiplication" means in this case that there is an imaginary quadratic number field $K$ acting on $V$ by endomorphisms of Hodge structures.
Why is is true?
My guess is that you can renumber the pieces of the Hodge decomposition to get a weight one Hodge structure of dimension 2, which should correspond to an elliptic curve.
1) Does this show that the Hodge structure $V$ is automatically polarizable?
2) Why does the elliptic curve constructed in that way have complex multiplication?
3) Does there always exists a K3 surface $X$ such that $V$ is the transcendental part of the singular cohomology $H^2(X(\mathbb{C}, \mathbb{Q})$?