Why can't you always write the anti-symmetric state as slater determinant? For example, consider the following state
$$|\Psi\rangle_a = [|r_1 r_2\rangle -|r_2 r_1\rangle ]\otimes \left[|\uparrow \downarrow\rangle +|\downarrow \uparrow \rangle \right] $$
You can't write this as a slater determinant but can write as sum of two slater determinants.

What does this show? From what I understand is that, The anti-symmetric Projector
$$A\equiv \frac{1}{N!}\sum_\alpha \epsilon_\alpha \mathcal{P}_\alpha$$
projector only basic set of Anti-symmetric subspace of vector space $\mathcal{V}$ and one need to form a linear combination to get the whole states. I'm not able to make this idea more rigorous. Is that true or always true? What's the better way to say it?
 A: Single Slater determinant states only make up a small fraction of the possible antisymmetric states in a many-body Hilbert space. Indeed, you've written down one such state that is not expressible as a single Slater determinant. However, Slater determinant states do form a basis for the antisymmetric subspace of a many-body Hilbert space, meaning that any antisymmetric state is expressible as a linear combination of Slater determinant states. (For more on this, see here.) This is also what you're seeing when you observe that your state is expressible as a sum of single Slater determinant states.
Note that this is equivalent to the statement that fermionic Fock states are a basis for a fermionic many-body Hilbert space, as fermionic Fock states are single Slater determinant states.
As far as the projector you've written down, this acts trivially on antisymmetric states, acts as a Slater determinant on product states, and annihilates symmetric states. The full many-body Hilbert space includes orthogonal subspaces of states which are symmetric and anti-symmetric, and all bosonic and fermionic many-body states lie in these subspaces, respectively. There is more discussion of this here.
