Fundamental and antifundamental representations, and hilbrt spaces

I am studying quarks $$u$$, $$d$$ and $$s$$. I know they can be represented by 3-vectors:

We also have the antiquarks that are represented by the same 3-vectors:

The difference between them is that when acting with ladder operators one has to use the Gell-Mann matrices $$\lambda_i$$ for the quarks and the antifundamental representation $$\bar{\lambda}_i$$ when acting on antiquarks.

Writing all this in Dirac notation we could say that $$\hat{\Lambda}_i |u\rangle \sim \lambda_i u \hspace{40pt} \hat{\Lambda}_i |\bar{u}\rangle \sim \bar{\lambda}_i \bar{u}$$ so $$\hat{\Lambda}_i$$ is the abstract operator that acts over the elements of the Hilbert space $$|u\rangle$$ and $$|\bar{u}\rangle$$.

My question is: because $$\hat{\Lambda}_i$$ can act over $$|u\rangle$$ and $$|\bar{u}\rangle$$ then I am tempted to say that they both live in the same Hilbert space. However if that were the case and if the 3-vector representation is correct, then they are the same state... So $$|u\rangle$$ and $$|\bar{u}\rangle$$ must be different but are represented by the same 3-vectors... How does this work? Does $$|u\rangle$$ and $$|\bar{u}\rangle$$ live in two different Hilbert spaces?

Also, a meson is composed by a quark and an antiquark. Lets consider the meson $$\bar{u} u$$. If I were to represent this meson in Dirac notation, would it be $$|u\rangle \otimes |\bar{u}\rangle$$?

The quarks and antiquarks transform by conjugate representations, respectively $$(1,0)$$ and $$(0,1)$$ in the Dynkin notation. In particular, the eigenvalues of the diagonal operators in the $$(0,1)$$ irrep are the negative of those in the $$(1,0)$$ since, if $$e^{i\theta \Lambda}$$ is diagonal (and hermitian) in $$(1,0)$$, then $$(e^{i\theta \Lambda})^* = e^{-i\theta \Lambda}$$ in $$(0,1)$$, illustrating how eigenvalues in an irrep are the negatives of the eigenvalues in its conjugate irrep.
Clearly, since no similarity transformation can change the eigenvalues, the irrep $$(1,0)$$ and $$(0,1)$$ must be distinct.
In particular, the standard Gell-Mann matrices act in the defining $$(1,0)$$ irrep, but the matrix elements of the generators in the $$(0,1)$$ will have sign differences from those in $$(1,0)$$ and thus will NOT be the standard Gell-Mann matrices (even though they will be $$3\times 3$$ matrices since $$(0,1)$$ is of dimension $$3$$).