Why can we write an arbitrary object $v_{a \dot{b} }$ our transformations in this basis act on as

$$ v_{a \dot{b} } = v_{\nu} \sigma^{ \nu}_{a \dot{b} } = v^0 \begin{pmatrix} 1&0 \\ 0&1 \end{pmatrix} + v^1 \begin{pmatrix} 0&1 \\ 1&0 \end{pmatrix} +v^2 \begin{pmatrix} 0&-i \\ i&0 \end{pmatrix} + v^3 \begin{pmatrix} 1&0\\0&-1 \end{pmatrix} $$

Formulated differently: How do we know that the vector space for the $(\frac{1}{2},\frac{1}{2})= (\frac{1}{2},0) \otimes (0,\frac{1}{2})$ representation of the Lorentz group is the space of hermitian $2\times2$ matrices? The vector space for the $(\frac{1}{2},0)$ representation is $\mathbb{C}^2$ and I guess the same is true for the $(0,\frac{1}{2})$ representation, but I can't put it together to end up with hermitian matrices.

EDIT: I found in the book Symmetry and the Standard Model: Mathematics and Particle Physics by Matthew Robinson the following explanation.

Recall that just as any real matrix can be written as the sum of a symmetric matrix and an antisymmetric matrix, any complex matrix can be written as the sum of a Hermitian matrix and an anti-Hermitian matrix. However, the two indices on our matrix $v^{a \dot b}$ transform under representations of $SU(2)$. Notice that in the generators of these copies of $SU(2)$, both sets of generators $N^-$ and $N^+$ are Hermitian (cf. (3.229)). So, we’ll limit our discussion to the case where $v^{a \dot b}$ is a Hermitian $2 \times 2$ matrix.

If anyone could help me understand this line of thought my problem would be solved.

Why does this allow us to " limit our discussion to the case where $v^{a \dot b}$ is a Hermitian $2 \times 2$ matrix"?

I understand that our representation here acts on complex $2\times2$ matrices. But I don't understand why we can restrict to hermitian matrices.