I know this post is old, but I just wanted to remark that it is possible to use matrix multiplication here, provided one is somewhat careful. The equation

$(\bar{\sigma}^\mu)^{\dot{\alpha}\alpha} = \epsilon^{\alpha\beta} \epsilon^{\dot{\alpha}\dot{\beta}} (\sigma^\mu)_{\beta\dot{\beta}} = -\epsilon^{\alpha\beta} (\sigma^\mu)_{\beta\dot{\beta}} \,\epsilon^{\beta \alpha}$

can be written using matrix multiplication as

$(\bar{\sigma}^\mu)^T = -\epsilon (\sigma^\mu) \epsilon$.

A short calculation shows that, if

$A = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$

then

$-\epsilon A \epsilon = \left( \begin{array}{cc} d & -c \\ -b & a \end{array} \right)$

whose transpose is

$\left( \begin{array}{cc} d & -b \\ -c & a \end{array} \right)$.

Thus $\bar{\sigma}^0 = I$ and $\bar{\sigma}^i = - \sigma^i$ using matrix notation.

I know this post is old, but I just wanted to remark that it is possible to use matrix multiplication here, provided one is somewhat careful. The equation

$(\bar{\sigma}^\mu)^{\dot{\alpha}\alpha} = \epsilon^{\alpha\beta} \epsilon^{\dot{\alpha}\dot{\beta}} (\sigma^\mu)_{\beta\dot{\beta}} = -\epsilon^{\alpha\beta} (\sigma^\mu)_{\beta\dot{\beta}} \,\epsilon^{\beta \alpha}$

can be written using matrix multiplication as

$(\bar{\sigma}^\mu)^T = -\epsilon (\sigma^\mu) \epsilon$

or, equivalently,

$\bar{\sigma}^\mu = -\epsilon (\sigma^\mu)^T \epsilon$

since $(A^T)^T = A$, $(ABC)^T = C^T B^T A^T$ and $\epsilon^T = -\epsilon$ ($\epsilon$ is skew-symmetric) for any square matrices $A$, $B$ and $C$ of the same size.

A short calculation shows that, if

$A = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$

then

$-\epsilon A^T \epsilon = \left( \begin{array}{cc} d & -b \\ -c & a \end{array} \right)$.

Thus $\bar{\sigma}^0 = I$ and $\bar{\sigma}^i = - \sigma^i$ using matrix notation.