Definition 1: Given linear programming problem (LP) $\max\left\{\left \langle c,x\right \rangle| Ax=b, x \geq 0\right\}$. Then its dual is $\min\left\{\left \langle b,u \right \rangle | A^Tu \geq c\right\}$
Definition 2: Given (LP) $\max\left\{\left \langle c,x \right \rangle | A_1x=b_1, \,\, A_2x \leq b_2, \,\, x \geq 0\right\}$. Then its dual is $\min\left\{\left \langle \begin{pmatrix} b_1\\ b_2 \end{pmatrix}, \begin{pmatrix} v\\ w \end{pmatrix} \right \rangle | A_1^Tv + A_2^Tw \geq c, \,\, w \geq 0\right\}$
I'm trying to show that both definitions are equivalent but I don't know how to do it? If I look at both, the only slight difference seems to be that in Def 1 you have a whole matrix $A$ but in Def 2 you have several submatrices $A_1, A_2$ that will form one matrix together (probably the same one as in Def 1.
But really no idea how this can be "shown"..? : / Maybe it's sufficient to choose an example and apply both definitions on it and if you get the same result, it's equivalent? Or this is not enough to "show" something?
Given a maximization problem of the form in definition $1$. Let us try to convert it into a maximization problem in the second form, and check the duality form definition $2$ is the same as duality in definition $1$.
We let $A_1=A$, $b_1=b$, $A_2=0$, $b_2$ be any nonnegative vector, then the dual accoding to Definition $2$ is
$$\min\left\{\left \langle \begin{pmatrix} b_1\\ b_2 \end{pmatrix}, \begin{pmatrix} v\\ w \end{pmatrix} \right \rangle | A_1^Tv + 0^Tw \geq c, \,\, w \geq 0\right\}$$
Also recall that $b_2$ is nonnegative and letting $w$ taking any positive value would only increases the objective function of the minimization problem. Hence the optimal $w$ must take value $0$. Hence it reduces to the dual in the first definition.
Similarly, given a maximization problem of the form in definition $2$. Let us try to convert it into a maximization in the form of the first definition by introducing slack variable $s$.
$$\max\left\{\left \langle c,x \right \rangle | A_1x=b_1, \,\, A_2x \leq b_2, \,\, x \geq 0\right\}\\=\max\left\{\left \langle \begin{pmatrix}c\\0 \end{pmatrix},\begin{pmatrix}x \\s\end{pmatrix} \right \rangle | \begin{pmatrix}A_1 & 0 \\ A_2 & I\end{pmatrix}\begin{pmatrix}x \\ s\end{pmatrix}=\begin{pmatrix}b_1 \\ b_2 \end{pmatrix} \,\, \begin{pmatrix}x \\s\end{pmatrix} \geq 0\right\}$$
Now, we use the dual in the first definition, we obtain
$$\min\left\{\left \langle \begin{pmatrix}b_1\\b_2 \end{pmatrix},\begin{pmatrix}v \\w\end{pmatrix} \right \rangle | \begin{pmatrix}A_1^T & A_2^T \\ 0 & I\end{pmatrix}\begin{pmatrix}v \\ w\end{pmatrix}\ge\begin{pmatrix}c \\ 0 \end{pmatrix} \right\}$$
which is equivalent to the dual stated in definition $2$.