Show $se(3)$ is the tangent space of $SE(3)$ the matrix group of $3D$ rigid motions 
$se(3)=\begin{pmatrix}A&c\\0&0\\\end{pmatrix}:A\in so(3), c\in \mathbb{R}^3\}$


Show that $se(3)$ is the tangent space of $SE(3)=\begin{pmatrix}A&c\\0&1\\\end{pmatrix}:A\in SE(3), c\in\mathbb{R}^3\}$

$so(3)$ is the set of skew symmetric matrices.
I have a definition for the tangent space as $T_p(M)=\{X\in M:\exists \gamma:(-\epsilon,\epsilon)\to M\text{ which is differentiable and } \gamma(0)=p, \gamma'(0)=X\}$
But I'm not sure how I would show $se(3)$ would satisfy this for $SE(3)$
 A: I'll assume you meant $A\in{\rm SO}(3)$ in your definition of ${\rm SE}(3)$.
Anyway, fix an element
$$M = \begin{bmatrix}A_0 & c_0 \\ 0 & 1\end{bmatrix}$$
in ${\rm SE}(3)$.
Let $\gamma(t)$ be a differentiable path in ${\rm SE}(3)$ defined near $t=0$ with $\gamma(0)=M$ and $\gamma'(0) = X$ for some matrix $X$. We see that $\gamma$ must have the form
$$\gamma(t) = \begin{bmatrix}A(t) & c(t) \\ 0 & 1\end{bmatrix}$$
where $A(0)=A_0$ and $c(0)=0$. Thus
$$X = \gamma'(0) = \begin{bmatrix}A'(0) & c'(0) \\ 0 & 0\end{bmatrix}.$$
Since $A(t)$ was a differentiable path in ${\rm SO}(3)$, $A'(0)\in T_{A}{\rm SO}(3) = {\rm so}(3)$, the set of skew-symmetric matrices. The vector $c'(0)$ is an element of $\mathbb{R}^3$, so we conclude $X\in {\rm se}(3)$. Thus $T_M{\rm SE}(3)\subset {\rm se}(3)$. On the other hand, the dimension of ${\rm SE}(3)$ and hence its tangent space is $\dim{\rm SO}(3)+3 = 3+3=6$ while the dimension of ${\rm se}(3)$ is also $3+3=6$, so we get equality $T_M{\rm SE}(3) = {\rm se}(3)$.
