For matrix Lie groups, the exponential map is usually defined as a mapping from the set of $n \times n$ matrices onto itself.
However, sometimes it is useful to have a minimal parametrization of our Lie algebra elements. Therefore, there is the hat operator $\hat{\cdot}$ which maps a $m$-vector onto the corresponding Lie algebra element
$ \hat{\cdot}: \mathbb R^m \rightarrow g, \quad \hat{\mathbf x} = \sum_{i=0}^m x_i \mathtt G_i, $
where $\mathtt G_i$ is called the $i$th generator of the matrix Lie algebra $g\subset \mathtt R^{n\times n}$ (especially by physicists). I am very much interested in the properties of this function.
Example for SO3: $\hat{\mathbf x} = \begin{bmatrix} 0&-x_3& x_2\\ x_3&0,&-x_1\\-x_2&x_1&0\end{bmatrix}$
Question 1: Does the function $\hat{\cdot}$ has a common name? Would "change of basis" transformation (from the standard basis to the Lie algebra basis) be the right name?
I guess in this context it make sense to assume that the family $\mathtt G_1,...,\mathtt G_m$ is linear independent. From the linear independence it follows that $\hat{\cdot}$ is bijective, right? Thus, there is an inverse function $v$ which maps a matrix Lie algebra onto the corresponding Lie algebra:
$ v(\cdot): g \rightarrow \mathbb R^m$ Example for SO3: $v(\mathtt R)= \begin{bmatrix}R_{3,2}\\R_{1,3}\\R_{2,1}\end{bmatrix}=\frac{1}{2} \begin{bmatrix}R_{3,2}-R_{2,3}\\R_{1,3}-R_{3,1}\\R_{2,1}-R_{1,2}\end{bmatrix} = -\begin{bmatrix}R_{2,3}\\R_{3,1}\\R_{1,2}\end{bmatrix} $
Question 2: Is there a closed form solution to write down/calculate the inverse of $\hat{\cdot}$?
I find it a bit confusing that the standard basis vectors are element of $\mathbb R^m$ while $\mathtt G_i$ are element of $\mathbb R^{n^2}$.
(Edit: Heavy changes after I developed a better understanding of the problem.)
(Edit 2: Replaced $[\cdot]$ by the common hat notation $\hat{\cdot}$.)