What means $\mathbb{R}^{n}$ in the equation $\mathbb{R}^{n}$ = $\operatorname{Ran}(A)+ \operatorname{Ker}(A^T)$? What means $\mathbb{R}^{n}$ in the equation $\mathbb{R}^{n}=\operatorname{Ran}(A)+ \operatorname{Ker}(A^T)$? Is it the number of the columns of $A$ $\in \mathbb{R}^{m \times n}$? And what is the difference between row-rank and column-rank?
 A: $\Bbb R^n$ is the $n$-dimensional real vector space. The row space and columns space are the spans of the row vectors and column vectors of a matrix (respectively), but row rank and column rank (the dimensions of these spaces) are equal.
So, what this equation means is that every $n$-dimensional real row vector can be obtained as a linear combination of elements of the row space of $A$ and elements of the null space of $A^T.$ Had we been considering column vectors instead of row vectors, then we would have $\Bbb R^m,$ instead.
A: A better question is: what does the right hand side mean?
$\operatorname{Ran}(A)$ is the column space—the span of the columns of $A$.
$\operatorname{Ker}(A^T)$ is the null space of $A^T$.
$+$ means to add together all pairs of vectors one from each space: $V + W = \{v + w : v \in V, w \in W\}$.
For example, let $A = \begin{pmatrix}1 & 1 & 2 \\ 0 & 1 & -1 \\ 1 & 2 & 1\end{pmatrix}$. Then
$$\operatorname{Ran}(A) = \operatorname{span}\left\{ \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}, \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} \right\} = \operatorname{span}\left\{ \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \right\}$$
(Use row reduction to find a basis.)
$$\operatorname{Ker}(A^T) = \operatorname{span}\left\{ \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} \right\}$$
(Because I picked $A$ such that $(1,1,2) + (0,1,-1) = (1,2,1)$—you can also use row reduction to find this.)
And finally
\begin{align} \operatorname{Ran}(A) + \operatorname{Ker}(A^T) &= \operatorname{span}\left\{ \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \right\} + \operatorname{span}\left\{ \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} \right\} \\ &= \operatorname{span}\left\{ \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} \right\} = \mathbf{R}^3. \end{align}
