rank($A$)=rank($A^T$) Is there an elementary explanation of why the row-rank of a matrix equals its column-rank (without using adjoint maps, resp. lots of technical computations)? What is the geometric intuition behind this?
 A: Since you seem to want a "geometrical" proof: the transpose of a matrix $A \in M_{m,n}(\mathbb{R})$ correspond to the adjoint of a morphism $u : \mathbb{R}^n \to \mathbb{R}^m$ for the standard inner products on $\mathbb{R}^n$ and $\mathbb{R}^m$. In other words:
$$\langle Ax, y \rangle = \langle x, A^T y \rangle$$
Adjoint morphisms have various properties; for example, $\operatorname{ker}(A^T) = \operatorname{im}(A)^\perp$ is supplementary to $\operatorname{im}(A)$. The rank-nullity theorem then yields $\operatorname{rk}(A) = \operatorname{rk}(A^T)$.
A: I don't know if you consider working with vectors and linear forms too technical, but some form of duality seems inevitable if you want a geometric interpretation of the transpose.
Given a $n\times m$ matrix$~M$, let it operate on column vectors of size $m$ by left multiplication as usual; the rank is the dimension of the image of this linear map$~f$ (this is the column rank, since images are linear combinations of the columns of$~M$). If instead one takes the operation$~g$ of right multiplication by$~M$ on row vectors of size$~n$, then the matrix of$~g$ is the transpose $M^\top$, and we must argue that the dimension of the image of this map (which is the row rank of$~M$, as images by$~g$ are linear combinations of the rows of$~M$) is the same as the column rank of$~M$.
If $r$ is the dimension of the image space$~W\subseteq K^n$ of$~f$, then we can find $r$ vectors $b_1,\ldots,b_r\in K^m$ such that $f(b_1),\ldots,f(b_r)$ form a basis of $W$. The coordinate functions for this basis form $r$ linearly independent linear forms $W\to K$, which can be extended to linear forms $\alpha_1,\ldots,\alpha_r$ defined on$~K^n$ (row vectors). Then by construction $\alpha_i(f(b_j))=\delta_{i,j}$ for $i,j\in\{1,\ldots,r\}$; moreover $r$ is the largest values for which this can be achieved (for any choice of the $\alpha_i$ and $b_j$). The linear forms $x\mapsto\alpha_i(f(x))$ on $K^m$ are the images$~g(\alpha_i)$, so we have also found column vectors $b_1,\ldots,b_r$ for which their evaluations, as functions on the image space of$~g$, form a maximal linearly independent set of linear forms on that space; therefore $r$ is also the rank of$~g$.
In algebraic terms, if $A$ is the matrix with rows $\alpha_1,\ldots,\alpha_r$, and $B$ is the matrix with columns $b_1,\ldots,b_r$, then $AMB=I_r$ and this is the largest identity matrix that can be so obtained. Transposing the equation we see that the maximum size is the same as for $M^\top$.
A: If you row reduce a matrix $A$ to RREF, the number of pivots (leading ones) is the rank. On the other hand, the rank theorem tells you that the column vectors of the original matrix corresponding to those pivots form a basis of the column space of the matrix. That is, the number of linearly independent columns is equal to the number of pivots. 
But the number of linearly independent columns of $A$ is the number of pivots you get by bringing $A^\top$ to RREF, and hence the rank of $A^\top$.
So rank$(A) = $rank$(A^\top)$.
A: The row-rank is equal to the dimension of the subspace created by the row-vectors. If you apply Gauss elimination you will see that the number of linearly independent vectors remains the same after transposition.
