I'm teaching linear algebra to first year students, and I was recently asked why is the rank of a matrix, representing a linear application in a given basis, equal to the dimension of the image space of this application.
If I think in terms of columns it is easy to see that applying it to the canonical basis vectors gives me a span of the image where every vector has for coordinates one column of the first matrix, thus the dimension of the image is the number of linearly independent columns.
But is there an equivalent short argument for rows? I mean without using the fact that the rank is invariant by applying transposition, and possibly without introducing a dual space which could make things only more confusing for my students.
Best Answer
The following intuitive explanation can be given.
Each row of the matrix $A$ of a linear map $T:V\to W$ describes one coordinate, with respect to the chosen basis of$~W$, of the images $T(v)$ of vectors in$~V$. (This is a linear form on $V$, hence an element of $V^*$, but there is no need to stress that point.) If one row is a linear combination of other rows, then this information allows reconstructing the corresponding coordinate from those other coordinates, given only the knowledge that $T(v)$ belongs to the image subspace $\def\Im{\operatorname{Im}}\Im(T)\subseteq W$. Therefore if one selects a maximal independent subset of $r$ rows of $A$, the selection defines a subset of $r$ coordinates, such that for $w\in\Im(T)$ all other coordinates of$~w$ can be recovered from those coordinates of$~w$ (by forming appropriate linear combinations).
It should be fairly intuitive that for a subspace of dimension $d$, it requires knowing $d$ (properly chosen) coordinates of a vector of the subspace to know exactly which vector it is. Formally this can be shown as follows. Our $r$ coordinates define a projection $p:\def\R{\Bbb R}W\to\R^r$, and our reconstruction of coordinates defines a linear map $s:\R^r\to W$ (keeping the given coordinates and reconstructing the remaining ones) such that $s(p(w))=w$ for all $w\in\Im(T)$ (we don't care what happens to elements outside $\Im(T)$). Thus the restriction of $p$ to $\Im(T)$ is certainly injective, and if we can show it to be surjective then $r=\dim(\Im(T))$, which is $\operatorname{rk}(T)$, will follow. But if it were not surjective, then the would be at least one nontrivial equation satisfied by all elements of $p(\Im(T))\subset\R^r$, a relation contradicting the independence of our set of $r$ chosen coordinates.