# Rank of a matrix and dimension of the image

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.

• You want to show by an elementary argument that the rank of a matrix is the same as that of its transpose, without using (or proving) that the rank of a matrix is the same as that of its transpose? Seems a tough task. Oct 22, 2014 at 9:48
• It would also be good to mention in the question exactly which definitions of rank (of a linear map and of a matrix) you are using. I suppose it is "dimension of the image" respectively "maximal number of independent columns". And what you want to show is that the rank of a linear map is also the maximal number of independent rows of any matrix representing it (i.e., the rank of the transpose matrix)? Oct 22, 2014 at 9:54
• Yes, sorry if I wasn't clear. Oct 22, 2014 at 11:00

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.

I find the explaination here pretty useful, it might not be doable in class, but a self-learner can find some use of it. Here is the copy

In a way, these are tautological, but it’s worth spelling out the tautology. Denote $$𝐴=[𝑎_{𝑖𝑗}]^{𝑚,𝑛}_{𝑖,𝑗=1}$$, so its column vectors are $$𝐜_𝐣=(𝑎_{1𝑗},…,𝑎_{𝑚𝑗})$$, or in more confusing but concise notation, $$𝐜_𝐣=[𝑎_{𝑖𝑗}]^𝑚_{𝑖=1}$$. The definition of the product of this matrix with a vector $$𝐯=[𝑣_𝑗]^𝑛_{𝑗=1}$$ is:

$$𝐴𝐯=[𝑏_𝑖]^𝑚_{𝑖=1},𝑏_𝑖=\sum_{j=1}^𝑛a_{𝑖𝑗}𝑣_𝑗$$

This has a nice interpretation in terms of linear combinations: the matrix product is a linear combination of the columns of the matrix, where the coefficients of the combination are the entries of the input vector. That is,

$$𝐴𝐯=\sum^𝑛_{𝑗=1}𝑣_𝑗𝐜_𝐣$$

Since the image of 𝐴 is, by definition, the set of all possible values of the left-hand side, and since the $$𝐜_𝐣$$ are the same in all of these expressions regardless of what $$𝐯$$ is, every element of the image is a linear combination of the $$𝐜𝐣$$; conversely, for any linear combination $$𝐰=\sum^𝑛_{𝑗=1}𝑢_𝑗𝐜_𝑗$$, we can write $$𝐰=𝐴𝐮$$, where $$𝐮=(𝑢_1,…,𝑢_𝑛)$$.

Therefore the image is exactly the set of all linear combinations of the matrix columns, i.e. their span.

For the implication about column rank, we have to agree on a definition of the column rank. I can’t think of one that is significantly different from “the dimension of the space spanned by the columns”, which is just a rephrasing of “the dimension of the image” in light of the previous point.