# The Unapologetic Mathematician

## Column Rank

Let’s go back and consider a linear map $T:V\rightarrow W$. Remember that we defined its rank to be the dimension of its image. Let’s consider this a little more closely.

Any vector in the image of $T$ can be written as $T(v)$ for some vector $v\in V$. If we pick a basis $\left\{e_i\right\}$ of $V$, then we can write $T(v)=T(v^ie_i)=v^iT(e_i)$. Thus the vectors $T(e_i)\in W$ span the image of $T$. And thus they contain a basis for the image.

More specifically, we can get a basis for the image by throwing out some of these vectors until those that remain are linearly independent. The number that remain must be the dimension of the image — the rank — and so must be independent of which vectors we throw out. Looking back at the maximality property of a basis, we can state a new characterization of the rank: it is the cardinality of the largest linearly independent subset of $\left\{T(e_i)\right\}$.

Now let’s consider in particular a linear transformation $T:\mathbb{F}^m\rightarrow\mathbb{F}^n$. Remember that these spaces of column vectors come with built-in bases $\left\{e_i\right\}$ and $\left\{f_j\right\}$ (respectively), and we have a matrix $T(e_i)=t_i^jf_j$. For each index $i$, then, we have the column vector

$\displaystyle T(e_i)=\begin{pmatrix}t_i^1\\t_i^2\\\vdots\\t_i^n\end{pmatrix}$

appearing as a column in the matrix $\left(t_i^j\right)$.

So what is the rank of $T$? It’s the maximum number of linearly independent columns in the matrix of $T$. This quantity we will call the “column rank” of the matrix.

July 1, 2008 - Posted by | Algebra, Linear Algebra

1. [...] Yesterday we defined the column rank of a matrix to be the maximal number of linearly independent columns. Flipping over, we can consider the [...]

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2. [...] , and we’ve got column vectors to consider. If all are linearly independent, then the column rank of the matrix is . Then the dimension of the image of is , and thus is [...]

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