The Unapologetic Mathematician

Mathematics for the interested outsider

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.

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July 1, 2008 Posted by | Algebra, Linear Algebra | 2 Comments