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.

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

   

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