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For an $m\times n$ matrix $A$, let $k$ be the number of variables $x_i$ in $\vec{x}=(x_1,x_2,\dots,x_n)$ for which $x_i$ must equal $0$ in the solution to $A\vec{x}=\vec{0}$. For instance the following matrix has $k=1$.

$$\begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$$

Is there anything known about $k$ (or even a name)? Is there any general technique to find $k$ for an arbitrary matrix? As far as I can see, $k\le \text{rank}(A)$ but I can't say much else beyond that.

A equivalent formulation might be, for a subspace $V\subseteq\mathbb{R}^n$, what is the maximum $k$ for which there exists a basis $B$ of $V$ such that $\{e_{i_1},e_{i_2},\dots,e_{i_k}\}\subseteq B$, where $e_{i_j}$ is a standard basis vector in $\mathbb{R}^n$. This reduces to the above when $V$ is the row space of $A$.

Frumple
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  • I take it you mean the minimum $k$? Perhaps check out http://math.stackexchange.com/questions/315259/sparse-basis-for-linear-subspace and http://mathoverflow.net/questions/179095/find-base-of-kernel-with-as-many-0-as-possible – Louis Deaett Dec 11 '15 at 14:57

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Actually, this idea does have a name. One way to think about this is that a solution gives you a linear dependence relations between the columns of $A$. The combinatorial study of these relations is formalized by matroid theory. In matroid theory terms, one of the entries in the solution vector must be 0 if the corresponding column in the matrix is a coloop in the matroid of columns in the matrix.

One useful observation: a column is a coloop if and only if deleting it causes the rank of the matrix to drop. That is, the last column will be a coloop if and only if when you row reduce to reduced echelon form, there's a pivot (i.e. a 1) in the last column. For other columns, you need to move them to the end (or row reduce, saving them for last).

Ben Webster
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  • In particular, the combinatorial study of such relations amounts to the study of representable matroids. – Louis Deaett Dec 11 '15 at 14:57
  • @LouisDeaett Right; I didn't mean to leave out non-representable matroids (well, actually often I do...); I changed the wording a little. – Ben Webster Dec 11 '15 at 19:42
  • I just wanted to point out that if you are interested only in dependence relations among columns of a matrix, then you do want to leave out the non-representable matroids! That is, the matroids that capture those relations are precisely the representable ones. But other matroids exist, and hence there is more to combinatorial linear algebra than just matroid theory. (Although, of course, they are intimately related.) – Louis Deaett Dec 11 '15 at 23:08