Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given is an almost-square matrix $A$ with $n$ columns and $n-1$ rows with maximum rank. The solutions of the homogeneous system $Ax = 0$ form a 1-dimensional subspace of $\mathbb{R}^n$.

I've discovered the following which I believe to be true but I can't prove: the components of the vector $x$ that spans the (1D) solution space are given by:

$x_i = (-1)^{i-1} |A_i|$

in which $|A_i|$ is the determinant of the square submatrix of A obtained by removing the i-th column from A. For example, in $\mathbb{R^3}$, $A$ is a 2x3 matrix, and $x$ as defined above turns out to be the crossproduct of the two row vectors of $A$.

Is this true, and if so, how can it be proved?

share|improve this question
    
Thanks, that makes total sense... very neat! –  user6216 Jan 25 '11 at 20:09
    
I merged your two accounts. Registering will help mitigate this problem and allow you to properly comment on questions (instead of posting a comment as an answer). –  Willie Wong Jan 25 '11 at 20:12
    
You may also want to mark ulvi's answer as accepted by clicking the grey check mark to the left of the answer. –  Rahul Jan 25 '11 at 20:18

1 Answer 1

You only need to show that the $x$ you defined is orthogonal to every row of $A$. You can see that's the case by using the row-expansion of the determinant: given any row $j$ of $A$, consider the $n \times n$ matrix obtained from $A$ by duplicating that row. This new square matrix $A'$ has determinant $0$, and you can compute its determinant by expanding along the newly created duplicate row: $$ 0={\rm det} (A') = \sum_i (-1)^{j-1} (-1)^i A_{ji} |A_i | = (-1)^{j-1} \sum_i A_{ji} x_i$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.