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.

I was wondering if the following lemma was easy to prove. I got a little tripped up when I saw the explicit condition that the matrix could have elements in any field not necessarily finite, so I didn't know if a standard proof from linear algebra would still apply.

Let $M$ be a matrix with entries in a (possibly infinite) field $F$.

Suppose that there exists a minor $m_n$ of order $n$ for $M$ that is nonzero and such that all minors of order $n+1$ which contain $m_n$ are zero.

How do you show the rank of the matrix is $n$?

share|improve this question
add comment

1 Answer 1

up vote 6 down vote accepted

That there exists a minor of order $n$ which is non-zero suggests that there exists at least $n$ linearly independent rows/columns of the matrix (namely the rows/columns of the submatrix corresponding to the minor). The rank is thus at least $n$. If the rank is larger than $n$ then there exists at least $n+1$ linearly independent rows, meaning the $n+1$ order submatrix taken using those rows will be invertible and have non-zero determinant contrary to the fact that the all $n+1$ order minors are zero. This shows that the rank must be $n$.

Edit: As Jyrki pointed out, only the minors which contain $m_n$ are zero. So let us proceed another way. The whole matrix cannot have full rank since it must contain $m_n$. That means there is at least one linearly dependent row and column which is not in $m_n$ since all rows and columns in $m_n$ are linearly independent. Removing the row and the column produces a smaller matrix which again contains $m_n$. We can repeat this procedure until all that remains is $m_n$, proving the rank is $n$.

share|improve this answer
    
Well... there is this minor detail that the assumption only said that those minors of size $n+1$ that contain the given minor of size $n$ vanish. So how do you rule out the possibility that there might be a non-vanishing minor of size $n+1$ that does not contain the given minor ;-> –  Jyrki Lahtonen Oct 26 '11 at 7:26
    
@Jyrki Thanks for pointing that out. I have provided an alternative argument. Please see if this argument is sufficient. –  EuYu Oct 26 '11 at 8:10
    
Much improved, but it sounds like you are assuming that the matrix is square? –  Jyrki Lahtonen Oct 26 '11 at 9:01
    
I don't think I ever assumed that. The wording on the last sentence is perhaps a little confusing. We can keep removing columns/rows until we exhaust one or the other. At that the point because $m_n$ contains either all rows or all columns of the matrix, the rank has to be $n$. –  EuYu Oct 26 '11 at 13:58
1  
As an example, suppose we have a $M\times N$ matrix with $N > M$. Then we would begin by taking the largest possible minor, a $M \times M$ minor which contains $m_n$. Then this minor has a linearly dependent column which we can remove to produce a $M \times N-1$ matrix. This proceed is repeated until we have just a $M \times M$ matrix, whereby we can start removing columns and rows until all we're left with is $m_n$. –  EuYu Oct 26 '11 at 14:08
show 1 more comment

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.