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Suppose we have a $m\times n$ matrix $A$ such that $\text{rank}(A) = n-1$ and the nullspace of $A$ is 1-dimensional and contains the all-ones vector, $e$. That is, $Ae=0$. I'd like to show that if we delete any column of $A$, the rank of the resulting matrix, call it $A^\prime$, still is $n-1$.

Note: $A$ is an oriented-incident matrix; i.e. each row contains precisely one 1 and one -1.

Strategy: Perhaps this can be shown by contradiction? That is, suppose $\text{rank}(A^\prime)=n-2$. Then this implies that the nullity of $A^\prime$ is 2. Then its nullspace must contain a vector with at least one zero-entry. I don't see how to arrive at the conclusion that the vector of all ones, $e$, cannot be in the nullspace of $A$ (hence the contradiction).

Any help/suggestions would be very much appreciated.

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1 Answer 1

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The sum of the columns of the matrix is zero. Hence every column lies in the span of the $n-1$ other columns. So if you delete any of them, you do not change the dimension of the space spanned by the remaining columns, which means that the resulting matrix has rank $n-1$.

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  • $\begingroup$ Thanks for the response. So my strategy is somewhat of an overkill? Also, would you consider your response to be complete (should I be writing it more formerly; i.e. using mathematical notation)? $\endgroup$
    – Dan P.
    Commented Feb 5, 2019 at 18:32
  • $\begingroup$ I would consider my response to be a complete Mathematical argument. As for submitting class-homework, I would probably increase dramatically some of the verbosity, which I leave to those who need to do so. $\endgroup$ Commented Feb 5, 2019 at 18:55

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