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A matrix $B$ is called a submatrix of $A$ if $B$ is obtained from $A$ by deleting some rows and columns.

Let $A$ be an $m \times n $ matrix with entries $0,1$ or $-1$ such that $1$ and $-1$ appears exactly once in each column of $A$. Let $B$ be a square submatrix of $A$. Show that $B$ has determinant $0,1$ or $-1$.

I want to show that this proposition is true...

For proving, I used Cofactor method..... However, I don't know how to express every $A$'s cases...

Any help is appreciated... I want to get some motivation about this problem..

Thank you!

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You can prove this by induction:

(1) It is clear for submatrix $B$ of dimension $1\times1$

(2) Suppose the claim is true for all submatrices of size $k\times k$. Now take a submatrix $B$ of size $(k+1) \times (k+1)$.

(2a) If $B$ has a zero column, its determinant is zero. If $B$ has a column with exactly one non-zero entry, we can develop its determinant with respect to this column, apply induction assumption, done.

(2b) It remains to consider matrices $B$, where each column contains both values $+1$ and $-1$. Let $e$ denote the vector with all entries equal to one. Then $B^Te=0$, hence $B$ has not full rank, and its determinant is zero.

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  • $\begingroup$ Could you explain about $B^T e=0$??? $\endgroup$ – bluejyellow Dec 15 '17 at 19:30
  • $\begingroup$ sum of all elements in each column of $B$ is zero, which means $e^TB=0$ or equivalently $B^Te=0$ $\endgroup$ – daw Dec 15 '17 at 21:30
  • $\begingroup$ If B has a column containining 1,1,-1, then we don't know $B^Te =0$... $\endgroup$ – bluejyellow Dec 16 '17 at 9:29
  • $\begingroup$ This could not happen as $A$ has at most one $1$ and one $-1$ per column $\endgroup$ – daw Dec 16 '17 at 13:25

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