# Show that all of these numbers are zero

• Dec 17th 2009, 08:35 PM
Bruno J.
Show that all of these numbers are zero
Suppose you have $\displaystyle 2n$ numbers $\displaystyle (n > 1)$ which have the following property : whenever you remove one of them (any one), you can split the remaining ones into two sets having equal sum.

Show that all of these numbers are zero.
• Dec 17th 2009, 08:42 PM
NonCommAlg
Quote:

Originally Posted by Bruno J.
Suppose you have $\displaystyle 2n-1$ numbers $\displaystyle (n > 1)$ which have the following property : whenever you remove one of them (any one), you can split the remaining ones into two sets having equal sum.

Show that all of these numbers are zero.

by "zero" did you mean "equal"? also, each of those two sets with equal sum must have exactly $\displaystyle n-1$ elements. otherwise, the claim would be false.
• Dec 17th 2009, 09:33 PM
Bruno J.
Oh, yeah. I'm sorry! I made a mistake in the statement of the problem! I have fixed my post. (Giggle)
• Dec 17th 2009, 11:23 PM
NonCommAlg
Quote:

Originally Posted by Bruno J.
Suppose you have $\displaystyle 2n$ numbers $\displaystyle (n > 1)$ which have the following property : whenever you remove one of them (any one), you can split the remaining ones into two sets having equal sum.

Show that all of these numbers are zero.

a weaker property: whenever you remove one of them (any one), either the sum of the remaining ones is zero or you can split the remaining ones into two sets having equal sum.

this problem is equivalen to this claim that the $\displaystyle 2n \times 2n$ matrix $\displaystyle A=[a_{ij}]$ with $\displaystyle a_{ii}=0, \ a_{ij}=\pm 1, \ \forall i \neq j,$ is invertible. to prove this claim we'll show that $\displaystyle \det A \neq 0$:

$\displaystyle \det A = \sum_{\sigma \in S_{2n}} \text{sign}(\sigma) \prod_{i=1}^{2n} a_{i \sigma(i)}=\sum_{\sigma \in D} \text{sign}(\sigma) \prod_{i=1}^{2n} a_{i \sigma(i)},$ where $\displaystyle D = \{\sigma \in S_{2n}: \ \ \sigma(i) \neq i, \ \forall i \},$ because we're given that $\displaystyle a_{ii}=0.$ so $\displaystyle D$ is the set of derangements of $\displaystyle \{1,2, \cdots , 2n \}.$

but we know that the number of derangements of a set with even number of elements is odd. so $\displaystyle \det A$ is a sum of odd number of terms where each term is $\displaystyle \pm 1.$ clearly this sum can

never be zero and hence $\displaystyle A$ is invertible. Q.E.D.