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Matric adding up to zero

 
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Dec6-09, 02:19 PM   #1
 

Matric adding up to zero


Suppose we put numbers (real numbers, not just integers) in a matrix like this:

a11 a12 a13 ... a1h
a21 a22 a23 ... a2h
.
.
.
an1 an2 an3 ... anh

Now this scheme of numbers we know that

a11 + a12 + a13 + ... + a1h = 0
a21 + a22 + a23 + ... + a2h = 0

For all rows and for all columns:
a11 + a21 + a31 + ... + an1 = 0
a12 + a22 + a32 + ... + an2 = 0

How many numbers can we delete from the scheme and still be able to know what the deleted numbers should be? For instance:

A = [ 1 -1 ; -1 1]

This matrix has this property. We can delete the a22 and have:

A = [1 -1 ; -1 a22] and it is obvious that a22 = 1, now delete one more and:
A = [1 a12 ; -1 a22]

Now we know that:

1 + a12 = 0
a12 + a22 = 0

Så we know that a12 = -1 and a22 = 1, so all information is retained if we delete these two numbers. Now delete one more:

A = [a11 a12 ; -1 a22] and we get:

a11 + a12 = 0
a11 + (-1) = 0
-1 + a22 = 0
a12 + a22 = 0

And we see a11 = 1, a22 = 1, a12 = -1 so the matrix is

A = [1 -1 ; -1 1]

And no information is lost. But if we delete the remain number we have:

A = [a11 a12 ; a21 a22] and there are a lot of matrices that have this property. For instance:

C * [1 -1 ; -1 1] where C is any constant will do. So for this matrix we could delete 1, 2 and 3 numbers. If we denote the maximum number of numbers we can delete by rm(A) (for "remove A"), what is the maximum of rm(A) given that A is a matrix that has these properties? For this matrix rm(A) = 3, it should of course depend on the size, so:

rm(A) = some function of h and n.

And I figure (but this is a wild guess) that it doesn't depend on what matrix A it is, just it's size. Can we give a formula for rm(A) in terms of h and n? For instance if A is a column/row matrix

A = [a1 a2 a3 ... an] if we delete one number we can still find it, but not if we delete two numnbers so rm(A) = 1 for a column or row matrix. But can we find a general formula for rm(A)?
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