1

This is my first question of SO so sorry if I made a mistake somewhere.

I have this matrices problem which I'm not able to solve. I asked many of my friends but they just don't get the question. This is the question:

$\text{Let n} \geq 3. \text{Let A} = ((a_{ij}))_{1 \leq i, j \leq n} \text{be an n} \times \text{n matrix such that a}_{ij} \in \{1, -1\} \text{for all 0} \leq i, j \leq n. \text{Suppose a}_{k1} = \text{1 for all 1} \leq k \leq \text{n. } \\ \text{And} \sum_{k = 1}^{n} a_{ki}\cdot a_{kj}\text{= 0. For all i} \neq j.\\ \text{Show that n is a multiple of 4.}$

(sorry I can't add an image and I'm not so good at latex too)

I think I've understood the summation part: when k = 1, the argument of summation becomes $a_{1i} \cdot a_{1j}$. Which is the sum of product of elements of 1st row taken 2 at a time. (that's what I interpreted $a_{1i} \cdot a_{1j}$ as. Is this right?) I'm not able to utilize the summation to get an equation in n.

Thanx in advance!

YashKandalkar
  • 13

1 Answers1

1

Clearly $n$ can't be odd as the inner product of the first column, which solely consists of ones, must be zero. Hence in every other column we must have an equal number of ones and minus ones, let's call that number $m$. (That is $n=2m$.)

Now consider two column except the first one. Have a look at the rows in the first column with element one (there are $m$ such rows). Let the number of ones in the corresponding rows in the second column be $k$, clearly $0\leq k\leq m$. Their product equals obviously $k-(m-k)=2k-m$. Similarly, the sum of the product of the other rows is $2k-m$ as well.

So in total the sum of all products is $4k-2m=2(2k-m)$. As that sum is zero, $m$ must be even, hence $n$ is divisible by $4$.

Michael Hoppe
  • 18,614