Let $A$ is a symmetric matrix with all diagonal element zero.

How to show that the dot product of $x$ with $Ax$ is divisible by $2$ ?

I can only verify for matrix of order (size) $2, 3, 4 $ but not sure about general $n×n$ matrix.

Edit: Here, $x$ is any vector in $\Bbb{Z}^n$.

If $A$ is real symmetric matrix with all diagonal elements $0$, then prove that $x^{T}Ax$ is divisible by $2$. ( Here $x$ is any vector in $\mathbb{Z}^{n}$, and $x^T$ denotes transpose of $x$ )

$2$nd edit: If $A$ is integer entry symmetric matrix with all diagonal elements $0$, then prove that $x^{T}Ax$ is divisible by $2$.

  • 1
    $\begingroup$ Your question needs mor information, such as: What ist $x$, what is $A$? In particular, are the entries of $A$ integers? What do you mean by "divisible"? Also, use MathJax to typeset mathematical content. $\endgroup$
    – Toffomat
    Aug 1, 2022 at 8:18
  • $\begingroup$ Sorry, I am posting for first time, not even sure how to type mathematics character, x is any vector in R^n $\endgroup$
    – noidea
    Aug 1, 2022 at 8:20
  • $\begingroup$ There's a mathjax tutorial here: math.meta.stackexchange.com/questions/5020/… $\endgroup$
    – Toffomat
    Aug 1, 2022 at 8:21
  • $\begingroup$ thank you, so much I got it. $\endgroup$
    – noidea
    Aug 1, 2022 at 8:23
  • 1
    $\begingroup$ Your specifications do have some problems. If $x\in\mathbb R^n$ the result of $\langle Ax, x\rangle$ must not even be integral. I believe you want to have $x\in\mathbb Z^n$ and $A\in\mathbb Z^{n\times n}$. $\endgroup$
    – Lazy
    Aug 1, 2022 at 9:07

1 Answer 1


Let $A=(a_{ij})$ be a symmetric matrix with $a_{ii}=0$.

Then by symmetry $a_{ij}=a_{ji}$

Let $Q_A:\Bbb{R}^n\to \Bbb{R}$ is the quadratic form associated with $A$.

$\begin{align}Q(x) =x^TAx&=\sum_{i,j}a_{ij}x_ix_j\\&=2\sum_{i\neq j, i>j}a_{ij}x_ix_j\end{align}$

If $A\in M_n(\Bbb{Z})$ and $x\in \Bbb{Z}^n$ , then clearly $2\mid Q(x) $

Let $A=\begin{pmatrix}0&{\frac{1}{2}}\\ {\frac{1}{2}}&0\end{pmatrix}$

Then $\begin{align}Q(x) =x^TAx&= \frac{1}{2}x_1x_2+\frac{1}{2}x_2x_1\\&=x_1x_2\end{align}$

Clearly $2\mid Q(x) $ for all $x\in \Bbb{R^2}\setminus \{0\}$ is false.

$A=\begin{pmatrix}0&1\\ {1}&0\end{pmatrix}$

Then $Q(x) =2x_1x_2$

Let $x=\begin{pmatrix}\frac{1}{2}\\\frac{1}{2}\end{pmatrix}$

Then $Q(x) =\frac{1}{2}$ and $2\nmid Q(x) $.

Hence $2\mid Q(x) $ if $a_{ij},x_i\in \Bbb{Z} $

  • $\begingroup$ Are you sure you have shown (in general) that $2|Q(x)$ implies that $a_{ij}, x_i\in\mathbb{Z}$? $\endgroup$
    – Thomas
    Aug 1, 2022 at 10:29
  • $\begingroup$ @Thomas My bad. I have shown if part. $\endgroup$ Aug 1, 2022 at 10:31

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