A positive definite, symmetric $n\times n$ matrix $G$ can be diagonalized as $G=Q^TDQ$, where $Q$ is an orthogonal matrix (i.e. $QQ^T=I$) and $D$ is a diagonal matrix, say $D=Diag(a_1,a_2,\cdots,a_n)$.
I am interested in finding out some of the properties of the matrix $Q$ (whose columns are eigenvectors of $G$) in the case where $G$ is an integer matrix (i.e. its entries are integer numbers). In particular, is the matrix $Q$ integer itself?
Edit: This question arises in connection to positive definite symmetric bilinear forms and whether they are integrally equivalent to a diagonal bilinear forms.
Edit: I have been looking on line at properties of each of the individual characteristics of the matrix $G$ that I mentioned (integer, symmetric, positive-definite) in terms of eigenvalues (the $a_i$'s), which are precisely the diagonal entries of $D$. I can't find information on the eigenvectors of $G$, which form the columns of $Q$. I was hoping to derive properties of $Q$ from those of $D$.
What I have so far is:
1) the matrix is integral, so its eigenvalues are algebraic integers.
2) the matrix is symmetric, so its eigenvalues are real.
3) the matrix is positive definite, so its eigenvalues are positive.
I can't seem to find some stronger results if all three properties hold.
Best Answer
Here is a hint. Since the matrix $Q$ is orthogonal $Q^TQ = I$. Hence,
$$ 1 = \sum_{j=1}^n q_{ij}^2 $$
for all $i = 1,\ldots,n$. If $Q$ is integral, then it follows from this equation that $Q$ has exactly one nonzero element in each column which is equal to $\pm 1$. Moreover, since $Q$ has full rank each row contains exactly one nonzero element, otherwise $Q$ would have linearly dependent columns. Thus, $Q$ has the structure of a permutation matrix with nonzero elements in $\{-1,1\}$. Now assume that $G$ and $Q$ are integral and find $D$.