# Diagonalization of positive definite symmetric integer matrix

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.

• Not just the permutation matrices, but the signed permutation matrices (which allow -1s as well). They are precisely the ones arising from K. Miller's argument. – Andreea M Oct 10 '16 at 16:11
• Whoops, that comment is very redundant. – Omnomnomnom Oct 10 '16 at 16:13
• No worries, much appreciated anyway. – Andreea M Oct 10 '16 at 16:15

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 $D$.

• I'm not sure how this ties in with the question, which is precisely whether and when I can assume that the matrix $Q$ is integral, given that $G$ is integral, symmetric and positive definite. – Andreea M Oct 10 '16 at 15:52
• Well, you now know the only possible form for an integral orthogonal matrix. So if you assume that G is integral and Q is integral, then you can find D. Once you have found D you know all possible integral spd matrices G for which Q is also integral. – K. Miller Oct 10 '16 at 15:55
• So, in other words, $Q$ will only be integral if $G$ was already diagonal. – Omnomnomnom Oct 10 '16 at 16:14
• Yes, I believe so. – K. Miller Oct 10 '16 at 16:16

The main thing is that there is a rational matrix $P$ of determinant $1$ such that $P^T G P = D$ is diagonal with rational entries. This $P$ is not orthogonal, but is very easy to find.

Given that you mention integral equivalence, you ought to look up the concept of genus of (integral) quadratic forms.

• Thanks, Jay. Would you be able to provide some references for the existence of such a $P$? I am trying to solve a multi-dimensional integral of a vector-variable $X$, involving bilinear forms, by changing coordinates to $QX$ and using a diagonal bilinear form instead. This allows me to separate the integral into a product of one-dimensional integrals over all the individual coordinates of $X$, making it more manageable. – Andreea M Oct 10 '16 at 16:55
• @AndreeaM see math.stackexchange.com/questions/1388421/… – Will Jagy Oct 10 '16 at 17:38