# Can a rational symmetric matrix $A$ be diagonalized as $A = P \Lambda S$ for some $P, S$ in the general linear group?

Let $$A$$ be a $$3 \times 3$$ symmetric matrix with rational entries. Does there exists a unique pair of matrices $$P, S \in \text{GL}_3(\mathbb{Z} )$$ depending on $$A$$ such that $$A = P \Lambda S$$, where $$\Lambda$$ is a diagonal matrix with rational entries?

In more details, I wonder whether, (if the answer is yes) this might be a consequence of the uniqueness of the Hermite normal form of $$A$$. Let $$M_1 A = H_1$$, where $$H_1$$ is upper triangular in Hermite normal form and $$M_1 \in \text{GL}_3(\mathbb{Z})$$. Then $$H_1^T = A M_1^T$$, where $$M_1^T$$ denotes the transpose of $$M_1$$. There exists a unique matrix $$M_2 \in \text{GL}_3(\mathbb{Z})$$ such that $$M_2 H_1^T = M_2 A M_1^T = H_2$$, where $$H_2$$ is upper triangular in Hermite normal form. We now have $$H_2^T = M_1 A M_2^T$$ and there exists a unique matrix $$M_3 \in \text{GL}_3(\mathbb{Z})$$ such that $$M_3 H_2^T = M_3 M_1 A M_2^T = H_3$$. Continuing in this manner, if $$k$$ is odd, then $$H_k = M_k M_{k-2} M_{k-4} \dots M_1 A M_2^T M_4^T \dots M_{k - 1}^T$$; if $$k$$ is even, then $$H_k = M_k M_{k-2} M_{k-4} \dots M_2 A M_1^T M_3^T \dots M_{k - 1}^T$$.

Question: Does this process terminate? That is, eventually, for some $$k \geq 1$$, do we have $$H_k = \Lambda$$, where $$\Lambda$$ is a diagonal matrix? Is any such result known? And if so, where can it be found? If the answer is no, then a counter-example would be appreciated.

• Google for Smith normal form. (And notice that you cannot guarantee uniqueness, as one can at least permute the diagonal elements of $\Lambda$) – Ilya Bogdanov Oct 31 '18 at 4:36