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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.

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

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