Diagonalizing matrices over $\mathbb{Z}/ p^k$

Question

Let $p$ be an odd prime and $A$ a symmetric matrix in $M_{n\times n}(\mathbb{Z}/ p^k)$. How does one prove there exists a matrix $M \in GL_n(\mathbb{Z}/ p^k)$ such that $M^tAM$ is diagonal?

I do not need a full proof, but rather some clue or idea of how to start.

I mean $\mathbb{Z}/p^k$. I will change the notation yo make it more clear. — Siegmeyer of Catarina, May 21 '23 at 10:23
(your definition of diagonalizable has a transpose, but normally it has an inverse there... My comment above is for the usual meaning of the word) — Mariano Suárez-Álvarez, May 21 '23 at 10:31
If we take $M=\begin{pmatrix}1 & 0 & 2\ 0 & 1 & 0\ 2 & 0 & 1\end{pmatrix}$ then $M^tAM=\begin{pmatrix}0 & 0 & 0\ 0 & 1 & 0\ 0 & 0 & 0\end{pmatrix}$, so it can be diagonalized. — Siegmeyer of Catarina, May 21 '23 at 10:41
@MarianoSuárez-Álvarez Why do we care about the definition with the inverse? I do not see the motivation behind your example — Siegmeyer of Catarina, May 21 '23 at 11:05
$A$ is called diagonalizable if there exists an invertible $M$ such that $MAM^{-1}$ is diagonal. You wrote something different. — Dietrich Burde, May 21 '23 at 11:26
@DietrichBurde The question is as intended, with the definition I give. I will eliminate the word "diagonalizable" — Siegmeyer of Catarina, May 21 '23 at 11:31

score 1 · Answer 1 · edited Jun 19 '23 at 22:26

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The question means "diagonalizable by congruence," not by similarity. Any square symmetric matrix over a field of characteristic $\ne$2 is diagonalizable by congruence. Here's an outline. The best reference I know is W. L. Ferrar's old book Algebra (Oxford U. P. 1957.)

Use the correspondence between symmetric matrices and quadratic forms. Successively apply linear transformations (equivalent to multiplying on the left and right by a by a non-singular matrix and its transpose) of the following types (i) if there is at least one diagonal term: new variable = linear combination of old variables to remove off-diagonal terms (ii) if there are some off-diagonal terms but no diagonal terms: old variable=new variable + another old variable. This transformation will force a diagonal term in the new variables. You can then apply another transformation of type (i). By applying these transformations successively, you can diagonalize the matrix by congruence.

edited Jun 19 '23 at 22:26

hbghlyj

2,115

answered May 21 '23 at 12:07

P. Lawrence

5,674

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@DietrichBurde My previous $M$ was indeed incorrect, but taking $M=\begin{pmatrix} 1 & 0 & 1\ 1 & 1 & 0\ 1 & 0 & 2\end{pmatrix}$ gives $M^tAM$ diagonal, and this $M$ is in $GL_n(\mathbb{Z}/p^k)$ – Siegmeyer of Catarina May 21 '23 at 12:43
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@SiegmeyerofCatarina Yes, now it is correct. I was thinking of "orthogonal diagonalization", but here $M^{-1}=M^T$ need not be true. – Dietrich Burde May 21 '23 at 12:47
@DietrichBurde Thank you both. This has been very helpful. – Siegmeyer of Catarina May 21 '23 at 12:50
By the way, does this method immediately work for $p^k$ with $k>1$? – Siegmeyer of Catarina May 21 '23 at 12:54
Ferrar's method works for any field of characteristic $\ne$2, in particular over $\mathbb Z/p^k$, whenever $p$ is an odd prime and $k \ge$ 1. – P. Lawrence May 22 '23 at 04:12

Diagonalizing matrices over $\mathbb{Z}/ p^k$

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