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Let $F$ be a field and let $A\in M_n(F)$ be a matrix with $det(A) = \pm 1 $. How can I show that $A$ is a product of involutions ? Of course the converse is true and clear.

By involution I mean a matrix whose square is the identity matrix.

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  • $\begingroup$ cf. Linear Algebra and its Applications Volume 13, Issues 1–2, 1976, Pages 157–162, product of involutions. sciencedirect.com/science/article/pii/0024379576900549 $\endgroup$ – loup blanc Feb 14 '16 at 20:44
  • $\begingroup$ @loupblanc Thanks. In fact they proved more, It is proved that any such matrix is a product of at most four involutions and the number four is sharp. The fact that I proposed the problem here is that I am looking for some elementary proof. And I do not mind if someone can prove a weaker form that (over any field) any such matrix is a product of a finite number of involutions. Sampson has proved this for real number field in 1974. $\endgroup$ – T.KM Feb 14 '16 at 21:14
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This looks trivial. Using elementary row/column operations, one can always decompose a square matrix into a product of the form $PDQ$, where $P$ and $Q$ are products of shear matrices and row/column exchange matrices, and $D$ is a diagonal matrix. Clearly, every row/column exchange matrix is an involution. And every shear matrix is a product of two involutions: $$ \pmatrix{1&x\\ 0&1}=\pmatrix{1&-x\\ 0&-1}\pmatrix{1&0\\ 0&-1}. $$ So, we only need to show that every diagonal matrix with determinant $\pm 1$ is a product of involutions. It's easy to see that $D=D_0\prod_{j=1}^{n-1} D_j$, where $D_0$ is either the identity matrix or $\operatorname{diag}(1,\ldots,1,-1)$ (hence $D_0$ is an involution) and every other $D_j$ is a diagonal matrix of determinant $1$ with at most two diagonal entries (that are reciprocal to each other) unequal to $1$. Yet each $D_j$ is a product of two involutions because $$ \pmatrix{x&0\\ 0&1/x}=\pmatrix{0&x\\ 1/x&0}\pmatrix{0&1\\ 1&0}. $$

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