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Dec 13, 2018 at 20:38 vote accept Alice
Dec 13, 2018 at 1:35 comment added David E Speyer @user44191 It's a nontrivial extension, not a direct sum. The diagonal zero symmetric matrices are a subspace and the "diagonal only" are a quotient.
Dec 12, 2018 at 23:59 comment added spin Perhaps it is better to express this in terms of vector spaces: $X$ corresponds to a symmetric bilinear form $b$ on a vector space $V$ over $\mathbb{F}_2$, and $X_0$ corresponds to the linear map $v \mapsto b(v,v)$. Further $A^tXA$ corresponds to the bilinear form $b'$ defined by $b'(v,w) = b(Av, Aw)$, with the associated linear map $v \mapsto b'(v,v)$ etc. Looking at it this way, the equality seems clear.
Dec 12, 2018 at 23:44 answer added dhy timeline score: 6
Dec 12, 2018 at 23:27 history edited user44191
Retagged to the specific characteristic, added rt for reasons explained in comment
Dec 12, 2018 at 23:27 comment added user44191 This likely can be seen as a statement about the representation theory of $G = SL_n(\mathbb{F}_2)$ over $\mathbb{F}_2$. If $\rho$ is the obvious representation, then this says that the symmetric component of $\rho \otimes \rho^t$ further decomposes to a diagonal-only component and a zero-diagonal component.
Dec 12, 2018 at 22:10 history edited Alice CC BY-SA 4.0
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Dec 12, 2018 at 21:40 review First posts
Dec 12, 2018 at 21:48
Dec 12, 2018 at 21:37 history asked Alice CC BY-SA 4.0