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All matrices and vectors in this post have entries in the field $\mathbb{F}_2$.

Fix some $n \geq 1$. For an $n \times n$ matrix $X$, write $X_0$ for the column vector whose entries are the diagonal entries in $X$. The following curious fact arose in a paper I am writing:

Fact: Let $X$ be a symmetric $n \times n$ matrix and let $A$ be an arbitrary $n \times n$ matrix. Then $(AXA^t)_0 = A (X_0)$. Here $A (X_0)$ means that we are multiplying the column vector $X_0$ by $A$.

This is easy enough to prove in a grungy way, but to me it basically comes out of nowhere. Does anyone know a conceptual reason for it? Or perhaps a bigger picture in which it sits?

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    $\begingroup$ 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. $\endgroup$
    – user44191
    Commented Dec 12, 2018 at 23:27
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    $\begingroup$ 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. $\endgroup$
    – spin
    Commented Dec 12, 2018 at 23:59
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    $\begingroup$ @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. $\endgroup$
    – David E Speyer
    Commented Dec 13, 2018 at 1:35

1 Answer 1

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I think one way of explaining this is via quadratic forms. The usual correspondence sends $X$ to the quadratic form $X(v)=vXv^t,$ $v$ a row vector. But in characteristic $2$, this formula simplifies to

$$X(v)=(vX_0)^2.$$

Now examine what happens to $AXA^t$. For any characteristic, we get $AXA^t(v)=vAXA^tv^t=(vA)X(vA)^t=X(vA).$ In characteristic $2$, this implies that $$(v(AXA^t)_0)^2=(vA(X_0))^2,$$

so

$$(AXA^t)_0=A(X_0)$$

as desired.

You can also phrase this in a representation-theoretic language (following the suggestion of user44191). Take the algebraic group $\operatorname{GL}(n)$ over $\mathbb{Z}.$ It has representations $U$ and $V$ (again over $\mathbb{Z}$) corresponding to $n\times n$ symmetric matrices and quadratic forms on $\mathbb{Z}^n.$ The natural map $U\rightarrow V$ is an isomorphism once you localize away from $2$, but let us examine the map $U/2U\rightarrow V/2V.$ This map is our map $X\mapsto(vX_0)^2,$ which shows that $U/2U$ has the desired quotient.

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