# Conceptual explanation for curious linear-algebra fact in characteristic $2$

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?

• 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. – user44191 Dec 12 '18 at 23:27
• 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. – spin Dec 12 '18 at 23:59
• @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. – David E Speyer Dec 13 '18 at 1:35

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.

• This is very helpful. Thanks! – Alice Dec 13 '18 at 20:38