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?



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


