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Let $A, B$ be two orthogonal matrices over a field $F$ of characteristic $2$ such that $$\det (A) + \det (B) = 0.$$ Is $(A+B)$ necessarily a singular matrix?

I have proved the result to be true for real matrices anf the result also holds for complex matrices. I even proved the result over any field of characteristic $\neq 2.$ Can it hold for matrices over a field of characteristic $2$?

I am asking this question because at the fag end of the proof of this result for real matrices I got a relation $2 \det (A + B) = 0,$ since $2 \neq 0$ over $\Bbb R$ we have the required result. But for any field $F$ of characteristic $2$ we have $2 = 0$ and hence we can't say whether or not $\det (A+B) = 0$ so that $(A+B)$ is a singular matrix.

Any help or suggestion in this regard will be highly appreciated. Thanks in advance.

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Here's a partial answer, namely specifically for $F=\Bbb F_2$: Here, $A^TA=I$ means that every column of $A$ has an odd number of $1$-entries. That is, with $v=\sum_ie_i$ as the all-1-vector, we have $v^TAe_i=1$ for all base vectors $e_i$. Same for $B$. But then $v^T(A+B)e_i=0$ for all base vectors, i.e., the image of $A+B$ is a proper subspace, meaning $A+B$ is singular.

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  • $\begingroup$ Very nice explanation @Hagen von Eitzen. Can we generalize this result for any finite field of characteristic $2$? $\endgroup$ Jul 11, 2020 at 19:04
  • $\begingroup$ For a field of characteristic $2$ we have $v^T A e_i = 1.$ What is $v^t A e_i$? This is nothing but the sum of the entries of the i-th column of $A.$ Am I right? What we know is that since $A^t A = I$ so sum of the squares of the entries in the $i$-th column is $1.$ But that means the sum of the entries of the $i$-th column is also equal to $1.$ $\endgroup$ Jul 12, 2020 at 5:52
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    $\begingroup$ Let us invoke the characteristic criterion of the underlying field. Let $a_{ki}$'s be the entries of $A$ in the $i$-th column. Then we know that $$\sum_k a_{ki}^2 = 1 \implies \left (\sum_k a_{ki} \right )^2 = 1.$$ But this in turn implies $$\sum_k a_{ki} = 1.$$ This follows from the fact that the identity $x^2 + 1 = (x+1)^2$ over a field of characteristic $2.$ $\endgroup$ Jul 12, 2020 at 5:56
  • $\begingroup$ @mathmaniac. I think you should flesh that out to an answer :-) $\endgroup$ Jul 12, 2020 at 8:38
  • $\begingroup$ @Jyrki Lahtonen sir have I done any mistake? Please let me know sir. $\endgroup$ Jul 12, 2020 at 8:58

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