# Is $(A+B)$ necessarily singular?

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.

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.
• Very nice explanation @Hagen von Eitzen. Can we generalize this result for any finite field of characteristic $2$? Jul 11, 2020 at 19:04
• 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 $1.$ Jul 12, 2020 at 5:52
• 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 $2.$ Jul 12, 2020 at 5:56