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If $A$ is any real matrix, then $A^tA$ and $A$ have same rank.

This result is not necessarily true for other fields; one can see various nice comments on this here.

The difference comes from following thing:

If $(a_1,a_2,\ldots, a_k)$ is a vector with entries in the field $\mathbb{R}$, then the pointwise (dot) product of this vector with itself is $0$ only if the vector is $\mathbf{0}$. This is not true for vectors with complex entries, such as $(1,i)$.

I came to the following natural question:

Question: Are there non-real fields $F$, which have property that $$ \left(a_1^2+a_2^2+ \dots + a_k^2 =0\right) \Rightarrow \left(a_1=a_2=\dots = a_k=0\right) $$ for some $k\ge 2$, or for all $k\ge 2$.

user26857
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    https://en.wikipedia.org/wiki/Ordered_field And also note that when using complex vectors, it is much more common to consider the conjugate transpose rather than just the transpose. It behaves a lot more like the real transpose does, including respecting this result. – Arthur Jul 01 '21 at 06:09
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    For example, the field of rational functions with real coefficients. – Gerry Myerson Jul 01 '21 at 07:54
  • Not quite sure what you mean by "non-real fields", but obviously any field which embeds into $\mathbb R$ has this property. Maybe these "trivial" examples are exactly what you want to exclude, in which case (see other comments). Also, compare https://en.m.wikipedia.org/wiki/Stufe_(algebra). You are looking for precisely those fields whose "Stufe" is $\infty$. – Torsten Schoeneberg Jan 23 '22 at 20:48

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