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If a 2x2 matrix has a zero determinant, why can we express it as an (outer) product of two vectors? I'm working on the spinor-helicity formalism, and am curious as to the rigorous mathematical proof behind this. Any direction to literature would be very useful!

Thank you!

EDIT: See page 10 of https://arxiv.org/pdf/1308.1697.pdf for the kind of thing I'm interested in.

Brad
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  • Any rank $1$ matrix can be expressed as an outer product of two non-null vectors. See here for example. – StubbornAtom Dec 30 '18 at 14:39
  • @StubbornAtom thank you for the link. Would you by chance know of any linear algebra textbooks which might have a rigorous proof of this? That link is along the lines that I'm thinking. Why does zero determinant mean that it is then rank 1? – Brad Dec 30 '18 at 14:46
  • Check the several linked posts to that question; the arguments are rigorous. I guess there are standard texts that have this result as an exercise even if it is not worked out. And zero determinant of an $n\times n$ matrix means the rank is less than $n$, which in your case means the rank is $0$ (for null matrix only) or $1$. – StubbornAtom Dec 30 '18 at 14:54
  • @StubbornAtom that last line was what I needed. Thank you very much! If you put that as an answer, I'll accept it. – Brad Dec 30 '18 at 14:55

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Thanks to @StubbornAtom's response, I found the answer that I was needing. Specifically relating to my project;

Since a zero determinant of any $n$ x $n$ matrix implies that the rank must be less than $n$, the rank for a 2x2 matrix must be 0 (null matrix) or 1. As a standard exercise in linear algebra, we can show that any rank-1 matrix may be written as the outer product of two vectors, a well-documented result in textbooks.

Brad
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