Zero Determinant Rank 1 (2x2) matrix being expressed as an outer product of two commuting vectors (spinors)
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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.
matrices determinant
asked Dec 30 '18 at 14:28
BradBrad
103
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add a comment |
$begingroup$
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.
matrices determinant
asked Dec 30 '18 at 14:28
BradBrad
103
$endgroup$
$begingroup$
Any rank $1$ matrix can be expressed as an outer product of two non-null vectors. See here for example.
$endgroup$
– StubbornAtom
Dec 30 '18 at 14:39
$begingroup$
@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?
$endgroup$
– Brad
Dec 30 '18 at 14:46
$begingroup$
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 $ntimes n$ matrix means the rank is less than $n$, which in your case means the rank is $0$ (for null matrix only) or $1$.
$endgroup$
– StubbornAtom
Dec 30 '18 at 14:54
$begingroup$
@StubbornAtom that last line was what I needed. Thank you very much! If you put that as an answer, I'll accept it.
$endgroup$
– Brad
Dec 30 '18 at 14:55
add a comment |
$begingroup$
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.
matrices determinant
asked Dec 30 '18 at 14:28
BradBrad
103
$endgroup$
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.
matrices determinant
matrices determinant
asked Dec 30 '18 at 14:28
BradBrad
103
asked Dec 30 '18 at 14:28
BradBrad
103
asked Dec 30 '18 at 14:28
BradBrad
103
asked Dec 30 '18 at 14:28
BradBrad
103
asked Dec 30 '18 at 14:28
BradBrad
103
103
$begingroup$
Any rank $1$ matrix can be expressed as an outer product of two non-null vectors. See here for example.
$endgroup$
– StubbornAtom
Dec 30 '18 at 14:39
$begingroup$
@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?
$endgroup$
– Brad
Dec 30 '18 at 14:46
$begingroup$
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 $ntimes n$ matrix means the rank is less than $n$, which in your case means the rank is $0$ (for null matrix only) or $1$.
$endgroup$
– StubbornAtom
Dec 30 '18 at 14:54
$begingroup$
@StubbornAtom that last line was what I needed. Thank you very much! If you put that as an answer, I'll accept it.
$endgroup$
– Brad
Dec 30 '18 at 14:55
add a comment |
$begingroup$
Any rank $1$ matrix can be expressed as an outer product of two non-null vectors. See here for example.
$endgroup$
– StubbornAtom
Dec 30 '18 at 14:39
$begingroup$
@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?
$endgroup$
– Brad
Dec 30 '18 at 14:46
$begingroup$
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 $ntimes n$ matrix means the rank is less than $n$, which in your case means the rank is $0$ (for null matrix only) or $1$.
$endgroup$
– StubbornAtom
Dec 30 '18 at 14:54
$begingroup$
@StubbornAtom that last line was what I needed. Thank you very much! If you put that as an answer, I'll accept it.
$endgroup$
– Brad
Dec 30 '18 at 14:55
$begingroup$
Any rank $1$ matrix can be expressed as an outer product of two non-null vectors. See here for example.
$endgroup$
– StubbornAtom
Dec 30 '18 at 14:39
$begingroup$
Any rank $1$ matrix can be expressed as an outer product of two non-null vectors. See here for example.
$endgroup$
– StubbornAtom
Dec 30 '18 at 14:39
$begingroup$
@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?
$endgroup$
– Brad
Dec 30 '18 at 14:46
$begingroup$
@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?
$endgroup$
– Brad
Dec 30 '18 at 14:46
$begingroup$
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 $ntimes n$ matrix means the rank is less than $n$, which in your case means the rank is $0$ (for null matrix only) or $1$.
$endgroup$
– StubbornAtom
Dec 30 '18 at 14:54
$begingroup$
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 $ntimes n$ matrix means the rank is less than $n$, which in your case means the rank is $0$ (for null matrix only) or $1$.
$endgroup$
– StubbornAtom
Dec 30 '18 at 14:54
$begingroup$
@StubbornAtom that last line was what I needed. Thank you very much! If you put that as an answer, I'll accept it.
$endgroup$
– Brad
Dec 30 '18 at 14:55
$begingroup$
@StubbornAtom that last line was what I needed. Thank you very much! If you put that as an answer, I'll accept it.
$endgroup$
– Brad
Dec 30 '18 at 14:55
add a comment |
1 Answer
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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.
answered Jan 14 at 16:48
BradBrad
103
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add a comment |
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$begingroup$
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.
answered Jan 14 at 16:48
BradBrad
103
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 14 at 16:48
BradBrad
103
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 14 at 16:48
BradBrad
103
$endgroup$
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.
answered Jan 14 at 16:48
BradBrad
103
answered Jan 14 at 16:48
BradBrad
103
answered Jan 14 at 16:48
BradBrad
103
answered Jan 14 at 16:48
BradBrad
103
103
add a comment |
add a comment |
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$begingroup$
Any rank $1$ matrix can be expressed as an outer product of two non-null vectors. See here for example.
$endgroup$
– StubbornAtom
Dec 30 '18 at 14:39
$begingroup$
@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?
$endgroup$
– Brad
Dec 30 '18 at 14:46
$begingroup$
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 $ntimes n$ matrix means the rank is less than $n$, which in your case means the rank is $0$ (for null matrix only) or $1$.
$endgroup$
– StubbornAtom
Dec 30 '18 at 14:54
$begingroup$
@StubbornAtom that last line was what I needed. Thank you very much! If you put that as an answer, I'll accept it.
$endgroup$
– Brad
Dec 30 '18 at 14:55