Nonnegative determinant matrix as a sum of two squares
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As one can see in this MSE question, we have $det(A^{2} + B^{2})geq 0$ for any real $ntimes n$ matrices $A, B$. Is the converse true? i.e. if a real $ntimes n$ matrix with $det Cgeq 0$, is given, can we always find $A, B$ such that $C = A^{2} + B^{2}$?
Also, I'm curious if $det(A^{2} + B^{2} +C^{2})geq 0$ is also true for any matrices $A, B, C$, since in this case, we can't apply the same trick as in the previous question.
linear-algebra
asked Dec 9 '18 at 23:02
Seewoo LeeSeewoo Lee
7,124927
$endgroup$
add a comment |
$begingroup$
As one can see in this MSE question, we have $det(A^{2} + B^{2})geq 0$ for any real $ntimes n$ matrices $A, B$. Is the converse true? i.e. if a real $ntimes n$ matrix with $det Cgeq 0$, is given, can we always find $A, B$ such that $C = A^{2} + B^{2}$?
Also, I'm curious if $det(A^{2} + B^{2} +C^{2})geq 0$ is also true for any matrices $A, B, C$, since in this case, we can't apply the same trick as in the previous question.
linear-algebra
asked Dec 9 '18 at 23:02
Seewoo LeeSeewoo Lee
7,124927
$endgroup$
1
$begingroup$
The question you have quoted does not tell you that $det(A^{2}+B^{2}) geq 0$ for all $A$ and $B$. If $AB=BA$ then this is correct.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 23:24
$begingroup$
@KaviRamaMurthy Oh I see...but I can't find any counterexample for now
$endgroup$
– Seewoo Lee
Dec 9 '18 at 23:39
add a comment |
$begingroup$
As one can see in this MSE question, we have $det(A^{2} + B^{2})geq 0$ for any real $ntimes n$ matrices $A, B$. Is the converse true? i.e. if a real $ntimes n$ matrix with $det Cgeq 0$, is given, can we always find $A, B$ such that $C = A^{2} + B^{2}$?
Also, I'm curious if $det(A^{2} + B^{2} +C^{2})geq 0$ is also true for any matrices $A, B, C$, since in this case, we can't apply the same trick as in the previous question.
linear-algebra
asked Dec 9 '18 at 23:02
Seewoo LeeSeewoo Lee
7,124927
$endgroup$
As one can see in this MSE question, we have $det(A^{2} + B^{2})geq 0$ for any real $ntimes n$ matrices $A, B$. Is the converse true? i.e. if a real $ntimes n$ matrix with $det Cgeq 0$, is given, can we always find $A, B$ such that $C = A^{2} + B^{2}$?
Also, I'm curious if $det(A^{2} + B^{2} +C^{2})geq 0$ is also true for any matrices $A, B, C$, since in this case, we can't apply the same trick as in the previous question.
linear-algebra
linear-algebra
asked Dec 9 '18 at 23:02
Seewoo LeeSeewoo Lee
7,124927
asked Dec 9 '18 at 23:02
Seewoo LeeSeewoo Lee
7,124927
edited Dec 9 '18 at 23:40
Seewoo Lee
asked Dec 9 '18 at 23:02
Seewoo LeeSeewoo Lee
7,124927
asked Dec 9 '18 at 23:02
Seewoo LeeSeewoo Lee
7,124927
asked Dec 9 '18 at 23:02
Seewoo LeeSeewoo Lee
7,124927
7,124927
1
$begingroup$
The question you have quoted does not tell you that $det(A^{2}+B^{2}) geq 0$ for all $A$ and $B$. If $AB=BA$ then this is correct.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 23:24
$begingroup$
@KaviRamaMurthy Oh I see...but I can't find any counterexample for now
$endgroup$
– Seewoo Lee
Dec 9 '18 at 23:39
add a comment |
1
$begingroup$
The question you have quoted does not tell you that $det(A^{2}+B^{2}) geq 0$ for all $A$ and $B$. If $AB=BA$ then this is correct.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 23:24
$begingroup$
@KaviRamaMurthy Oh I see...but I can't find any counterexample for now
$endgroup$
– Seewoo Lee
Dec 9 '18 at 23:39
1
1
$begingroup$
The question you have quoted does not tell you that $det(A^{2}+B^{2}) geq 0$ for all $A$ and $B$. If $AB=BA$ then this is correct.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 23:24
$begingroup$
The question you have quoted does not tell you that $det(A^{2}+B^{2}) geq 0$ for all $A$ and $B$. If $AB=BA$ then this is correct.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 23:24
$begingroup$
@KaviRamaMurthy Oh I see...but I can't find any counterexample for now
$endgroup$
– Seewoo Lee
Dec 9 '18 at 23:39
$begingroup$
@KaviRamaMurthy Oh I see...but I can't find any counterexample for now
$endgroup$
– Seewoo Lee
Dec 9 '18 at 23:39
add a comment |
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1
$begingroup$
The question you have quoted does not tell you that $det(A^{2}+B^{2}) geq 0$ for all $A$ and $B$. If $AB=BA$ then this is correct.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 23:24
$begingroup$
@KaviRamaMurthy Oh I see...but I can't find any counterexample for now
$endgroup$
– Seewoo Lee
Dec 9 '18 at 23:39