A linear combination of the set ${ {bf A},{bf A}^T}$












2












$begingroup$


Consider $mathbf{A}$, an $n times n$ matrix over $mathbb{F}_q$, the finite field with $q$ elements. The transpose of $mathbf{A}$ is denoted with $mathbf{A}^T$.
Let $mathbf{I}_n$ denote the identity matrix of order $n$.
Assume that the matrix $bf A$ has the following property over $mathbb{F}_q$,
$
mathbf{A}mathbf{A}^T=mathbf{I}_n.
$



We say $mathbf{B}$ is a linear combination of the set
${mathbf{A},mathbf{A}^T}$, if
$mathbf{B}=alpha_1mathbf{A}+alpha_2mathbf{A}^T$
where $alpha_i$ with $1leq i leq 2$ are elements in $mathbb{F}_q$.



My Question:
How to find an $ntimes n$ matrix $mathbf{B}$ such that
$mathbf{B}mathbf{B}^T=-mathbf{I}_n$ over $mathbb{F}_q$ whithout restriction on $q$?



Example: Consider $alpha in mathbb{F}_q$ such that $alpha^2=-1$, then
by considering $mathbf{B}=alphamathbf{A}$, we get $mathbf{B}mathbf{B}^T=-mathbf{I}_n$.
But due to the condition over $alpha$, we have $q equiv1pmod{4}$.






linear-algebra matrices finite-fields







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    If $A^2=I$, then $A=A^{-1}$, and so there is no need to include $alpha_3$. However, your question doesn't actually make reference to $A$. Do you need it to, i.e., do you require that $B$ is of the form $alpha_1 A + alpha_2 A^T$ for some $A$ satisfying $A^2=I$?
    $endgroup$
    – Aaron
    Dec 1 '18 at 22:44












  • $begingroup$
    @Aaron You right. Based on your comment i edit the question.
    $endgroup$
    – Amin235
    Dec 1 '18 at 22:51












  • $begingroup$
    @Aaron The condition $A=A^{-1}$ was not necessary and because of this I omitted it.
    $endgroup$
    – Amin235
    Dec 1 '18 at 22:58










  • $begingroup$
    @egreg I appreciate you for your nice edition.
    $endgroup$
    – Amin235
    Dec 1 '18 at 23:21
















2












$begingroup$


Consider $mathbf{A}$, an $n times n$ matrix over $mathbb{F}_q$, the finite field with $q$ elements. The transpose of $mathbf{A}$ is denoted with $mathbf{A}^T$.
Let $mathbf{I}_n$ denote the identity matrix of order $n$.
Assume that the matrix $bf A$ has the following property over $mathbb{F}_q$,
$
mathbf{A}mathbf{A}^T=mathbf{I}_n.
$



We say $mathbf{B}$ is a linear combination of the set
${mathbf{A},mathbf{A}^T}$, if
$mathbf{B}=alpha_1mathbf{A}+alpha_2mathbf{A}^T$
where $alpha_i$ with $1leq i leq 2$ are elements in $mathbb{F}_q$.



My Question:
How to find an $ntimes n$ matrix $mathbf{B}$ such that
$mathbf{B}mathbf{B}^T=-mathbf{I}_n$ over $mathbb{F}_q$ whithout restriction on $q$?



Example: Consider $alpha in mathbb{F}_q$ such that $alpha^2=-1$, then
by considering $mathbf{B}=alphamathbf{A}$, we get $mathbf{B}mathbf{B}^T=-mathbf{I}_n$.
But due to the condition over $alpha$, we have $q equiv1pmod{4}$.






linear-algebra matrices finite-fields







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    If $A^2=I$, then $A=A^{-1}$, and so there is no need to include $alpha_3$. However, your question doesn't actually make reference to $A$. Do you need it to, i.e., do you require that $B$ is of the form $alpha_1 A + alpha_2 A^T$ for some $A$ satisfying $A^2=I$?
    $endgroup$
    – Aaron
    Dec 1 '18 at 22:44












  • $begingroup$
    @Aaron You right. Based on your comment i edit the question.
    $endgroup$
    – Amin235
    Dec 1 '18 at 22:51












  • $begingroup$
    @Aaron The condition $A=A^{-1}$ was not necessary and because of this I omitted it.
    $endgroup$
    – Amin235
    Dec 1 '18 at 22:58










  • $begingroup$
    @egreg I appreciate you for your nice edition.
    $endgroup$
    – Amin235
    Dec 1 '18 at 23:21














2












2








2


1



$begingroup$


Consider $mathbf{A}$, an $n times n$ matrix over $mathbb{F}_q$, the finite field with $q$ elements. The transpose of $mathbf{A}$ is denoted with $mathbf{A}^T$.
Let $mathbf{I}_n$ denote the identity matrix of order $n$.
Assume that the matrix $bf A$ has the following property over $mathbb{F}_q$,
$
mathbf{A}mathbf{A}^T=mathbf{I}_n.
$



We say $mathbf{B}$ is a linear combination of the set
${mathbf{A},mathbf{A}^T}$, if
$mathbf{B}=alpha_1mathbf{A}+alpha_2mathbf{A}^T$
where $alpha_i$ with $1leq i leq 2$ are elements in $mathbb{F}_q$.



My Question:
How to find an $ntimes n$ matrix $mathbf{B}$ such that
$mathbf{B}mathbf{B}^T=-mathbf{I}_n$ over $mathbb{F}_q$ whithout restriction on $q$?



Example: Consider $alpha in mathbb{F}_q$ such that $alpha^2=-1$, then
by considering $mathbf{B}=alphamathbf{A}$, we get $mathbf{B}mathbf{B}^T=-mathbf{I}_n$.
But due to the condition over $alpha$, we have $q equiv1pmod{4}$.






linear-algebra matrices finite-fields







share|cite|improve this question











$endgroup$




Consider $mathbf{A}$, an $n times n$ matrix over $mathbb{F}_q$, the finite field with $q$ elements. The transpose of $mathbf{A}$ is denoted with $mathbf{A}^T$.
Let $mathbf{I}_n$ denote the identity matrix of order $n$.
Assume that the matrix $bf A$ has the following property over $mathbb{F}_q$,
$
mathbf{A}mathbf{A}^T=mathbf{I}_n.
$



We say $mathbf{B}$ is a linear combination of the set
${mathbf{A},mathbf{A}^T}$, if
$mathbf{B}=alpha_1mathbf{A}+alpha_2mathbf{A}^T$
where $alpha_i$ with $1leq i leq 2$ are elements in $mathbb{F}_q$.



My Question:
How to find an $ntimes n$ matrix $mathbf{B}$ such that
$mathbf{B}mathbf{B}^T=-mathbf{I}_n$ over $mathbb{F}_q$ whithout restriction on $q$?



Example: Consider $alpha in mathbb{F}_q$ such that $alpha^2=-1$, then
by considering $mathbf{B}=alphamathbf{A}$, we get $mathbf{B}mathbf{B}^T=-mathbf{I}_n$.
But due to the condition over $alpha$, we have $q equiv1pmod{4}$.






linear-algebra matrices finite-fields




linear-algebra matrices finite-fields






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 '18 at 23:11









egreg

180k1485202




180k1485202










asked Dec 1 '18 at 22:34









Amin235Amin235

1,201621




1,201621








  • 2




    $begingroup$
    If $A^2=I$, then $A=A^{-1}$, and so there is no need to include $alpha_3$. However, your question doesn't actually make reference to $A$. Do you need it to, i.e., do you require that $B$ is of the form $alpha_1 A + alpha_2 A^T$ for some $A$ satisfying $A^2=I$?
    $endgroup$
    – Aaron
    Dec 1 '18 at 22:44












  • $begingroup$
    @Aaron You right. Based on your comment i edit the question.
    $endgroup$
    – Amin235
    Dec 1 '18 at 22:51












  • $begingroup$
    @Aaron The condition $A=A^{-1}$ was not necessary and because of this I omitted it.
    $endgroup$
    – Amin235
    Dec 1 '18 at 22:58










  • $begingroup$
    @egreg I appreciate you for your nice edition.
    $endgroup$
    – Amin235
    Dec 1 '18 at 23:21














  • 2




    $begingroup$
    If $A^2=I$, then $A=A^{-1}$, and so there is no need to include $alpha_3$. However, your question doesn't actually make reference to $A$. Do you need it to, i.e., do you require that $B$ is of the form $alpha_1 A + alpha_2 A^T$ for some $A$ satisfying $A^2=I$?
    $endgroup$
    – Aaron
    Dec 1 '18 at 22:44












  • $begingroup$
    @Aaron You right. Based on your comment i edit the question.
    $endgroup$
    – Amin235
    Dec 1 '18 at 22:51












  • $begingroup$
    @Aaron The condition $A=A^{-1}$ was not necessary and because of this I omitted it.
    $endgroup$
    – Amin235
    Dec 1 '18 at 22:58










  • $begingroup$
    @egreg I appreciate you for your nice edition.
    $endgroup$
    – Amin235
    Dec 1 '18 at 23:21








2




2




$begingroup$
If $A^2=I$, then $A=A^{-1}$, and so there is no need to include $alpha_3$. However, your question doesn't actually make reference to $A$. Do you need it to, i.e., do you require that $B$ is of the form $alpha_1 A + alpha_2 A^T$ for some $A$ satisfying $A^2=I$?
$endgroup$
– Aaron
Dec 1 '18 at 22:44






$begingroup$
If $A^2=I$, then $A=A^{-1}$, and so there is no need to include $alpha_3$. However, your question doesn't actually make reference to $A$. Do you need it to, i.e., do you require that $B$ is of the form $alpha_1 A + alpha_2 A^T$ for some $A$ satisfying $A^2=I$?
$endgroup$
– Aaron
Dec 1 '18 at 22:44














$begingroup$
@Aaron You right. Based on your comment i edit the question.
$endgroup$
– Amin235
Dec 1 '18 at 22:51






$begingroup$
@Aaron You right. Based on your comment i edit the question.
$endgroup$
– Amin235
Dec 1 '18 at 22:51














$begingroup$
@Aaron The condition $A=A^{-1}$ was not necessary and because of this I omitted it.
$endgroup$
– Amin235
Dec 1 '18 at 22:58




$begingroup$
@Aaron The condition $A=A^{-1}$ was not necessary and because of this I omitted it.
$endgroup$
– Amin235
Dec 1 '18 at 22:58












$begingroup$
@egreg I appreciate you for your nice edition.
$endgroup$
– Amin235
Dec 1 '18 at 23:21




$begingroup$
@egreg I appreciate you for your nice edition.
$endgroup$
– Amin235
Dec 1 '18 at 23:21










1 Answer
1






active

oldest

votes


















1












$begingroup$

This isn't always possible. For example, if $A=I$, then the problem collapses to $(alpha A)(alpha A)^T=alpha^2 I=-I$, but $alpha^2=-1$ does not always have solutions.



However, if $B=alpha_1 A + alpha_2 A^T$, then $BB^T=(alpha_1^2+alpha_2^2)I+alpha_1 alpha_2 (A^2+(A^T)^2)$. If $alpha^2=-1$ has no solutions in our field, then the only way we can hope to solve is if $A^2+(A^T)^2$ is a scalar.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    "$A^2+(A^T)^2$ be a scalar" is an interesting note. Thanks.
    $endgroup$
    – Amin235
    Dec 2 '18 at 8:24













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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

This isn't always possible. For example, if $A=I$, then the problem collapses to $(alpha A)(alpha A)^T=alpha^2 I=-I$, but $alpha^2=-1$ does not always have solutions.



However, if $B=alpha_1 A + alpha_2 A^T$, then $BB^T=(alpha_1^2+alpha_2^2)I+alpha_1 alpha_2 (A^2+(A^T)^2)$. If $alpha^2=-1$ has no solutions in our field, then the only way we can hope to solve is if $A^2+(A^T)^2$ is a scalar.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    "$A^2+(A^T)^2$ be a scalar" is an interesting note. Thanks.
    $endgroup$
    – Amin235
    Dec 2 '18 at 8:24


















1












$begingroup$

This isn't always possible. For example, if $A=I$, then the problem collapses to $(alpha A)(alpha A)^T=alpha^2 I=-I$, but $alpha^2=-1$ does not always have solutions.



However, if $B=alpha_1 A + alpha_2 A^T$, then $BB^T=(alpha_1^2+alpha_2^2)I+alpha_1 alpha_2 (A^2+(A^T)^2)$. If $alpha^2=-1$ has no solutions in our field, then the only way we can hope to solve is if $A^2+(A^T)^2$ is a scalar.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    "$A^2+(A^T)^2$ be a scalar" is an interesting note. Thanks.
    $endgroup$
    – Amin235
    Dec 2 '18 at 8:24
















1












1








1





$begingroup$

This isn't always possible. For example, if $A=I$, then the problem collapses to $(alpha A)(alpha A)^T=alpha^2 I=-I$, but $alpha^2=-1$ does not always have solutions.



However, if $B=alpha_1 A + alpha_2 A^T$, then $BB^T=(alpha_1^2+alpha_2^2)I+alpha_1 alpha_2 (A^2+(A^T)^2)$. If $alpha^2=-1$ has no solutions in our field, then the only way we can hope to solve is if $A^2+(A^T)^2$ is a scalar.






share|cite|improve this answer









$endgroup$



This isn't always possible. For example, if $A=I$, then the problem collapses to $(alpha A)(alpha A)^T=alpha^2 I=-I$, but $alpha^2=-1$ does not always have solutions.



However, if $B=alpha_1 A + alpha_2 A^T$, then $BB^T=(alpha_1^2+alpha_2^2)I+alpha_1 alpha_2 (A^2+(A^T)^2)$. If $alpha^2=-1$ has no solutions in our field, then the only way we can hope to solve is if $A^2+(A^T)^2$ is a scalar.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 1 '18 at 23:48









AaronAaron

15.9k22754




15.9k22754












  • $begingroup$
    "$A^2+(A^T)^2$ be a scalar" is an interesting note. Thanks.
    $endgroup$
    – Amin235
    Dec 2 '18 at 8:24




















  • $begingroup$
    "$A^2+(A^T)^2$ be a scalar" is an interesting note. Thanks.
    $endgroup$
    – Amin235
    Dec 2 '18 at 8:24


















$begingroup$
"$A^2+(A^T)^2$ be a scalar" is an interesting note. Thanks.
$endgroup$
– Amin235
Dec 2 '18 at 8:24






$begingroup$
"$A^2+(A^T)^2$ be a scalar" is an interesting note. Thanks.
$endgroup$
– Amin235
Dec 2 '18 at 8:24




















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