Assume $int^a_b operatornametr(A(t)B(t))~dt=0$ for any $B$, where $A,B$ are $n times n$ matrices. Does this imply $A=0$?

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Assume $displaystyleint^a_b operatornametr(A(t)B(t))~dt=0$ for any $B$, where $A,B$ are $n times n$ matrices.



Does this imply $A=0$?



If this is not true, can we add some conditions for $A, B$ to make the proposition true? For example, add some conditions like $A$ is symmetric, $A$ is skew-symmetric, both $A~textand~B in SO(n)$, etc.




calculus real-analysis linear-algebra functional-analysis lie-groups




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  • $A$ and $B$ are matricial functions, aren't they?
    – Gonzalo Benavides
    Jul 27 at 22:14










  • Hint: Set $B(t)=A(t)^T$.
    – Math Lover
    Jul 27 at 22:17







  • 1




    @MathLover Rather, set $B(t) = A(t)^T$.
    – mechanodroid
    Jul 27 at 22:19










  • @mechanodroid Thanks.
    – Math Lover
    Jul 27 at 22:21






  • 1




    @DionelJaime A matrix whose coefficients are (sufficiently regular) functions of a parameter $t$, presumably.
    – Clement C.
    Jul 27 at 22:31














up vote
4
down vote

favorite












Assume $displaystyleint^a_b operatornametr(A(t)B(t))~dt=0$ for any $B$, where $A,B$ are $n times n$ matrices.



Does this imply $A=0$?



If this is not true, can we add some conditions for $A, B$ to make the proposition true? For example, add some conditions like $A$ is symmetric, $A$ is skew-symmetric, both $A~textand~B in SO(n)$, etc.




calculus real-analysis linear-algebra functional-analysis lie-groups




share|cite|improve this question





















  • $A$ and $B$ are matricial functions, aren't they?
    – Gonzalo Benavides
    Jul 27 at 22:14










  • Hint: Set $B(t)=A(t)^T$.
    – Math Lover
    Jul 27 at 22:17







  • 1




    @MathLover Rather, set $B(t) = A(t)^T$.
    – mechanodroid
    Jul 27 at 22:19










  • @mechanodroid Thanks.
    – Math Lover
    Jul 27 at 22:21






  • 1




    @DionelJaime A matrix whose coefficients are (sufficiently regular) functions of a parameter $t$, presumably.
    – Clement C.
    Jul 27 at 22:31












up vote
4
down vote

favorite









up vote
4
down vote

favorite











Assume $displaystyleint^a_b operatornametr(A(t)B(t))~dt=0$ for any $B$, where $A,B$ are $n times n$ matrices.



Does this imply $A=0$?



If this is not true, can we add some conditions for $A, B$ to make the proposition true? For example, add some conditions like $A$ is symmetric, $A$ is skew-symmetric, both $A~textand~B in SO(n)$, etc.




calculus real-analysis linear-algebra functional-analysis lie-groups




share|cite|improve this question













Assume $displaystyleint^a_b operatornametr(A(t)B(t))~dt=0$ for any $B$, where $A,B$ are $n times n$ matrices.



Does this imply $A=0$?



If this is not true, can we add some conditions for $A, B$ to make the proposition true? For example, add some conditions like $A$ is symmetric, $A$ is skew-symmetric, both $A~textand~B in SO(n)$, etc.





calculus real-analysis linear-algebra functional-analysis lie-groups





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share|cite|improve this question




share|cite|improve this question








edited Jul 27 at 23:11









Xander Henderson

13.1k83150




13.1k83150









asked Jul 27 at 22:13









XYZ

1038




1038











  • $A$ and $B$ are matricial functions, aren't they?
    – Gonzalo Benavides
    Jul 27 at 22:14










  • Hint: Set $B(t)=A(t)^T$.
    – Math Lover
    Jul 27 at 22:17







  • 1




    @MathLover Rather, set $B(t) = A(t)^T$.
    – mechanodroid
    Jul 27 at 22:19










  • @mechanodroid Thanks.
    – Math Lover
    Jul 27 at 22:21






  • 1




    @DionelJaime A matrix whose coefficients are (sufficiently regular) functions of a parameter $t$, presumably.
    – Clement C.
    Jul 27 at 22:31
















  • $A$ and $B$ are matricial functions, aren't they?
    – Gonzalo Benavides
    Jul 27 at 22:14










  • Hint: Set $B(t)=A(t)^T$.
    – Math Lover
    Jul 27 at 22:17







  • 1




    @MathLover Rather, set $B(t) = A(t)^T$.
    – mechanodroid
    Jul 27 at 22:19










  • @mechanodroid Thanks.
    – Math Lover
    Jul 27 at 22:21






  • 1




    @DionelJaime A matrix whose coefficients are (sufficiently regular) functions of a parameter $t$, presumably.
    – Clement C.
    Jul 27 at 22:31















$A$ and $B$ are matricial functions, aren't they?
– Gonzalo Benavides
Jul 27 at 22:14




$A$ and $B$ are matricial functions, aren't they?
– Gonzalo Benavides
Jul 27 at 22:14












Hint: Set $B(t)=A(t)^T$.
– Math Lover
Jul 27 at 22:17





Hint: Set $B(t)=A(t)^T$.
– Math Lover
Jul 27 at 22:17





1




1




@MathLover Rather, set $B(t) = A(t)^T$.
– mechanodroid
Jul 27 at 22:19




@MathLover Rather, set $B(t) = A(t)^T$.
– mechanodroid
Jul 27 at 22:19












@mechanodroid Thanks.
– Math Lover
Jul 27 at 22:21




@mechanodroid Thanks.
– Math Lover
Jul 27 at 22:21




1




1




@DionelJaime A matrix whose coefficients are (sufficiently regular) functions of a parameter $t$, presumably.
– Clement C.
Jul 27 at 22:31




@DionelJaime A matrix whose coefficients are (sufficiently regular) functions of a parameter $t$, presumably.
– Clement C.
Jul 27 at 22:31










1 Answer
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Following Math Lover's hint, consider $B(t)$ being $A(t)^top$. Then check that the integrand $texttr(A(t) A(t)^top)=sum_i sum_j A(t)_ij^2$ is nonnegative.






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  • 1




    Or better yet $B=A^dagger$, so the argument works even for complex matrices.
    – J.G.
    Jul 27 at 22:23










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1 Answer
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1 Answer
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active

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oldest

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up vote
1
down vote



accepted










Following Math Lover's hint, consider $B(t)$ being $A(t)^top$. Then check that the integrand $texttr(A(t) A(t)^top)=sum_i sum_j A(t)_ij^2$ is nonnegative.






share|cite|improve this answer

















  • 1




    Or better yet $B=A^dagger$, so the argument works even for complex matrices.
    – J.G.
    Jul 27 at 22:23














up vote
1
down vote



accepted










Following Math Lover's hint, consider $B(t)$ being $A(t)^top$. Then check that the integrand $texttr(A(t) A(t)^top)=sum_i sum_j A(t)_ij^2$ is nonnegative.






share|cite|improve this answer

















  • 1




    Or better yet $B=A^dagger$, so the argument works even for complex matrices.
    – J.G.
    Jul 27 at 22:23












up vote
1
down vote



accepted







up vote
1
down vote



accepted






Following Math Lover's hint, consider $B(t)$ being $A(t)^top$. Then check that the integrand $texttr(A(t) A(t)^top)=sum_i sum_j A(t)_ij^2$ is nonnegative.






share|cite|improve this answer













Following Math Lover's hint, consider $B(t)$ being $A(t)^top$. Then check that the integrand $texttr(A(t) A(t)^top)=sum_i sum_j A(t)_ij^2$ is nonnegative.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 27 at 22:20









angryavian

34.5k12874




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  • 1




    Or better yet $B=A^dagger$, so the argument works even for complex matrices.
    – J.G.
    Jul 27 at 22:23












  • 1




    Or better yet $B=A^dagger$, so the argument works even for complex matrices.
    – J.G.
    Jul 27 at 22:23







1




1




Or better yet $B=A^dagger$, so the argument works even for complex matrices.
– J.G.
Jul 27 at 22:23




Or better yet $B=A^dagger$, so the argument works even for complex matrices.
– J.G.
Jul 27 at 22:23












 

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