Eigenvalues of $L^dagger L$ are positive












-1














Let $L$ be a linear operator on a complex vector space $V$, how to show that eigenvalues of $L^dagger L$ are positive? I guess this might be a statement taught in a functional analysis class or an advanced linear algebra class?






linear-algebra functional-analysis







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  • Is $dagger$ the same Hermitian?
    – Mostafa Ayaz
    Nov 29 at 13:09










  • @MostafaAyaz Right.
    – user546106
    Nov 29 at 13:10










  • Is $V$ a Hilbert space? And is $L$ a bounded linear operator?
    – Olivier Moschetta
    Nov 29 at 13:10
















-1














Let $L$ be a linear operator on a complex vector space $V$, how to show that eigenvalues of $L^dagger L$ are positive? I guess this might be a statement taught in a functional analysis class or an advanced linear algebra class?






linear-algebra functional-analysis







share|cite|improve this question
























  • Is $dagger$ the same Hermitian?
    – Mostafa Ayaz
    Nov 29 at 13:09










  • @MostafaAyaz Right.
    – user546106
    Nov 29 at 13:10










  • Is $V$ a Hilbert space? And is $L$ a bounded linear operator?
    – Olivier Moschetta
    Nov 29 at 13:10














-1












-1








-1







Let $L$ be a linear operator on a complex vector space $V$, how to show that eigenvalues of $L^dagger L$ are positive? I guess this might be a statement taught in a functional analysis class or an advanced linear algebra class?






linear-algebra functional-analysis







share|cite|improve this question















Let $L$ be a linear operator on a complex vector space $V$, how to show that eigenvalues of $L^dagger L$ are positive? I guess this might be a statement taught in a functional analysis class or an advanced linear algebra class?






linear-algebra functional-analysis




linear-algebra functional-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 at 13:07

























asked Nov 29 at 13:02









user546106

1318




1318












  • Is $dagger$ the same Hermitian?
    – Mostafa Ayaz
    Nov 29 at 13:09










  • @MostafaAyaz Right.
    – user546106
    Nov 29 at 13:10










  • Is $V$ a Hilbert space? And is $L$ a bounded linear operator?
    – Olivier Moschetta
    Nov 29 at 13:10


















  • Is $dagger$ the same Hermitian?
    – Mostafa Ayaz
    Nov 29 at 13:09










  • @MostafaAyaz Right.
    – user546106
    Nov 29 at 13:10










  • Is $V$ a Hilbert space? And is $L$ a bounded linear operator?
    – Olivier Moschetta
    Nov 29 at 13:10
















Is $dagger$ the same Hermitian?
– Mostafa Ayaz
Nov 29 at 13:09




Is $dagger$ the same Hermitian?
– Mostafa Ayaz
Nov 29 at 13:09












@MostafaAyaz Right.
– user546106
Nov 29 at 13:10




@MostafaAyaz Right.
– user546106
Nov 29 at 13:10












Is $V$ a Hilbert space? And is $L$ a bounded linear operator?
– Olivier Moschetta
Nov 29 at 13:10




Is $V$ a Hilbert space? And is $L$ a bounded linear operator?
– Olivier Moschetta
Nov 29 at 13:10










2 Answers
2






active

oldest

votes


















0














The property "self-adjoint" is only defined for operators on inner product spaces.



Now suppose that $V$ is an inner product space with inner product $( cdot, cdot)$.



Then we have for $u,v in V: (L^dagger Lu,v)=(Lu,Lv)=(u, L^dagger Lv).$ This shows that $L^dagger L$ is selfadjoint.



If $ mu$ is an eigenvalue of $L^dagger L$, then there is $u in V$ with $L^dagger Lu = mu u$. Thus



$ mu (u,u)=( mu u,u)= (L^dagger Lu,u)=(Lu,Lu).$



This gives $ mu ge 0$.






share|cite|improve this answer



















  • 1




    I think you mean "this shows that $L^{dagger}L$ is self-adjoint".
    – Olivier Moschetta
    Nov 29 at 13:13












  • Could you please explain where the self-adjointness of $L^dagger L$ was used? Since $mu=|Lu|^2/|u|^2$, do we also have that $mu>0$?
    – Cm7F7Bb
    Nov 29 at 13:15





















0














Using the singular value decomposition (SVD) https://en.wikipedia.org/wiki/Singular_value_decomposition



we have $$L=UDVto L^dagger =V^dagger D^dagger U^dagger$$therefore $$L^dagger L=V^dagger D^dagger D V$$since $V$ is unitary, the eigenvalues of $Ldagger L$ are the same as $D^dagger D$ and since $D^dagger D$ is diagonal positive semi definite, the proof is complete.






share|cite|improve this answer























  • en.wikipedia.org/wiki/… Here is a proof which shows that the eigenvalues of a Hermitian matrix are positive, but what if the space $V$ is of infinite dimension?
    – user546106
    Nov 29 at 13:13










  • It matters not since $v^dagger L^dagger Lv=|Lv|^2ge 0$....
    – Mostafa Ayaz
    Nov 29 at 13:19











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0














The property "self-adjoint" is only defined for operators on inner product spaces.



Now suppose that $V$ is an inner product space with inner product $( cdot, cdot)$.



Then we have for $u,v in V: (L^dagger Lu,v)=(Lu,Lv)=(u, L^dagger Lv).$ This shows that $L^dagger L$ is selfadjoint.



If $ mu$ is an eigenvalue of $L^dagger L$, then there is $u in V$ with $L^dagger Lu = mu u$. Thus



$ mu (u,u)=( mu u,u)= (L^dagger Lu,u)=(Lu,Lu).$



This gives $ mu ge 0$.






share|cite|improve this answer



















  • 1




    I think you mean "this shows that $L^{dagger}L$ is self-adjoint".
    – Olivier Moschetta
    Nov 29 at 13:13












  • Could you please explain where the self-adjointness of $L^dagger L$ was used? Since $mu=|Lu|^2/|u|^2$, do we also have that $mu>0$?
    – Cm7F7Bb
    Nov 29 at 13:15


















0














The property "self-adjoint" is only defined for operators on inner product spaces.



Now suppose that $V$ is an inner product space with inner product $( cdot, cdot)$.



Then we have for $u,v in V: (L^dagger Lu,v)=(Lu,Lv)=(u, L^dagger Lv).$ This shows that $L^dagger L$ is selfadjoint.



If $ mu$ is an eigenvalue of $L^dagger L$, then there is $u in V$ with $L^dagger Lu = mu u$. Thus



$ mu (u,u)=( mu u,u)= (L^dagger Lu,u)=(Lu,Lu).$



This gives $ mu ge 0$.






share|cite|improve this answer



















  • 1




    I think you mean "this shows that $L^{dagger}L$ is self-adjoint".
    – Olivier Moschetta
    Nov 29 at 13:13












  • Could you please explain where the self-adjointness of $L^dagger L$ was used? Since $mu=|Lu|^2/|u|^2$, do we also have that $mu>0$?
    – Cm7F7Bb
    Nov 29 at 13:15
















0












0








0






The property "self-adjoint" is only defined for operators on inner product spaces.



Now suppose that $V$ is an inner product space with inner product $( cdot, cdot)$.



Then we have for $u,v in V: (L^dagger Lu,v)=(Lu,Lv)=(u, L^dagger Lv).$ This shows that $L^dagger L$ is selfadjoint.



If $ mu$ is an eigenvalue of $L^dagger L$, then there is $u in V$ with $L^dagger Lu = mu u$. Thus



$ mu (u,u)=( mu u,u)= (L^dagger Lu,u)=(Lu,Lu).$



This gives $ mu ge 0$.






share|cite|improve this answer














The property "self-adjoint" is only defined for operators on inner product spaces.



Now suppose that $V$ is an inner product space with inner product $( cdot, cdot)$.



Then we have for $u,v in V: (L^dagger Lu,v)=(Lu,Lv)=(u, L^dagger Lv).$ This shows that $L^dagger L$ is selfadjoint.



If $ mu$ is an eigenvalue of $L^dagger L$, then there is $u in V$ with $L^dagger Lu = mu u$. Thus



$ mu (u,u)=( mu u,u)= (L^dagger Lu,u)=(Lu,Lu).$



This gives $ mu ge 0$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 29 at 13:34

























answered Nov 29 at 13:11









Fred

44.1k1644




44.1k1644








  • 1




    I think you mean "this shows that $L^{dagger}L$ is self-adjoint".
    – Olivier Moschetta
    Nov 29 at 13:13












  • Could you please explain where the self-adjointness of $L^dagger L$ was used? Since $mu=|Lu|^2/|u|^2$, do we also have that $mu>0$?
    – Cm7F7Bb
    Nov 29 at 13:15
















  • 1




    I think you mean "this shows that $L^{dagger}L$ is self-adjoint".
    – Olivier Moschetta
    Nov 29 at 13:13












  • Could you please explain where the self-adjointness of $L^dagger L$ was used? Since $mu=|Lu|^2/|u|^2$, do we also have that $mu>0$?
    – Cm7F7Bb
    Nov 29 at 13:15










1




1




I think you mean "this shows that $L^{dagger}L$ is self-adjoint".
– Olivier Moschetta
Nov 29 at 13:13






I think you mean "this shows that $L^{dagger}L$ is self-adjoint".
– Olivier Moschetta
Nov 29 at 13:13














Could you please explain where the self-adjointness of $L^dagger L$ was used? Since $mu=|Lu|^2/|u|^2$, do we also have that $mu>0$?
– Cm7F7Bb
Nov 29 at 13:15






Could you please explain where the self-adjointness of $L^dagger L$ was used? Since $mu=|Lu|^2/|u|^2$, do we also have that $mu>0$?
– Cm7F7Bb
Nov 29 at 13:15













0














Using the singular value decomposition (SVD) https://en.wikipedia.org/wiki/Singular_value_decomposition



we have $$L=UDVto L^dagger =V^dagger D^dagger U^dagger$$therefore $$L^dagger L=V^dagger D^dagger D V$$since $V$ is unitary, the eigenvalues of $Ldagger L$ are the same as $D^dagger D$ and since $D^dagger D$ is diagonal positive semi definite, the proof is complete.






share|cite|improve this answer























  • en.wikipedia.org/wiki/… Here is a proof which shows that the eigenvalues of a Hermitian matrix are positive, but what if the space $V$ is of infinite dimension?
    – user546106
    Nov 29 at 13:13










  • It matters not since $v^dagger L^dagger Lv=|Lv|^2ge 0$....
    – Mostafa Ayaz
    Nov 29 at 13:19
















0














Using the singular value decomposition (SVD) https://en.wikipedia.org/wiki/Singular_value_decomposition



we have $$L=UDVto L^dagger =V^dagger D^dagger U^dagger$$therefore $$L^dagger L=V^dagger D^dagger D V$$since $V$ is unitary, the eigenvalues of $Ldagger L$ are the same as $D^dagger D$ and since $D^dagger D$ is diagonal positive semi definite, the proof is complete.






share|cite|improve this answer























  • en.wikipedia.org/wiki/… Here is a proof which shows that the eigenvalues of a Hermitian matrix are positive, but what if the space $V$ is of infinite dimension?
    – user546106
    Nov 29 at 13:13










  • It matters not since $v^dagger L^dagger Lv=|Lv|^2ge 0$....
    – Mostafa Ayaz
    Nov 29 at 13:19














0












0








0






Using the singular value decomposition (SVD) https://en.wikipedia.org/wiki/Singular_value_decomposition



we have $$L=UDVto L^dagger =V^dagger D^dagger U^dagger$$therefore $$L^dagger L=V^dagger D^dagger D V$$since $V$ is unitary, the eigenvalues of $Ldagger L$ are the same as $D^dagger D$ and since $D^dagger D$ is diagonal positive semi definite, the proof is complete.






share|cite|improve this answer














Using the singular value decomposition (SVD) https://en.wikipedia.org/wiki/Singular_value_decomposition



we have $$L=UDVto L^dagger =V^dagger D^dagger U^dagger$$therefore $$L^dagger L=V^dagger D^dagger D V$$since $V$ is unitary, the eigenvalues of $Ldagger L$ are the same as $D^dagger D$ and since $D^dagger D$ is diagonal positive semi definite, the proof is complete.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 29 at 13:13

























answered Nov 29 at 13:10









Mostafa Ayaz

13.6k3836




13.6k3836












  • en.wikipedia.org/wiki/… Here is a proof which shows that the eigenvalues of a Hermitian matrix are positive, but what if the space $V$ is of infinite dimension?
    – user546106
    Nov 29 at 13:13










  • It matters not since $v^dagger L^dagger Lv=|Lv|^2ge 0$....
    – Mostafa Ayaz
    Nov 29 at 13:19


















  • en.wikipedia.org/wiki/… Here is a proof which shows that the eigenvalues of a Hermitian matrix are positive, but what if the space $V$ is of infinite dimension?
    – user546106
    Nov 29 at 13:13










  • It matters not since $v^dagger L^dagger Lv=|Lv|^2ge 0$....
    – Mostafa Ayaz
    Nov 29 at 13:19
















en.wikipedia.org/wiki/… Here is a proof which shows that the eigenvalues of a Hermitian matrix are positive, but what if the space $V$ is of infinite dimension?
– user546106
Nov 29 at 13:13




en.wikipedia.org/wiki/… Here is a proof which shows that the eigenvalues of a Hermitian matrix are positive, but what if the space $V$ is of infinite dimension?
– user546106
Nov 29 at 13:13












It matters not since $v^dagger L^dagger Lv=|Lv|^2ge 0$....
– Mostafa Ayaz
Nov 29 at 13:19




It matters not since $v^dagger L^dagger Lv=|Lv|^2ge 0$....
– Mostafa Ayaz
Nov 29 at 13:19


















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