A question about the eigenvactors of positive semi-definite matrices

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Assume that $AinmathbbC^ntimes n$ is a Hermitian positive semi-definite matrix and $zinmathbbC^n$ such that $zneq 0$. If $z^*Az=0$, is it true to conclude that $z$ is an eigenvector of $A$ associated with eigenvalue $0$?






matrices positive-semidefinite







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  • 1




    Is $A$ symmetric?
    – A. Pongrácz
    Aug 31 at 5:58






  • 2




    Not if $z=0$, and even if $zneq 0$, the only possible eigenvalue you can get is $0$.
    – Kusma
    Aug 31 at 6:05






  • 1




    May be that is exactly what he wants, finding the eigenvectors associated with $0$ ?
    – nicomezi
    Aug 31 at 6:20






  • 1




    More to the point, the vector $z=0$ is not an eigenvector, but satisfies $z^*Az=0$. So the conclusion is wrong without further assumptions.
    – Kusma
    Aug 31 at 6:30






  • 1




    If you assume $zne 0$, then you are correct.
    – G_0_pi_i_e
    Aug 31 at 6:52














up vote
0
down vote

favorite












Assume that $AinmathbbC^ntimes n$ is a Hermitian positive semi-definite matrix and $zinmathbbC^n$ such that $zneq 0$. If $z^*Az=0$, is it true to conclude that $z$ is an eigenvector of $A$ associated with eigenvalue $0$?






matrices positive-semidefinite







share|cite|improve this question



















  • 1




    Is $A$ symmetric?
    – A. Pongrácz
    Aug 31 at 5:58






  • 2




    Not if $z=0$, and even if $zneq 0$, the only possible eigenvalue you can get is $0$.
    – Kusma
    Aug 31 at 6:05






  • 1




    May be that is exactly what he wants, finding the eigenvectors associated with $0$ ?
    – nicomezi
    Aug 31 at 6:20






  • 1




    More to the point, the vector $z=0$ is not an eigenvector, but satisfies $z^*Az=0$. So the conclusion is wrong without further assumptions.
    – Kusma
    Aug 31 at 6:30






  • 1




    If you assume $zne 0$, then you are correct.
    – G_0_pi_i_e
    Aug 31 at 6:52












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Assume that $AinmathbbC^ntimes n$ is a Hermitian positive semi-definite matrix and $zinmathbbC^n$ such that $zneq 0$. If $z^*Az=0$, is it true to conclude that $z$ is an eigenvector of $A$ associated with eigenvalue $0$?






matrices positive-semidefinite







share|cite|improve this question















Assume that $AinmathbbC^ntimes n$ is a Hermitian positive semi-definite matrix and $zinmathbbC^n$ such that $zneq 0$. If $z^*Az=0$, is it true to conclude that $z$ is an eigenvector of $A$ associated with eigenvalue $0$?






matrices positive-semidefinite




matrices positive-semidefinite






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 31 at 7:48

























asked Aug 31 at 5:54









like_math

28217




28217







  • 1




    Is $A$ symmetric?
    – A. Pongrácz
    Aug 31 at 5:58






  • 2




    Not if $z=0$, and even if $zneq 0$, the only possible eigenvalue you can get is $0$.
    – Kusma
    Aug 31 at 6:05






  • 1




    May be that is exactly what he wants, finding the eigenvectors associated with $0$ ?
    – nicomezi
    Aug 31 at 6:20






  • 1




    More to the point, the vector $z=0$ is not an eigenvector, but satisfies $z^*Az=0$. So the conclusion is wrong without further assumptions.
    – Kusma
    Aug 31 at 6:30






  • 1




    If you assume $zne 0$, then you are correct.
    – G_0_pi_i_e
    Aug 31 at 6:52












  • 1




    Is $A$ symmetric?
    – A. Pongrácz
    Aug 31 at 5:58






  • 2




    Not if $z=0$, and even if $zneq 0$, the only possible eigenvalue you can get is $0$.
    – Kusma
    Aug 31 at 6:05






  • 1




    May be that is exactly what he wants, finding the eigenvectors associated with $0$ ?
    – nicomezi
    Aug 31 at 6:20






  • 1




    More to the point, the vector $z=0$ is not an eigenvector, but satisfies $z^*Az=0$. So the conclusion is wrong without further assumptions.
    – Kusma
    Aug 31 at 6:30






  • 1




    If you assume $zne 0$, then you are correct.
    – G_0_pi_i_e
    Aug 31 at 6:52







1




1




Is $A$ symmetric?
– A. Pongrácz
Aug 31 at 5:58




Is $A$ symmetric?
– A. Pongrácz
Aug 31 at 5:58




2




2




Not if $z=0$, and even if $zneq 0$, the only possible eigenvalue you can get is $0$.
– Kusma
Aug 31 at 6:05




Not if $z=0$, and even if $zneq 0$, the only possible eigenvalue you can get is $0$.
– Kusma
Aug 31 at 6:05




1




1




May be that is exactly what he wants, finding the eigenvectors associated with $0$ ?
– nicomezi
Aug 31 at 6:20




May be that is exactly what he wants, finding the eigenvectors associated with $0$ ?
– nicomezi
Aug 31 at 6:20




1




1




More to the point, the vector $z=0$ is not an eigenvector, but satisfies $z^*Az=0$. So the conclusion is wrong without further assumptions.
– Kusma
Aug 31 at 6:30




More to the point, the vector $z=0$ is not an eigenvector, but satisfies $z^*Az=0$. So the conclusion is wrong without further assumptions.
– Kusma
Aug 31 at 6:30




1




1




If you assume $zne 0$, then you are correct.
– G_0_pi_i_e
Aug 31 at 6:52




If you assume $zne 0$, then you are correct.
– G_0_pi_i_e
Aug 31 at 6:52










1 Answer
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This is true. Using diagonalization we can construct a positive semi-definite matrix $B$ such that $B^2=A$. Then $z^*Az=|Bz||^2$ so we get $Bz=0$ which implies $Az=0$.






share|cite|improve this answer
















  • 1




    How is $z^*Az=|Bz||^2$ ? I think it would have been true if $B^*B=A$.
    – nicomezi
    Aug 31 at 7:06






  • 1




    @nicomezi For complex scalars $A$ positive semi definite implies $A=A^*$. This is a standard fact in operator theory (but this is not true for real scalars).
    – Kavi Rama Murthy
    Aug 31 at 7:07







  • 1




    I know, but my doubts are about having $B^2=A$. $A$ being a positive definite complex matrix is equivalent to having $B$ such that $B^*B=A$, not $B^2=A$. Furthermore, if you have $B^2=A$, the link between $z^*Az$ and $|Bz|^2$ is not clear.
    – nicomezi
    Aug 31 at 7:16







  • 1




    @nicomezi $B$ is also positive semi-definite, so what is difference between $B^2$ and $B^*B$?
    – Kavi Rama Murthy
    Aug 31 at 7:18






  • 1




    Nothing indeed. I am really tired these days.
    – nicomezi
    Aug 31 at 7:20










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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes








up vote
1
down vote













This is true. Using diagonalization we can construct a positive semi-definite matrix $B$ such that $B^2=A$. Then $z^*Az=|Bz||^2$ so we get $Bz=0$ which implies $Az=0$.






share|cite|improve this answer
















  • 1




    How is $z^*Az=|Bz||^2$ ? I think it would have been true if $B^*B=A$.
    – nicomezi
    Aug 31 at 7:06






  • 1




    @nicomezi For complex scalars $A$ positive semi definite implies $A=A^*$. This is a standard fact in operator theory (but this is not true for real scalars).
    – Kavi Rama Murthy
    Aug 31 at 7:07







  • 1




    I know, but my doubts are about having $B^2=A$. $A$ being a positive definite complex matrix is equivalent to having $B$ such that $B^*B=A$, not $B^2=A$. Furthermore, if you have $B^2=A$, the link between $z^*Az$ and $|Bz|^2$ is not clear.
    – nicomezi
    Aug 31 at 7:16







  • 1




    @nicomezi $B$ is also positive semi-definite, so what is difference between $B^2$ and $B^*B$?
    – Kavi Rama Murthy
    Aug 31 at 7:18






  • 1




    Nothing indeed. I am really tired these days.
    – nicomezi
    Aug 31 at 7:20














up vote
1
down vote













This is true. Using diagonalization we can construct a positive semi-definite matrix $B$ such that $B^2=A$. Then $z^*Az=|Bz||^2$ so we get $Bz=0$ which implies $Az=0$.






share|cite|improve this answer
















  • 1




    How is $z^*Az=|Bz||^2$ ? I think it would have been true if $B^*B=A$.
    – nicomezi
    Aug 31 at 7:06






  • 1




    @nicomezi For complex scalars $A$ positive semi definite implies $A=A^*$. This is a standard fact in operator theory (but this is not true for real scalars).
    – Kavi Rama Murthy
    Aug 31 at 7:07







  • 1




    I know, but my doubts are about having $B^2=A$. $A$ being a positive definite complex matrix is equivalent to having $B$ such that $B^*B=A$, not $B^2=A$. Furthermore, if you have $B^2=A$, the link between $z^*Az$ and $|Bz|^2$ is not clear.
    – nicomezi
    Aug 31 at 7:16







  • 1




    @nicomezi $B$ is also positive semi-definite, so what is difference between $B^2$ and $B^*B$?
    – Kavi Rama Murthy
    Aug 31 at 7:18






  • 1




    Nothing indeed. I am really tired these days.
    – nicomezi
    Aug 31 at 7:20












up vote
1
down vote










up vote
1
down vote









This is true. Using diagonalization we can construct a positive semi-definite matrix $B$ such that $B^2=A$. Then $z^*Az=|Bz||^2$ so we get $Bz=0$ which implies $Az=0$.






share|cite|improve this answer












This is true. Using diagonalization we can construct a positive semi-definite matrix $B$ such that $B^2=A$. Then $z^*Az=|Bz||^2$ so we get $Bz=0$ which implies $Az=0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 31 at 6:28









Kavi Rama Murthy

25.4k31435




25.4k31435







  • 1




    How is $z^*Az=|Bz||^2$ ? I think it would have been true if $B^*B=A$.
    – nicomezi
    Aug 31 at 7:06






  • 1




    @nicomezi For complex scalars $A$ positive semi definite implies $A=A^*$. This is a standard fact in operator theory (but this is not true for real scalars).
    – Kavi Rama Murthy
    Aug 31 at 7:07







  • 1




    I know, but my doubts are about having $B^2=A$. $A$ being a positive definite complex matrix is equivalent to having $B$ such that $B^*B=A$, not $B^2=A$. Furthermore, if you have $B^2=A$, the link between $z^*Az$ and $|Bz|^2$ is not clear.
    – nicomezi
    Aug 31 at 7:16







  • 1




    @nicomezi $B$ is also positive semi-definite, so what is difference between $B^2$ and $B^*B$?
    – Kavi Rama Murthy
    Aug 31 at 7:18






  • 1




    Nothing indeed. I am really tired these days.
    – nicomezi
    Aug 31 at 7:20












  • 1




    How is $z^*Az=|Bz||^2$ ? I think it would have been true if $B^*B=A$.
    – nicomezi
    Aug 31 at 7:06






  • 1




    @nicomezi For complex scalars $A$ positive semi definite implies $A=A^*$. This is a standard fact in operator theory (but this is not true for real scalars).
    – Kavi Rama Murthy
    Aug 31 at 7:07







  • 1




    I know, but my doubts are about having $B^2=A$. $A$ being a positive definite complex matrix is equivalent to having $B$ such that $B^*B=A$, not $B^2=A$. Furthermore, if you have $B^2=A$, the link between $z^*Az$ and $|Bz|^2$ is not clear.
    – nicomezi
    Aug 31 at 7:16







  • 1




    @nicomezi $B$ is also positive semi-definite, so what is difference between $B^2$ and $B^*B$?
    – Kavi Rama Murthy
    Aug 31 at 7:18






  • 1




    Nothing indeed. I am really tired these days.
    – nicomezi
    Aug 31 at 7:20







1




1




How is $z^*Az=|Bz||^2$ ? I think it would have been true if $B^*B=A$.
– nicomezi
Aug 31 at 7:06




How is $z^*Az=|Bz||^2$ ? I think it would have been true if $B^*B=A$.
– nicomezi
Aug 31 at 7:06




1




1




@nicomezi For complex scalars $A$ positive semi definite implies $A=A^*$. This is a standard fact in operator theory (but this is not true for real scalars).
– Kavi Rama Murthy
Aug 31 at 7:07





@nicomezi For complex scalars $A$ positive semi definite implies $A=A^*$. This is a standard fact in operator theory (but this is not true for real scalars).
– Kavi Rama Murthy
Aug 31 at 7:07





1




1




I know, but my doubts are about having $B^2=A$. $A$ being a positive definite complex matrix is equivalent to having $B$ such that $B^*B=A$, not $B^2=A$. Furthermore, if you have $B^2=A$, the link between $z^*Az$ and $|Bz|^2$ is not clear.
– nicomezi
Aug 31 at 7:16





I know, but my doubts are about having $B^2=A$. $A$ being a positive definite complex matrix is equivalent to having $B$ such that $B^*B=A$, not $B^2=A$. Furthermore, if you have $B^2=A$, the link between $z^*Az$ and $|Bz|^2$ is not clear.
– nicomezi
Aug 31 at 7:16





1




1




@nicomezi $B$ is also positive semi-definite, so what is difference between $B^2$ and $B^*B$?
– Kavi Rama Murthy
Aug 31 at 7:18




@nicomezi $B$ is also positive semi-definite, so what is difference between $B^2$ and $B^*B$?
– Kavi Rama Murthy
Aug 31 at 7:18




1




1




Nothing indeed. I am really tired these days.
– nicomezi
Aug 31 at 7:20




Nothing indeed. I am really tired these days.
– nicomezi
Aug 31 at 7:20

















 

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