Link between Hadamard's inequality for positive-definite matrices and general Hadamard's inequality












1














I've found two versions of Hadamard's inequality :



(1) If $P$ is a $ntimes n$ positive-semidefinite matrix, then :
$$det(P)leprod_{i=1}^n p_{ii} .$$
(2) For any $M$ is a $ntimes n$ matrix, then :
$$|det(M)|leprod_{i=1}^n ||m_i || =prod_{i=1}^n left(sum_{j=1}^n |a_{ij}|^2right)^{1/2}.$$
It's easy to prove that $(2)Rightarrow (1)$, but do we have $(1)Rightarrow (2)$ ?






matrices inequality determinant







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    Yes it does. Let $M$ be any matrix. Then $MM^t$ is a positive semi-definite matrix.
    – mouthetics
    yesterday
















1














I've found two versions of Hadamard's inequality :



(1) If $P$ is a $ntimes n$ positive-semidefinite matrix, then :
$$det(P)leprod_{i=1}^n p_{ii} .$$
(2) For any $M$ is a $ntimes n$ matrix, then :
$$|det(M)|leprod_{i=1}^n ||m_i || =prod_{i=1}^n left(sum_{j=1}^n |a_{ij}|^2right)^{1/2}.$$
It's easy to prove that $(2)Rightarrow (1)$, but do we have $(1)Rightarrow (2)$ ?






matrices inequality determinant







share|cite|improve this question







New contributor




Mishikumo2019 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 1




    Yes it does. Let $M$ be any matrix. Then $MM^t$ is a positive semi-definite matrix.
    – mouthetics
    yesterday














1












1








1







I've found two versions of Hadamard's inequality :



(1) If $P$ is a $ntimes n$ positive-semidefinite matrix, then :
$$det(P)leprod_{i=1}^n p_{ii} .$$
(2) For any $M$ is a $ntimes n$ matrix, then :
$$|det(M)|leprod_{i=1}^n ||m_i || =prod_{i=1}^n left(sum_{j=1}^n |a_{ij}|^2right)^{1/2}.$$
It's easy to prove that $(2)Rightarrow (1)$, but do we have $(1)Rightarrow (2)$ ?






matrices inequality determinant







share|cite|improve this question







New contributor




Mishikumo2019 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I've found two versions of Hadamard's inequality :



(1) If $P$ is a $ntimes n$ positive-semidefinite matrix, then :
$$det(P)leprod_{i=1}^n p_{ii} .$$
(2) For any $M$ is a $ntimes n$ matrix, then :
$$|det(M)|leprod_{i=1}^n ||m_i || =prod_{i=1}^n left(sum_{j=1}^n |a_{ij}|^2right)^{1/2}.$$
It's easy to prove that $(2)Rightarrow (1)$, but do we have $(1)Rightarrow (2)$ ?






matrices inequality determinant




matrices inequality determinant






share|cite|improve this question







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Mishikumo2019 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 1




    Yes it does. Let $M$ be any matrix. Then $MM^t$ is a positive semi-definite matrix.
    – mouthetics
    yesterday














  • 1




    Yes it does. Let $M$ be any matrix. Then $MM^t$ is a positive semi-definite matrix.
    – mouthetics
    yesterday








1




1




Yes it does. Let $M$ be any matrix. Then $MM^t$ is a positive semi-definite matrix.
– mouthetics
yesterday




Yes it does. Let $M$ be any matrix. Then $MM^t$ is a positive semi-definite matrix.
– mouthetics
yesterday










1 Answer
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Using comment of @mouthetics,



From (1) we know that $sqrt{det(MM^T)}leprod_{i=1}^n{MM^T}_{ii}^{1/2}=prod_{i=1}^n left(sum_{j=1}^n |m_{ij}|^2right)^{1/2}=prod_{i=1}^n ||m_i ||$



Since $det(MM^T)=det(M)det(M^T)=det(M)^2$



we have $|det(M)|=sqrt{det(MM^T)}leprod_{i=1}^n ||m_i ||$.






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    Using comment of @mouthetics,



    From (1) we know that $sqrt{det(MM^T)}leprod_{i=1}^n{MM^T}_{ii}^{1/2}=prod_{i=1}^n left(sum_{j=1}^n |m_{ij}|^2right)^{1/2}=prod_{i=1}^n ||m_i ||$



    Since $det(MM^T)=det(M)det(M^T)=det(M)^2$



    we have $|det(M)|=sqrt{det(MM^T)}leprod_{i=1}^n ||m_i ||$.






    share|cite|improve this answer


























      0














      Using comment of @mouthetics,



      From (1) we know that $sqrt{det(MM^T)}leprod_{i=1}^n{MM^T}_{ii}^{1/2}=prod_{i=1}^n left(sum_{j=1}^n |m_{ij}|^2right)^{1/2}=prod_{i=1}^n ||m_i ||$



      Since $det(MM^T)=det(M)det(M^T)=det(M)^2$



      we have $|det(M)|=sqrt{det(MM^T)}leprod_{i=1}^n ||m_i ||$.






      share|cite|improve this answer
























        0












        0








        0






        Using comment of @mouthetics,



        From (1) we know that $sqrt{det(MM^T)}leprod_{i=1}^n{MM^T}_{ii}^{1/2}=prod_{i=1}^n left(sum_{j=1}^n |m_{ij}|^2right)^{1/2}=prod_{i=1}^n ||m_i ||$



        Since $det(MM^T)=det(M)det(M^T)=det(M)^2$



        we have $|det(M)|=sqrt{det(MM^T)}leprod_{i=1}^n ||m_i ||$.






        share|cite|improve this answer












        Using comment of @mouthetics,



        From (1) we know that $sqrt{det(MM^T)}leprod_{i=1}^n{MM^T}_{ii}^{1/2}=prod_{i=1}^n left(sum_{j=1}^n |m_{ij}|^2right)^{1/2}=prod_{i=1}^n ||m_i ||$



        Since $det(MM^T)=det(M)det(M^T)=det(M)^2$



        we have $|det(M)|=sqrt{det(MM^T)}leprod_{i=1}^n ||m_i ||$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 21 hours ago









        Lee

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