Link between Hadamard's inequality for positive-definite matrices and general Hadamard's inequality
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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
asked Jan 3 at 13:26
Mishikumo2019Mishikumo2019
83
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add a comment |
$begingroup$
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
asked Jan 3 at 13:26
Mishikumo2019Mishikumo2019
83
$endgroup$
1
$begingroup$
Yes it does. Let $M$ be any matrix. Then $MM^t$ is a positive semi-definite matrix.
$endgroup$
– mouthetics
Jan 3 at 16:02
add a comment |
$begingroup$
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
asked Jan 3 at 13:26
Mishikumo2019Mishikumo2019
83
$endgroup$
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
asked Jan 3 at 13:26
Mishikumo2019Mishikumo2019
83
asked Jan 3 at 13:26
Mishikumo2019Mishikumo2019
83
asked Jan 3 at 13:26
Mishikumo2019Mishikumo2019
83
asked Jan 3 at 13:26
Mishikumo2019Mishikumo2019
83
asked Jan 3 at 13:26
Mishikumo2019Mishikumo2019
83
83
1
$begingroup$
Yes it does. Let $M$ be any matrix. Then $MM^t$ is a positive semi-definite matrix.
$endgroup$
– mouthetics
Jan 3 at 16:02
add a comment |
1
$begingroup$
Yes it does. Let $M$ be any matrix. Then $MM^t$ is a positive semi-definite matrix.
$endgroup$
– mouthetics
Jan 3 at 16:02
1
1
$begingroup$
Yes it does. Let $M$ be any matrix. Then $MM^t$ is a positive semi-definite matrix.
$endgroup$
– mouthetics
Jan 3 at 16:02
$begingroup$
Yes it does. Let $M$ be any matrix. Then $MM^t$ is a positive semi-definite matrix.
$endgroup$
– mouthetics
Jan 3 at 16:02
add a comment |
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 $|det(M)|=sqrt{det(MM^T)}leprod_{i=1}^n ||m_i ||$.
answered Jan 4 at 7:24
LeeLee
330111
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$begingroup$
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 $|det(M)|=sqrt{det(MM^T)}leprod_{i=1}^n ||m_i ||$.
answered Jan 4 at 7:24
LeeLee
330111
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add a comment |
$begingroup$
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 $|det(M)|=sqrt{det(MM^T)}leprod_{i=1}^n ||m_i ||$.
answered Jan 4 at 7:24
LeeLee
330111
$endgroup$
add a comment |
$begingroup$
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 $|det(M)|=sqrt{det(MM^T)}leprod_{i=1}^n ||m_i ||$.
answered Jan 4 at 7:24
LeeLee
330111
$endgroup$
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 $|det(M)|=sqrt{det(MM^T)}leprod_{i=1}^n ||m_i ||$.
answered Jan 4 at 7:24
LeeLee
330111
answered Jan 4 at 7:24
LeeLee
330111
answered Jan 4 at 7:24
LeeLee
330111
answered Jan 4 at 7:24
LeeLee
330111
330111
add a comment |
add a comment |
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Yes it does. Let $M$ be any matrix. Then $MM^t$ is a positive semi-definite matrix.
$endgroup$
– mouthetics
Jan 3 at 16:02