Hadamard's inequality proof
$begingroup$
I have the following inequality to prove.
With $A in M_n(R)$ show that:
$$
(det(A))^2 leq prod_{i=1}^nleft( sum_{k=1}^n A_{k,i}^2right)
$$
What I already have:
I found out that:
$$G(v_1,ldots,v_m) = det(A^T A)=det(A^T)cdotdet(A)=(det(A))^2 $$
Also that with $G(v_1,ldots,v_m) = (det(A))^2$ results that:
$$operatorname{Vol}(v_1,ldots,v_m)= left|det(A)right| =prod_{i=1}^n |s(v_k,u_i)|$$
I don't even now if this is right.. It would be really helpful if someone could help me with the proof..
linear-algebra inequality
edited May 29 '17 at 18:30
Omnomnomnom
127k790178
asked May 29 '17 at 15:58
PhysXPhysX
1328
$endgroup$
add a comment |
$begingroup$
I have the following inequality to prove.
With $A in M_n(R)$ show that:
$$
(det(A))^2 leq prod_{i=1}^nleft( sum_{k=1}^n A_{k,i}^2right)
$$
What I already have:
I found out that:
$$G(v_1,ldots,v_m) = det(A^T A)=det(A^T)cdotdet(A)=(det(A))^2 $$
Also that with $G(v_1,ldots,v_m) = (det(A))^2$ results that:
$$operatorname{Vol}(v_1,ldots,v_m)= left|det(A)right| =prod_{i=1}^n |s(v_k,u_i)|$$
I don't even now if this is right.. It would be really helpful if someone could help me with the proof..
linear-algebra inequality
edited May 29 '17 at 18:30
Omnomnomnom
127k790178
asked May 29 '17 at 15:58
PhysXPhysX
1328
$endgroup$
add a comment |
$begingroup$
I have the following inequality to prove.
With $A in M_n(R)$ show that:
$$
(det(A))^2 leq prod_{i=1}^nleft( sum_{k=1}^n A_{k,i}^2right)
$$
What I already have:
I found out that:
$$G(v_1,ldots,v_m) = det(A^T A)=det(A^T)cdotdet(A)=(det(A))^2 $$
Also that with $G(v_1,ldots,v_m) = (det(A))^2$ results that:
$$operatorname{Vol}(v_1,ldots,v_m)= left|det(A)right| =prod_{i=1}^n |s(v_k,u_i)|$$
I don't even now if this is right.. It would be really helpful if someone could help me with the proof..
linear-algebra inequality
edited May 29 '17 at 18:30
Omnomnomnom
127k790178
asked May 29 '17 at 15:58
PhysXPhysX
1328
$endgroup$
I have the following inequality to prove.
With $A in M_n(R)$ show that:
$$
(det(A))^2 leq prod_{i=1}^nleft( sum_{k=1}^n A_{k,i}^2right)
$$
What I already have:
I found out that:
$$G(v_1,ldots,v_m) = det(A^T A)=det(A^T)cdotdet(A)=(det(A))^2 $$
Also that with $G(v_1,ldots,v_m) = (det(A))^2$ results that:
$$operatorname{Vol}(v_1,ldots,v_m)= left|det(A)right| =prod_{i=1}^n |s(v_k,u_i)|$$
I don't even now if this is right.. It would be really helpful if someone could help me with the proof..
linear-algebra inequality
linear-algebra inequality
edited May 29 '17 at 18:30
Omnomnomnom
127k790178
asked May 29 '17 at 15:58
PhysXPhysX
1328
edited May 29 '17 at 18:30
Omnomnomnom
127k790178
asked May 29 '17 at 15:58
PhysXPhysX
1328
edited May 29 '17 at 18:30
Omnomnomnom
127k790178
edited May 29 '17 at 18:30
Omnomnomnom
127k790178
edited May 29 '17 at 18:30
Omnomnomnom
127k790178
127k790178
asked May 29 '17 at 15:58
PhysXPhysX
1328
asked May 29 '17 at 15:58
PhysXPhysX
1328
asked May 29 '17 at 15:58
PhysXPhysX
1328
1328
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There's a quick proof using the $QR$ decomposition. In particular, note that any matrix $A$ can be written as
$$
A = QR
$$
where $Q$ orthogonal and $R$ is upper triangular (in fact, the columns of $Q$ can be taken as the orthonormal basis attained via the Gram-Schmidt process). Let $q_j$ denote the columns of $Q$, let $a_j$ denote the columns of $j$ (for $j = 1,dots,n$), and let $r_{ij}$ denote the entries of $R$. Then
$$
a_j = sum_{i=1}^j r_{ij}q_j
$$
It follows that
$$
|a_j|^2 = sum_{i=1}^j |r_{ij}|^2|q_j| geq |r_{jj}|^2 implies |r_{jj}| leq |a_{j}|
$$
Finally, we have
$$
|det(A)| = |det(Q)| |det(R)| = 1 cdot left|prod_{j=1}^n r_{jj}right| leq
prod_{j=1}^n |a_j|
$$
as desired.
answered May 29 '17 at 18:19
OmnomnomnomOmnomnomnom
127k790178
$endgroup$
$begingroup$
Thanks! I did read on Wikipedia that you can use QR decomposition, however I was mainly worried because of the (det(A))^2
$endgroup$
– PhysX
May 29 '17 at 18:26
1
$begingroup$
We could have done it using $A^TA$ directly. In particular, note that $A^TA = R^TR$, and from there we can apply the same inequality to $det(R^TR) = det(R)^2$.
$endgroup$
– Omnomnomnom
May 29 '17 at 18:31
$begingroup$
Thanks for the information! It's always good to have an alternative.
$endgroup$
– PhysX
May 29 '17 at 18:33
$begingroup$
can you use QR if it's not being said that $A$ is a orthogonal matrix?
$endgroup$
– Jneven
Jul 16 '18 at 10:07
1
$begingroup$
@Jneven Yes. Any matrix $A$ will have a $QR$ decomposition, as I state in my answer. Perhaps you'd find the wiki page for QR-decomposition helpful.
$endgroup$
– Omnomnomnom
Jul 16 '18 at 10:10
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There's a quick proof using the $QR$ decomposition. In particular, note that any matrix $A$ can be written as
$$
A = QR
$$
where $Q$ orthogonal and $R$ is upper triangular (in fact, the columns of $Q$ can be taken as the orthonormal basis attained via the Gram-Schmidt process). Let $q_j$ denote the columns of $Q$, let $a_j$ denote the columns of $j$ (for $j = 1,dots,n$), and let $r_{ij}$ denote the entries of $R$. Then
$$
a_j = sum_{i=1}^j r_{ij}q_j
$$
It follows that
$$
|a_j|^2 = sum_{i=1}^j |r_{ij}|^2|q_j| geq |r_{jj}|^2 implies |r_{jj}| leq |a_{j}|
$$
Finally, we have
$$
|det(A)| = |det(Q)| |det(R)| = 1 cdot left|prod_{j=1}^n r_{jj}right| leq
prod_{j=1}^n |a_j|
$$
as desired.
answered May 29 '17 at 18:19
OmnomnomnomOmnomnomnom
127k790178
$endgroup$
$begingroup$
Thanks! I did read on Wikipedia that you can use QR decomposition, however I was mainly worried because of the (det(A))^2
$endgroup$
– PhysX
May 29 '17 at 18:26
1
$begingroup$
We could have done it using $A^TA$ directly. In particular, note that $A^TA = R^TR$, and from there we can apply the same inequality to $det(R^TR) = det(R)^2$.
$endgroup$
– Omnomnomnom
May 29 '17 at 18:31
$begingroup$
Thanks for the information! It's always good to have an alternative.
$endgroup$
– PhysX
May 29 '17 at 18:33
$begingroup$
can you use QR if it's not being said that $A$ is a orthogonal matrix?
$endgroup$
– Jneven
Jul 16 '18 at 10:07
1
$begingroup$
@Jneven Yes. Any matrix $A$ will have a $QR$ decomposition, as I state in my answer. Perhaps you'd find the wiki page for QR-decomposition helpful.
$endgroup$
– Omnomnomnom
Jul 16 '18 at 10:10
add a comment |
$begingroup$
There's a quick proof using the $QR$ decomposition. In particular, note that any matrix $A$ can be written as
$$
A = QR
$$
where $Q$ orthogonal and $R$ is upper triangular (in fact, the columns of $Q$ can be taken as the orthonormal basis attained via the Gram-Schmidt process). Let $q_j$ denote the columns of $Q$, let $a_j$ denote the columns of $j$ (for $j = 1,dots,n$), and let $r_{ij}$ denote the entries of $R$. Then
$$
a_j = sum_{i=1}^j r_{ij}q_j
$$
It follows that
$$
|a_j|^2 = sum_{i=1}^j |r_{ij}|^2|q_j| geq |r_{jj}|^2 implies |r_{jj}| leq |a_{j}|
$$
Finally, we have
$$
|det(A)| = |det(Q)| |det(R)| = 1 cdot left|prod_{j=1}^n r_{jj}right| leq
prod_{j=1}^n |a_j|
$$
as desired.
answered May 29 '17 at 18:19
OmnomnomnomOmnomnomnom
127k790178
$endgroup$
$begingroup$
Thanks! I did read on Wikipedia that you can use QR decomposition, however I was mainly worried because of the (det(A))^2
$endgroup$
– PhysX
May 29 '17 at 18:26
1
$begingroup$
We could have done it using $A^TA$ directly. In particular, note that $A^TA = R^TR$, and from there we can apply the same inequality to $det(R^TR) = det(R)^2$.
$endgroup$
– Omnomnomnom
May 29 '17 at 18:31
$begingroup$
Thanks for the information! It's always good to have an alternative.
$endgroup$
– PhysX
May 29 '17 at 18:33
$begingroup$
can you use QR if it's not being said that $A$ is a orthogonal matrix?
$endgroup$
– Jneven
Jul 16 '18 at 10:07
1
$begingroup$
@Jneven Yes. Any matrix $A$ will have a $QR$ decomposition, as I state in my answer. Perhaps you'd find the wiki page for QR-decomposition helpful.
$endgroup$
– Omnomnomnom
Jul 16 '18 at 10:10
add a comment |
$begingroup$
There's a quick proof using the $QR$ decomposition. In particular, note that any matrix $A$ can be written as
$$
A = QR
$$
where $Q$ orthogonal and $R$ is upper triangular (in fact, the columns of $Q$ can be taken as the orthonormal basis attained via the Gram-Schmidt process). Let $q_j$ denote the columns of $Q$, let $a_j$ denote the columns of $j$ (for $j = 1,dots,n$), and let $r_{ij}$ denote the entries of $R$. Then
$$
a_j = sum_{i=1}^j r_{ij}q_j
$$
It follows that
$$
|a_j|^2 = sum_{i=1}^j |r_{ij}|^2|q_j| geq |r_{jj}|^2 implies |r_{jj}| leq |a_{j}|
$$
Finally, we have
$$
|det(A)| = |det(Q)| |det(R)| = 1 cdot left|prod_{j=1}^n r_{jj}right| leq
prod_{j=1}^n |a_j|
$$
as desired.
answered May 29 '17 at 18:19
OmnomnomnomOmnomnomnom
127k790178
$endgroup$
There's a quick proof using the $QR$ decomposition. In particular, note that any matrix $A$ can be written as
$$
A = QR
$$
where $Q$ orthogonal and $R$ is upper triangular (in fact, the columns of $Q$ can be taken as the orthonormal basis attained via the Gram-Schmidt process). Let $q_j$ denote the columns of $Q$, let $a_j$ denote the columns of $j$ (for $j = 1,dots,n$), and let $r_{ij}$ denote the entries of $R$. Then
$$
a_j = sum_{i=1}^j r_{ij}q_j
$$
It follows that
$$
|a_j|^2 = sum_{i=1}^j |r_{ij}|^2|q_j| geq |r_{jj}|^2 implies |r_{jj}| leq |a_{j}|
$$
Finally, we have
$$
|det(A)| = |det(Q)| |det(R)| = 1 cdot left|prod_{j=1}^n r_{jj}right| leq
prod_{j=1}^n |a_j|
$$
as desired.
answered May 29 '17 at 18:19
OmnomnomnomOmnomnomnom
127k790178
edited May 29 '17 at 18:28
answered May 29 '17 at 18:19
OmnomnomnomOmnomnomnom
127k790178
answered May 29 '17 at 18:19
OmnomnomnomOmnomnomnom
127k790178
answered May 29 '17 at 18:19
OmnomnomnomOmnomnomnom
127k790178
127k790178
$begingroup$
Thanks! I did read on Wikipedia that you can use QR decomposition, however I was mainly worried because of the (det(A))^2
$endgroup$
– PhysX
May 29 '17 at 18:26
1
$begingroup$
We could have done it using $A^TA$ directly. In particular, note that $A^TA = R^TR$, and from there we can apply the same inequality to $det(R^TR) = det(R)^2$.
$endgroup$
– Omnomnomnom
May 29 '17 at 18:31
$begingroup$
Thanks for the information! It's always good to have an alternative.
$endgroup$
– PhysX
May 29 '17 at 18:33
$begingroup$
can you use QR if it's not being said that $A$ is a orthogonal matrix?
$endgroup$
– Jneven
Jul 16 '18 at 10:07
1
$begingroup$
@Jneven Yes. Any matrix $A$ will have a $QR$ decomposition, as I state in my answer. Perhaps you'd find the wiki page for QR-decomposition helpful.
$endgroup$
– Omnomnomnom
Jul 16 '18 at 10:10
add a comment |
$begingroup$
Thanks! I did read on Wikipedia that you can use QR decomposition, however I was mainly worried because of the (det(A))^2
$endgroup$
– PhysX
May 29 '17 at 18:26
1
$begingroup$
We could have done it using $A^TA$ directly. In particular, note that $A^TA = R^TR$, and from there we can apply the same inequality to $det(R^TR) = det(R)^2$.
$endgroup$
– Omnomnomnom
May 29 '17 at 18:31
$begingroup$
Thanks for the information! It's always good to have an alternative.
$endgroup$
– PhysX
May 29 '17 at 18:33
$begingroup$
can you use QR if it's not being said that $A$ is a orthogonal matrix?
$endgroup$
– Jneven
Jul 16 '18 at 10:07
1
$begingroup$
@Jneven Yes. Any matrix $A$ will have a $QR$ decomposition, as I state in my answer. Perhaps you'd find the wiki page for QR-decomposition helpful.
$endgroup$
– Omnomnomnom
Jul 16 '18 at 10:10
$begingroup$
Thanks! I did read on Wikipedia that you can use QR decomposition, however I was mainly worried because of the (det(A))^2
$endgroup$
– PhysX
May 29 '17 at 18:26
$begingroup$
Thanks! I did read on Wikipedia that you can use QR decomposition, however I was mainly worried because of the (det(A))^2
$endgroup$
– PhysX
May 29 '17 at 18:26
1
1
$begingroup$
We could have done it using $A^TA$ directly. In particular, note that $A^TA = R^TR$, and from there we can apply the same inequality to $det(R^TR) = det(R)^2$.
$endgroup$
– Omnomnomnom
May 29 '17 at 18:31
$begingroup$
We could have done it using $A^TA$ directly. In particular, note that $A^TA = R^TR$, and from there we can apply the same inequality to $det(R^TR) = det(R)^2$.
$endgroup$
– Omnomnomnom
May 29 '17 at 18:31
$begingroup$
Thanks for the information! It's always good to have an alternative.
$endgroup$
– PhysX
May 29 '17 at 18:33
$begingroup$
Thanks for the information! It's always good to have an alternative.
$endgroup$
– PhysX
May 29 '17 at 18:33
$begingroup$
can you use QR if it's not being said that $A$ is a orthogonal matrix?
$endgroup$
– Jneven
Jul 16 '18 at 10:07
$begingroup$
can you use QR if it's not being said that $A$ is a orthogonal matrix?
$endgroup$
– Jneven
Jul 16 '18 at 10:07
1
1
$begingroup$
@Jneven Yes. Any matrix $A$ will have a $QR$ decomposition, as I state in my answer. Perhaps you'd find the wiki page for QR-decomposition helpful.
$endgroup$
– Omnomnomnom
Jul 16 '18 at 10:10
$begingroup$
@Jneven Yes. Any matrix $A$ will have a $QR$ decomposition, as I state in my answer. Perhaps you'd find the wiki page for QR-decomposition helpful.
$endgroup$
– Omnomnomnom
Jul 16 '18 at 10:10
add a comment |
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