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Frobenius Norm of Hadamard Product and Trace
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I'm trying to relate the Frobenius Norm of a Hadamard Product to a trace that does not include another Hadamard Product, if possible. In other words, if A and B are (sxr) matrices, with not all positive values,
$left||Acirc B right||^2_F = $?
I'm trying to relate the sum of the squares of all the entries of the Hadamard product. I know that the sum of all the entries of the Hadamard product are
$sum_i sum_j (Acirc B)_ij = tr(A B^T) $
and also
$left||Acirc B right||^2_F = tr((Acirc B)^T(Acirc B)) $
But I am trying to get $left||Acirc B right||^2_F $ in the form of the trace of some combination of A and B, without a Hadamard product. Even an inequality would help. I've tried various identities and inequalities related to Hadamard Product and Frobenius Norm, but I am not having any luck.
Any suggestions would be appreciated.
matrices hadamard-product
asked Jul 16 at 1:44
Rich
112
add a comment |Â
up vote
1
down vote
favorite
I'm trying to relate the Frobenius Norm of a Hadamard Product to a trace that does not include another Hadamard Product, if possible. In other words, if A and B are (sxr) matrices, with not all positive values,
$left||Acirc B right||^2_F = $?
I'm trying to relate the sum of the squares of all the entries of the Hadamard product. I know that the sum of all the entries of the Hadamard product are
$sum_i sum_j (Acirc B)_ij = tr(A B^T) $
and also
$left||Acirc B right||^2_F = tr((Acirc B)^T(Acirc B)) $
But I am trying to get $left||Acirc B right||^2_F $ in the form of the trace of some combination of A and B, without a Hadamard product. Even an inequality would help. I've tried various identities and inequalities related to Hadamard Product and Frobenius Norm, but I am not having any luck.
Any suggestions would be appreciated.
matrices hadamard-product
asked Jul 16 at 1:44
Rich
112
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm trying to relate the Frobenius Norm of a Hadamard Product to a trace that does not include another Hadamard Product, if possible. In other words, if A and B are (sxr) matrices, with not all positive values,
$left||Acirc B right||^2_F = $?
I'm trying to relate the sum of the squares of all the entries of the Hadamard product. I know that the sum of all the entries of the Hadamard product are
$sum_i sum_j (Acirc B)_ij = tr(A B^T) $
and also
$left||Acirc B right||^2_F = tr((Acirc B)^T(Acirc B)) $
But I am trying to get $left||Acirc B right||^2_F $ in the form of the trace of some combination of A and B, without a Hadamard product. Even an inequality would help. I've tried various identities and inequalities related to Hadamard Product and Frobenius Norm, but I am not having any luck.
Any suggestions would be appreciated.
matrices hadamard-product
asked Jul 16 at 1:44
Rich
112
I'm trying to relate the Frobenius Norm of a Hadamard Product to a trace that does not include another Hadamard Product, if possible. In other words, if A and B are (sxr) matrices, with not all positive values,
$left||Acirc B right||^2_F = $?
I'm trying to relate the sum of the squares of all the entries of the Hadamard product. I know that the sum of all the entries of the Hadamard product are
$sum_i sum_j (Acirc B)_ij = tr(A B^T) $
and also
$left||Acirc B right||^2_F = tr((Acirc B)^T(Acirc B)) $
But I am trying to get $left||Acirc B right||^2_F $ in the form of the trace of some combination of A and B, without a Hadamard product. Even an inequality would help. I've tried various identities and inequalities related to Hadamard Product and Frobenius Norm, but I am not having any luck.
Any suggestions would be appreciated.
matrices hadamard-product
asked Jul 16 at 1:44
Rich
112
asked Jul 16 at 1:44
Rich
112
asked Jul 16 at 1:44
Rich
112
asked Jul 16 at 1:44
Rich
112
112
add a comment |Â
add a comment |Â
1 Answer
1
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oldest
votes
up vote
0
down vote
Let
$$eqalign
D_a &= rm Diag(rm vec(A)) cr
D_b &= rm Diag(rm vec(B)) cr
$$
Then
$$eqalign_F^2
&= $$
answered Jul 16 at 3:19
greg
5,7431715
Interesting solution, and that seems to work. Is there any way to relate this back to some multiple of the original A matrix? This is part of a larger problem I'm working.
– Rich
Jul 18 at 0:36
Note that every element of $A$ lies on the diagonal of $D_a$, and is therefore an eigenvalue of $D_a$. Functions like the determinant and trace are ultimately functions of the eigenvalues. The eigenvalues of a matrix are a different set than the elements of the matrix -- and smaller one since $(n<n^2)$. I don't see any obvious way to connect the eigenvalues with the individual elements. Another problem: simply rearranging the elements of $A$ will change its eigenvalues, whereas the eigenvalues of $D_a$ are unaffected by this.
– greg
Jul 18 at 2:07
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Let
$$eqalign
D_a &= rm Diag(rm vec(A)) cr
D_b &= rm Diag(rm vec(B)) cr
$$
Then
$$eqalign_F^2
&= $$
answered Jul 16 at 3:19
greg
5,7431715
Interesting solution, and that seems to work. Is there any way to relate this back to some multiple of the original A matrix? This is part of a larger problem I'm working.
– Rich
Jul 18 at 0:36
Note that every element of $A$ lies on the diagonal of $D_a$, and is therefore an eigenvalue of $D_a$. Functions like the determinant and trace are ultimately functions of the eigenvalues. The eigenvalues of a matrix are a different set than the elements of the matrix -- and smaller one since $(n<n^2)$. I don't see any obvious way to connect the eigenvalues with the individual elements. Another problem: simply rearranging the elements of $A$ will change its eigenvalues, whereas the eigenvalues of $D_a$ are unaffected by this.
– greg
Jul 18 at 2:07
add a comment |Â
up vote
0
down vote
Let
$$eqalign
D_a &= rm Diag(rm vec(A)) cr
D_b &= rm Diag(rm vec(B)) cr
$$
Then
$$eqalign_F^2
&= $$
answered Jul 16 at 3:19
greg
5,7431715
Interesting solution, and that seems to work. Is there any way to relate this back to some multiple of the original A matrix? This is part of a larger problem I'm working.
– Rich
Jul 18 at 0:36
Note that every element of $A$ lies on the diagonal of $D_a$, and is therefore an eigenvalue of $D_a$. Functions like the determinant and trace are ultimately functions of the eigenvalues. The eigenvalues of a matrix are a different set than the elements of the matrix -- and smaller one since $(n<n^2)$. I don't see any obvious way to connect the eigenvalues with the individual elements. Another problem: simply rearranging the elements of $A$ will change its eigenvalues, whereas the eigenvalues of $D_a$ are unaffected by this.
– greg
Jul 18 at 2:07
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let
$$eqalign
D_a &= rm Diag(rm vec(A)) cr
D_b &= rm Diag(rm vec(B)) cr
$$
Then
$$eqalign_F^2
&= $$
answered Jul 16 at 3:19
greg
5,7431715
Let
$$eqalign
D_a &= rm Diag(rm vec(A)) cr
D_b &= rm Diag(rm vec(B)) cr
$$
Then
$$eqalign_F^2
&= $$
answered Jul 16 at 3:19
greg
5,7431715
answered Jul 16 at 3:19
greg
5,7431715
answered Jul 16 at 3:19
greg
5,7431715
answered Jul 16 at 3:19
greg
5,7431715
5,7431715
Interesting solution, and that seems to work. Is there any way to relate this back to some multiple of the original A matrix? This is part of a larger problem I'm working.
– Rich
Jul 18 at 0:36
Note that every element of $A$ lies on the diagonal of $D_a$, and is therefore an eigenvalue of $D_a$. Functions like the determinant and trace are ultimately functions of the eigenvalues. The eigenvalues of a matrix are a different set than the elements of the matrix -- and smaller one since $(n<n^2)$. I don't see any obvious way to connect the eigenvalues with the individual elements. Another problem: simply rearranging the elements of $A$ will change its eigenvalues, whereas the eigenvalues of $D_a$ are unaffected by this.
– greg
Jul 18 at 2:07
add a comment |Â
Interesting solution, and that seems to work. Is there any way to relate this back to some multiple of the original A matrix? This is part of a larger problem I'm working.
– Rich
Jul 18 at 0:36
Note that every element of $A$ lies on the diagonal of $D_a$, and is therefore an eigenvalue of $D_a$. Functions like the determinant and trace are ultimately functions of the eigenvalues. The eigenvalues of a matrix are a different set than the elements of the matrix -- and smaller one since $(n<n^2)$. I don't see any obvious way to connect the eigenvalues with the individual elements. Another problem: simply rearranging the elements of $A$ will change its eigenvalues, whereas the eigenvalues of $D_a$ are unaffected by this.
– greg
Jul 18 at 2:07
Interesting solution, and that seems to work. Is there any way to relate this back to some multiple of the original A matrix? This is part of a larger problem I'm working.
– Rich
Jul 18 at 0:36
Interesting solution, and that seems to work. Is there any way to relate this back to some multiple of the original A matrix? This is part of a larger problem I'm working.
– Rich
Jul 18 at 0:36
Note that every element of $A$ lies on the diagonal of $D_a$, and is therefore an eigenvalue of $D_a$. Functions like the determinant and trace are ultimately functions of the eigenvalues. The eigenvalues of a matrix are a different set than the elements of the matrix -- and smaller one since $(n<n^2)$. I don't see any obvious way to connect the eigenvalues with the individual elements. Another problem: simply rearranging the elements of $A$ will change its eigenvalues, whereas the eigenvalues of $D_a$ are unaffected by this.
– greg
Jul 18 at 2:07
Note that every element of $A$ lies on the diagonal of $D_a$, and is therefore an eigenvalue of $D_a$. Functions like the determinant and trace are ultimately functions of the eigenvalues. The eigenvalues of a matrix are a different set than the elements of the matrix -- and smaller one since $(n<n^2)$. I don't see any obvious way to connect the eigenvalues with the individual elements. Another problem: simply rearranging the elements of $A$ will change its eigenvalues, whereas the eigenvalues of $D_a$ are unaffected by this.
– greg
Jul 18 at 2:07
add a comment |Â
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