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Are the same expression kronecker Product and vec operator representation when differentiating a matrix by a...
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When differentiating a matrix by a matrix it is true that dA(X)/dX is the same thing as dvec(A(X))/dvec(X) acorrording to this paper.
So, I compared kronecker product representation with vec operator representation following case.
$mathbf{Y}=(2,3)$, $mathbf{X}=(4,2)$
$$
begin{aligned}
frac{partial mathbf{Y}}{partial mathbf{X}} &= frac{partial }{partial mathbf{X}} otimes mathbf{Y} \[10pt]
&=left[begin{array}{cc}
dfrac{partial }{partial X_{11}} & dfrac{partial }{partial X_{12}} \
dfrac{partial }{partial X_{21}} & dfrac{partial }{partial X_{22}} \
dfrac{partial }{partial X_{31}} & dfrac{partial }{partial X_{32}} \
dfrac{partial }{partial X_{41}} & dfrac{partial }{partial X_{42}}
end{array}right] otimes mathbf{Y} \[10pt]
&= left[begin{array}{ccc|ccc}
dfrac{partial Y_{11}}{partial X_{11}} &
dfrac{partial Y_{12}}{partial X_{11}} &
dfrac{partial Y_{13}}{partial X_{11}} &
dfrac{partial Y_{11}}{partial X_{12}} &
dfrac{partial Y_{12}}{partial X_{12}} &
dfrac{partial Y_{13}}{partial X_{12}} \
dfrac{partial Y_{21}}{partial X_{11}} &
dfrac{partial Y_{22}}{partial X_{11}} &
dfrac{partial Y_{23}}{partial X_{11}} &
dfrac{partial Y_{21}}{partial X_{12}} &
dfrac{partial Y_{22}}{partial X_{12}} &
dfrac{partial Y_{23}}{partial X_{12}} \[10pt] hline
dfrac{partial Y_{11}}{partial X_{21}} &
dfrac{partial Y_{12}}{partial X_{21}} &
dfrac{partial Y_{13}}{partial X_{21}} &
dfrac{partial Y_{11}}{partial X_{22}} &
dfrac{partial Y_{12}}{partial X_{22}} &
dfrac{partial Y_{13}}{partial X_{22}} \
dfrac{partial Y_{21}}{partial X_{21}} &
dfrac{partial Y_{22}}{partial X_{21}} &
dfrac{partial Y_{23}}{partial X_{21}} &
dfrac{partial Y_{21}}{partial X_{22}} &
dfrac{partial Y_{22}}{partial X_{22}} &
dfrac{partial Y_{23}}{partial X_{22}} \[10pt] hline
dfrac{partial Y_{11}}{partial X_{31}} &
dfrac{partial Y_{12}}{partial X_{31}} &
dfrac{partial Y_{13}}{partial X_{31}} &
dfrac{partial Y_{11}}{partial X_{32}} &
dfrac{partial Y_{12}}{partial X_{32}} &
dfrac{partial Y_{13}}{partial X_{32}} \
dfrac{partial Y_{21}}{partial X_{31}} &
dfrac{partial Y_{22}}{partial X_{31}} &
dfrac{partial Y_{23}}{partial X_{31}} &
dfrac{partial Y_{21}}{partial X_{32}} &
dfrac{partial Y_{22}}{partial X_{32}} &
dfrac{partial Y_{23}}{partial X_{32}} \[10pt] hline
dfrac{partial Y_{11}}{partial X_{41}} &
dfrac{partial Y_{12}}{partial X_{41}} &
dfrac{partial Y_{13}}{partial X_{41}} &
dfrac{partial Y_{11}}{partial X_{42}} &
dfrac{partial Y_{12}}{partial X_{42}} &
dfrac{partial Y_{13}}{partial X_{42}} \
dfrac{partial Y_{21}}{partial X_{41}} &
dfrac{partial Y_{22}}{partial X_{41}} &
dfrac{partial Y_{23}}{partial X_{41}} &
dfrac{partial Y_{21}}{partial X_{42}} &
dfrac{partial Y_{22}}{partial X_{42}} &
dfrac{partial Y_{23}}{partial X_{42}}
end{array}
right]
end{aligned}
$$
and
$ text{vec}(mathbf{Y})= mathbf{y}= begin{bmatrix} y_{11} & y_{21} & y_{12} & y_{22} & y_{13} & y_{23} end{bmatrix}^T$
$ text{vec}(mathbf{X})= mathbf{x}= begin{bmatrix} x_{11} & x_{21} & x_{31} & x_{41} & x_{12} & x_{22} & x_{32} & x_{42} end{bmatrix}^T$
$$
begin{aligned}
frac{partial text{vec}(mathbf{Y})}{partial text{vec}(mathbf{X})} &= frac{partial mathbf{y}}{partial mathbf{x}} \
&= left[begin{array}{cccccc}
dfrac{partial Y_{11}}{partial X_{11}} &
dfrac{partial Y_{21}}{partial X_{11}} &
dfrac{partial Y_{12}}{partial X_{11}} &
dfrac{partial Y_{22}}{partial X_{11}} &
dfrac{partial Y_{13}}{partial X_{11}} &
dfrac{partial Y_{23}}{partial X_{11}} \[5pt]
dfrac{partial Y_{11}}{partial X_{21}} &
dfrac{partial Y_{21}}{partial X_{21}} &
dfrac{partial Y_{12}}{partial X_{21}} &
dfrac{partial Y_{22}}{partial X_{21}} &
dfrac{partial Y_{13}}{partial X_{21}} &
dfrac{partial Y_{23}}{partial X_{21}} \[5pt]
dfrac{partial Y_{11}}{partial X_{31}} &
dfrac{partial Y_{21}}{partial X_{31}} &
dfrac{partial Y_{12}}{partial X_{31}} &
dfrac{partial Y_{22}}{partial X_{31}} &
dfrac{partial Y_{13}}{partial X_{31}} &
dfrac{partial Y_{23}}{partial X_{31}} \[5pt]
dfrac{partial Y_{11}}{partial X_{41}} &
dfrac{partial Y_{21}}{partial X_{41}} &
dfrac{partial Y_{12}}{partial X_{41}} &
dfrac{partial Y_{22}}{partial X_{41}} &
dfrac{partial Y_{13}}{partial X_{41}} &
dfrac{partial Y_{23}}{partial X_{41}} \[5pt]
dfrac{partial Y_{11}}{partial X_{12}} &
dfrac{partial Y_{21}}{partial X_{12}} &
dfrac{partial Y_{12}}{partial X_{12}} &
dfrac{partial Y_{22}}{partial X_{12}} &
dfrac{partial Y_{13}}{partial X_{12}} &
dfrac{partial Y_{23}}{partial X_{12}} \[5pt]
dfrac{partial Y_{11}}{partial X_{22}} &
dfrac{partial Y_{21}}{partial X_{22}} &
dfrac{partial Y_{12}}{partial X_{22}} &
dfrac{partial Y_{22}}{partial X_{22}} &
dfrac{partial Y_{13}}{partial X_{22}} &
dfrac{partial Y_{23}}{partial X_{22}} \[5pt]
dfrac{partial Y_{11}}{partial X_{32}} &
dfrac{partial Y_{21}}{partial X_{32}} &
dfrac{partial Y_{12}}{partial X_{32}} &
dfrac{partial Y_{22}}{partial X_{32}} &
dfrac{partial Y_{13}}{partial X_{32}} &
dfrac{partial Y_{23}}{partial X_{32}} \[5pt]
dfrac{partial Y_{11}}{partial X_{42}} &
dfrac{partial Y_{21}}{partial X_{42}} &
dfrac{partial Y_{12}}{partial X_{42}} &
dfrac{partial Y_{22}}{partial X_{42}} &
dfrac{partial Y_{13}}{partial X_{42}} &
dfrac{partial Y_{23}}{partial X_{42}}
end{array}
right]
end{aligned}
$$
If these results are the same, how do I transpose $frac{partial }{partial mathbf{X}} otimes mathbf{Y}$ to make it equal to $frac{partial text{vec}(mathbf{Y})}{partial text{vec}(mathbf{X})}$?
matrix-calculus kronecker-product
asked Nov 12 at 6:47
metamath
235
add a comment |
up vote
0
down vote
favorite
When differentiating a matrix by a matrix it is true that dA(X)/dX is the same thing as dvec(A(X))/dvec(X) acorrording to this paper.
So, I compared kronecker product representation with vec operator representation following case.
$mathbf{Y}=(2,3)$, $mathbf{X}=(4,2)$
$$
begin{aligned}
frac{partial mathbf{Y}}{partial mathbf{X}} &= frac{partial }{partial mathbf{X}} otimes mathbf{Y} \[10pt]
&=left[begin{array}{cc}
dfrac{partial }{partial X_{11}} & dfrac{partial }{partial X_{12}} \
dfrac{partial }{partial X_{21}} & dfrac{partial }{partial X_{22}} \
dfrac{partial }{partial X_{31}} & dfrac{partial }{partial X_{32}} \
dfrac{partial }{partial X_{41}} & dfrac{partial }{partial X_{42}}
end{array}right] otimes mathbf{Y} \[10pt]
&= left[begin{array}{ccc|ccc}
dfrac{partial Y_{11}}{partial X_{11}} &
dfrac{partial Y_{12}}{partial X_{11}} &
dfrac{partial Y_{13}}{partial X_{11}} &
dfrac{partial Y_{11}}{partial X_{12}} &
dfrac{partial Y_{12}}{partial X_{12}} &
dfrac{partial Y_{13}}{partial X_{12}} \
dfrac{partial Y_{21}}{partial X_{11}} &
dfrac{partial Y_{22}}{partial X_{11}} &
dfrac{partial Y_{23}}{partial X_{11}} &
dfrac{partial Y_{21}}{partial X_{12}} &
dfrac{partial Y_{22}}{partial X_{12}} &
dfrac{partial Y_{23}}{partial X_{12}} \[10pt] hline
dfrac{partial Y_{11}}{partial X_{21}} &
dfrac{partial Y_{12}}{partial X_{21}} &
dfrac{partial Y_{13}}{partial X_{21}} &
dfrac{partial Y_{11}}{partial X_{22}} &
dfrac{partial Y_{12}}{partial X_{22}} &
dfrac{partial Y_{13}}{partial X_{22}} \
dfrac{partial Y_{21}}{partial X_{21}} &
dfrac{partial Y_{22}}{partial X_{21}} &
dfrac{partial Y_{23}}{partial X_{21}} &
dfrac{partial Y_{21}}{partial X_{22}} &
dfrac{partial Y_{22}}{partial X_{22}} &
dfrac{partial Y_{23}}{partial X_{22}} \[10pt] hline
dfrac{partial Y_{11}}{partial X_{31}} &
dfrac{partial Y_{12}}{partial X_{31}} &
dfrac{partial Y_{13}}{partial X_{31}} &
dfrac{partial Y_{11}}{partial X_{32}} &
dfrac{partial Y_{12}}{partial X_{32}} &
dfrac{partial Y_{13}}{partial X_{32}} \
dfrac{partial Y_{21}}{partial X_{31}} &
dfrac{partial Y_{22}}{partial X_{31}} &
dfrac{partial Y_{23}}{partial X_{31}} &
dfrac{partial Y_{21}}{partial X_{32}} &
dfrac{partial Y_{22}}{partial X_{32}} &
dfrac{partial Y_{23}}{partial X_{32}} \[10pt] hline
dfrac{partial Y_{11}}{partial X_{41}} &
dfrac{partial Y_{12}}{partial X_{41}} &
dfrac{partial Y_{13}}{partial X_{41}} &
dfrac{partial Y_{11}}{partial X_{42}} &
dfrac{partial Y_{12}}{partial X_{42}} &
dfrac{partial Y_{13}}{partial X_{42}} \
dfrac{partial Y_{21}}{partial X_{41}} &
dfrac{partial Y_{22}}{partial X_{41}} &
dfrac{partial Y_{23}}{partial X_{41}} &
dfrac{partial Y_{21}}{partial X_{42}} &
dfrac{partial Y_{22}}{partial X_{42}} &
dfrac{partial Y_{23}}{partial X_{42}}
end{array}
right]
end{aligned}
$$
and
$ text{vec}(mathbf{Y})= mathbf{y}= begin{bmatrix} y_{11} & y_{21} & y_{12} & y_{22} & y_{13} & y_{23} end{bmatrix}^T$
$ text{vec}(mathbf{X})= mathbf{x}= begin{bmatrix} x_{11} & x_{21} & x_{31} & x_{41} & x_{12} & x_{22} & x_{32} & x_{42} end{bmatrix}^T$
$$
begin{aligned}
frac{partial text{vec}(mathbf{Y})}{partial text{vec}(mathbf{X})} &= frac{partial mathbf{y}}{partial mathbf{x}} \
&= left[begin{array}{cccccc}
dfrac{partial Y_{11}}{partial X_{11}} &
dfrac{partial Y_{21}}{partial X_{11}} &
dfrac{partial Y_{12}}{partial X_{11}} &
dfrac{partial Y_{22}}{partial X_{11}} &
dfrac{partial Y_{13}}{partial X_{11}} &
dfrac{partial Y_{23}}{partial X_{11}} \[5pt]
dfrac{partial Y_{11}}{partial X_{21}} &
dfrac{partial Y_{21}}{partial X_{21}} &
dfrac{partial Y_{12}}{partial X_{21}} &
dfrac{partial Y_{22}}{partial X_{21}} &
dfrac{partial Y_{13}}{partial X_{21}} &
dfrac{partial Y_{23}}{partial X_{21}} \[5pt]
dfrac{partial Y_{11}}{partial X_{31}} &
dfrac{partial Y_{21}}{partial X_{31}} &
dfrac{partial Y_{12}}{partial X_{31}} &
dfrac{partial Y_{22}}{partial X_{31}} &
dfrac{partial Y_{13}}{partial X_{31}} &
dfrac{partial Y_{23}}{partial X_{31}} \[5pt]
dfrac{partial Y_{11}}{partial X_{41}} &
dfrac{partial Y_{21}}{partial X_{41}} &
dfrac{partial Y_{12}}{partial X_{41}} &
dfrac{partial Y_{22}}{partial X_{41}} &
dfrac{partial Y_{13}}{partial X_{41}} &
dfrac{partial Y_{23}}{partial X_{41}} \[5pt]
dfrac{partial Y_{11}}{partial X_{12}} &
dfrac{partial Y_{21}}{partial X_{12}} &
dfrac{partial Y_{12}}{partial X_{12}} &
dfrac{partial Y_{22}}{partial X_{12}} &
dfrac{partial Y_{13}}{partial X_{12}} &
dfrac{partial Y_{23}}{partial X_{12}} \[5pt]
dfrac{partial Y_{11}}{partial X_{22}} &
dfrac{partial Y_{21}}{partial X_{22}} &
dfrac{partial Y_{12}}{partial X_{22}} &
dfrac{partial Y_{22}}{partial X_{22}} &
dfrac{partial Y_{13}}{partial X_{22}} &
dfrac{partial Y_{23}}{partial X_{22}} \[5pt]
dfrac{partial Y_{11}}{partial X_{32}} &
dfrac{partial Y_{21}}{partial X_{32}} &
dfrac{partial Y_{12}}{partial X_{32}} &
dfrac{partial Y_{22}}{partial X_{32}} &
dfrac{partial Y_{13}}{partial X_{32}} &
dfrac{partial Y_{23}}{partial X_{32}} \[5pt]
dfrac{partial Y_{11}}{partial X_{42}} &
dfrac{partial Y_{21}}{partial X_{42}} &
dfrac{partial Y_{12}}{partial X_{42}} &
dfrac{partial Y_{22}}{partial X_{42}} &
dfrac{partial Y_{13}}{partial X_{42}} &
dfrac{partial Y_{23}}{partial X_{42}}
end{array}
right]
end{aligned}
$$
If these results are the same, how do I transpose $frac{partial }{partial mathbf{X}} otimes mathbf{Y}$ to make it equal to $frac{partial text{vec}(mathbf{Y})}{partial text{vec}(mathbf{X})}$?
matrix-calculus kronecker-product
asked Nov 12 at 6:47
metamath
235
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
When differentiating a matrix by a matrix it is true that dA(X)/dX is the same thing as dvec(A(X))/dvec(X) acorrording to this paper.
So, I compared kronecker product representation with vec operator representation following case.
$mathbf{Y}=(2,3)$, $mathbf{X}=(4,2)$
$$
begin{aligned}
frac{partial mathbf{Y}}{partial mathbf{X}} &= frac{partial }{partial mathbf{X}} otimes mathbf{Y} \[10pt]
&=left[begin{array}{cc}
dfrac{partial }{partial X_{11}} & dfrac{partial }{partial X_{12}} \
dfrac{partial }{partial X_{21}} & dfrac{partial }{partial X_{22}} \
dfrac{partial }{partial X_{31}} & dfrac{partial }{partial X_{32}} \
dfrac{partial }{partial X_{41}} & dfrac{partial }{partial X_{42}}
end{array}right] otimes mathbf{Y} \[10pt]
&= left[begin{array}{ccc|ccc}
dfrac{partial Y_{11}}{partial X_{11}} &
dfrac{partial Y_{12}}{partial X_{11}} &
dfrac{partial Y_{13}}{partial X_{11}} &
dfrac{partial Y_{11}}{partial X_{12}} &
dfrac{partial Y_{12}}{partial X_{12}} &
dfrac{partial Y_{13}}{partial X_{12}} \
dfrac{partial Y_{21}}{partial X_{11}} &
dfrac{partial Y_{22}}{partial X_{11}} &
dfrac{partial Y_{23}}{partial X_{11}} &
dfrac{partial Y_{21}}{partial X_{12}} &
dfrac{partial Y_{22}}{partial X_{12}} &
dfrac{partial Y_{23}}{partial X_{12}} \[10pt] hline
dfrac{partial Y_{11}}{partial X_{21}} &
dfrac{partial Y_{12}}{partial X_{21}} &
dfrac{partial Y_{13}}{partial X_{21}} &
dfrac{partial Y_{11}}{partial X_{22}} &
dfrac{partial Y_{12}}{partial X_{22}} &
dfrac{partial Y_{13}}{partial X_{22}} \
dfrac{partial Y_{21}}{partial X_{21}} &
dfrac{partial Y_{22}}{partial X_{21}} &
dfrac{partial Y_{23}}{partial X_{21}} &
dfrac{partial Y_{21}}{partial X_{22}} &
dfrac{partial Y_{22}}{partial X_{22}} &
dfrac{partial Y_{23}}{partial X_{22}} \[10pt] hline
dfrac{partial Y_{11}}{partial X_{31}} &
dfrac{partial Y_{12}}{partial X_{31}} &
dfrac{partial Y_{13}}{partial X_{31}} &
dfrac{partial Y_{11}}{partial X_{32}} &
dfrac{partial Y_{12}}{partial X_{32}} &
dfrac{partial Y_{13}}{partial X_{32}} \
dfrac{partial Y_{21}}{partial X_{31}} &
dfrac{partial Y_{22}}{partial X_{31}} &
dfrac{partial Y_{23}}{partial X_{31}} &
dfrac{partial Y_{21}}{partial X_{32}} &
dfrac{partial Y_{22}}{partial X_{32}} &
dfrac{partial Y_{23}}{partial X_{32}} \[10pt] hline
dfrac{partial Y_{11}}{partial X_{41}} &
dfrac{partial Y_{12}}{partial X_{41}} &
dfrac{partial Y_{13}}{partial X_{41}} &
dfrac{partial Y_{11}}{partial X_{42}} &
dfrac{partial Y_{12}}{partial X_{42}} &
dfrac{partial Y_{13}}{partial X_{42}} \
dfrac{partial Y_{21}}{partial X_{41}} &
dfrac{partial Y_{22}}{partial X_{41}} &
dfrac{partial Y_{23}}{partial X_{41}} &
dfrac{partial Y_{21}}{partial X_{42}} &
dfrac{partial Y_{22}}{partial X_{42}} &
dfrac{partial Y_{23}}{partial X_{42}}
end{array}
right]
end{aligned}
$$
and
$ text{vec}(mathbf{Y})= mathbf{y}= begin{bmatrix} y_{11} & y_{21} & y_{12} & y_{22} & y_{13} & y_{23} end{bmatrix}^T$
$ text{vec}(mathbf{X})= mathbf{x}= begin{bmatrix} x_{11} & x_{21} & x_{31} & x_{41} & x_{12} & x_{22} & x_{32} & x_{42} end{bmatrix}^T$
$$
begin{aligned}
frac{partial text{vec}(mathbf{Y})}{partial text{vec}(mathbf{X})} &= frac{partial mathbf{y}}{partial mathbf{x}} \
&= left[begin{array}{cccccc}
dfrac{partial Y_{11}}{partial X_{11}} &
dfrac{partial Y_{21}}{partial X_{11}} &
dfrac{partial Y_{12}}{partial X_{11}} &
dfrac{partial Y_{22}}{partial X_{11}} &
dfrac{partial Y_{13}}{partial X_{11}} &
dfrac{partial Y_{23}}{partial X_{11}} \[5pt]
dfrac{partial Y_{11}}{partial X_{21}} &
dfrac{partial Y_{21}}{partial X_{21}} &
dfrac{partial Y_{12}}{partial X_{21}} &
dfrac{partial Y_{22}}{partial X_{21}} &
dfrac{partial Y_{13}}{partial X_{21}} &
dfrac{partial Y_{23}}{partial X_{21}} \[5pt]
dfrac{partial Y_{11}}{partial X_{31}} &
dfrac{partial Y_{21}}{partial X_{31}} &
dfrac{partial Y_{12}}{partial X_{31}} &
dfrac{partial Y_{22}}{partial X_{31}} &
dfrac{partial Y_{13}}{partial X_{31}} &
dfrac{partial Y_{23}}{partial X_{31}} \[5pt]
dfrac{partial Y_{11}}{partial X_{41}} &
dfrac{partial Y_{21}}{partial X_{41}} &
dfrac{partial Y_{12}}{partial X_{41}} &
dfrac{partial Y_{22}}{partial X_{41}} &
dfrac{partial Y_{13}}{partial X_{41}} &
dfrac{partial Y_{23}}{partial X_{41}} \[5pt]
dfrac{partial Y_{11}}{partial X_{12}} &
dfrac{partial Y_{21}}{partial X_{12}} &
dfrac{partial Y_{12}}{partial X_{12}} &
dfrac{partial Y_{22}}{partial X_{12}} &
dfrac{partial Y_{13}}{partial X_{12}} &
dfrac{partial Y_{23}}{partial X_{12}} \[5pt]
dfrac{partial Y_{11}}{partial X_{22}} &
dfrac{partial Y_{21}}{partial X_{22}} &
dfrac{partial Y_{12}}{partial X_{22}} &
dfrac{partial Y_{22}}{partial X_{22}} &
dfrac{partial Y_{13}}{partial X_{22}} &
dfrac{partial Y_{23}}{partial X_{22}} \[5pt]
dfrac{partial Y_{11}}{partial X_{32}} &
dfrac{partial Y_{21}}{partial X_{32}} &
dfrac{partial Y_{12}}{partial X_{32}} &
dfrac{partial Y_{22}}{partial X_{32}} &
dfrac{partial Y_{13}}{partial X_{32}} &
dfrac{partial Y_{23}}{partial X_{32}} \[5pt]
dfrac{partial Y_{11}}{partial X_{42}} &
dfrac{partial Y_{21}}{partial X_{42}} &
dfrac{partial Y_{12}}{partial X_{42}} &
dfrac{partial Y_{22}}{partial X_{42}} &
dfrac{partial Y_{13}}{partial X_{42}} &
dfrac{partial Y_{23}}{partial X_{42}}
end{array}
right]
end{aligned}
$$
If these results are the same, how do I transpose $frac{partial }{partial mathbf{X}} otimes mathbf{Y}$ to make it equal to $frac{partial text{vec}(mathbf{Y})}{partial text{vec}(mathbf{X})}$?
matrix-calculus kronecker-product
asked Nov 12 at 6:47
metamath
235
When differentiating a matrix by a matrix it is true that dA(X)/dX is the same thing as dvec(A(X))/dvec(X) acorrording to this paper.
So, I compared kronecker product representation with vec operator representation following case.
$mathbf{Y}=(2,3)$, $mathbf{X}=(4,2)$
$$
begin{aligned}
frac{partial mathbf{Y}}{partial mathbf{X}} &= frac{partial }{partial mathbf{X}} otimes mathbf{Y} \[10pt]
&=left[begin{array}{cc}
dfrac{partial }{partial X_{11}} & dfrac{partial }{partial X_{12}} \
dfrac{partial }{partial X_{21}} & dfrac{partial }{partial X_{22}} \
dfrac{partial }{partial X_{31}} & dfrac{partial }{partial X_{32}} \
dfrac{partial }{partial X_{41}} & dfrac{partial }{partial X_{42}}
end{array}right] otimes mathbf{Y} \[10pt]
&= left[begin{array}{ccc|ccc}
dfrac{partial Y_{11}}{partial X_{11}} &
dfrac{partial Y_{12}}{partial X_{11}} &
dfrac{partial Y_{13}}{partial X_{11}} &
dfrac{partial Y_{11}}{partial X_{12}} &
dfrac{partial Y_{12}}{partial X_{12}} &
dfrac{partial Y_{13}}{partial X_{12}} \
dfrac{partial Y_{21}}{partial X_{11}} &
dfrac{partial Y_{22}}{partial X_{11}} &
dfrac{partial Y_{23}}{partial X_{11}} &
dfrac{partial Y_{21}}{partial X_{12}} &
dfrac{partial Y_{22}}{partial X_{12}} &
dfrac{partial Y_{23}}{partial X_{12}} \[10pt] hline
dfrac{partial Y_{11}}{partial X_{21}} &
dfrac{partial Y_{12}}{partial X_{21}} &
dfrac{partial Y_{13}}{partial X_{21}} &
dfrac{partial Y_{11}}{partial X_{22}} &
dfrac{partial Y_{12}}{partial X_{22}} &
dfrac{partial Y_{13}}{partial X_{22}} \
dfrac{partial Y_{21}}{partial X_{21}} &
dfrac{partial Y_{22}}{partial X_{21}} &
dfrac{partial Y_{23}}{partial X_{21}} &
dfrac{partial Y_{21}}{partial X_{22}} &
dfrac{partial Y_{22}}{partial X_{22}} &
dfrac{partial Y_{23}}{partial X_{22}} \[10pt] hline
dfrac{partial Y_{11}}{partial X_{31}} &
dfrac{partial Y_{12}}{partial X_{31}} &
dfrac{partial Y_{13}}{partial X_{31}} &
dfrac{partial Y_{11}}{partial X_{32}} &
dfrac{partial Y_{12}}{partial X_{32}} &
dfrac{partial Y_{13}}{partial X_{32}} \
dfrac{partial Y_{21}}{partial X_{31}} &
dfrac{partial Y_{22}}{partial X_{31}} &
dfrac{partial Y_{23}}{partial X_{31}} &
dfrac{partial Y_{21}}{partial X_{32}} &
dfrac{partial Y_{22}}{partial X_{32}} &
dfrac{partial Y_{23}}{partial X_{32}} \[10pt] hline
dfrac{partial Y_{11}}{partial X_{41}} &
dfrac{partial Y_{12}}{partial X_{41}} &
dfrac{partial Y_{13}}{partial X_{41}} &
dfrac{partial Y_{11}}{partial X_{42}} &
dfrac{partial Y_{12}}{partial X_{42}} &
dfrac{partial Y_{13}}{partial X_{42}} \
dfrac{partial Y_{21}}{partial X_{41}} &
dfrac{partial Y_{22}}{partial X_{41}} &
dfrac{partial Y_{23}}{partial X_{41}} &
dfrac{partial Y_{21}}{partial X_{42}} &
dfrac{partial Y_{22}}{partial X_{42}} &
dfrac{partial Y_{23}}{partial X_{42}}
end{array}
right]
end{aligned}
$$
and
$ text{vec}(mathbf{Y})= mathbf{y}= begin{bmatrix} y_{11} & y_{21} & y_{12} & y_{22} & y_{13} & y_{23} end{bmatrix}^T$
$ text{vec}(mathbf{X})= mathbf{x}= begin{bmatrix} x_{11} & x_{21} & x_{31} & x_{41} & x_{12} & x_{22} & x_{32} & x_{42} end{bmatrix}^T$
$$
begin{aligned}
frac{partial text{vec}(mathbf{Y})}{partial text{vec}(mathbf{X})} &= frac{partial mathbf{y}}{partial mathbf{x}} \
&= left[begin{array}{cccccc}
dfrac{partial Y_{11}}{partial X_{11}} &
dfrac{partial Y_{21}}{partial X_{11}} &
dfrac{partial Y_{12}}{partial X_{11}} &
dfrac{partial Y_{22}}{partial X_{11}} &
dfrac{partial Y_{13}}{partial X_{11}} &
dfrac{partial Y_{23}}{partial X_{11}} \[5pt]
dfrac{partial Y_{11}}{partial X_{21}} &
dfrac{partial Y_{21}}{partial X_{21}} &
dfrac{partial Y_{12}}{partial X_{21}} &
dfrac{partial Y_{22}}{partial X_{21}} &
dfrac{partial Y_{13}}{partial X_{21}} &
dfrac{partial Y_{23}}{partial X_{21}} \[5pt]
dfrac{partial Y_{11}}{partial X_{31}} &
dfrac{partial Y_{21}}{partial X_{31}} &
dfrac{partial Y_{12}}{partial X_{31}} &
dfrac{partial Y_{22}}{partial X_{31}} &
dfrac{partial Y_{13}}{partial X_{31}} &
dfrac{partial Y_{23}}{partial X_{31}} \[5pt]
dfrac{partial Y_{11}}{partial X_{41}} &
dfrac{partial Y_{21}}{partial X_{41}} &
dfrac{partial Y_{12}}{partial X_{41}} &
dfrac{partial Y_{22}}{partial X_{41}} &
dfrac{partial Y_{13}}{partial X_{41}} &
dfrac{partial Y_{23}}{partial X_{41}} \[5pt]
dfrac{partial Y_{11}}{partial X_{12}} &
dfrac{partial Y_{21}}{partial X_{12}} &
dfrac{partial Y_{12}}{partial X_{12}} &
dfrac{partial Y_{22}}{partial X_{12}} &
dfrac{partial Y_{13}}{partial X_{12}} &
dfrac{partial Y_{23}}{partial X_{12}} \[5pt]
dfrac{partial Y_{11}}{partial X_{22}} &
dfrac{partial Y_{21}}{partial X_{22}} &
dfrac{partial Y_{12}}{partial X_{22}} &
dfrac{partial Y_{22}}{partial X_{22}} &
dfrac{partial Y_{13}}{partial X_{22}} &
dfrac{partial Y_{23}}{partial X_{22}} \[5pt]
dfrac{partial Y_{11}}{partial X_{32}} &
dfrac{partial Y_{21}}{partial X_{32}} &
dfrac{partial Y_{12}}{partial X_{32}} &
dfrac{partial Y_{22}}{partial X_{32}} &
dfrac{partial Y_{13}}{partial X_{32}} &
dfrac{partial Y_{23}}{partial X_{32}} \[5pt]
dfrac{partial Y_{11}}{partial X_{42}} &
dfrac{partial Y_{21}}{partial X_{42}} &
dfrac{partial Y_{12}}{partial X_{42}} &
dfrac{partial Y_{22}}{partial X_{42}} &
dfrac{partial Y_{13}}{partial X_{42}} &
dfrac{partial Y_{23}}{partial X_{42}}
end{array}
right]
end{aligned}
$$
If these results are the same, how do I transpose $frac{partial }{partial mathbf{X}} otimes mathbf{Y}$ to make it equal to $frac{partial text{vec}(mathbf{Y})}{partial text{vec}(mathbf{X})}$?
matrix-calculus kronecker-product
matrix-calculus kronecker-product
asked Nov 12 at 6:47
metamath
235
asked Nov 12 at 6:47
metamath
235
asked Nov 12 at 6:47
metamath
235
asked Nov 12 at 6:47
metamath
235
asked Nov 12 at 6:47
metamath
235
235
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The derivative operation is a distraction, so let's omit it.
Given the matrices
$$eqalign{
X &in {mathbb R}^{mtimes n} cr
Y &in {mathbb R}^{ptimes q} cr
}$$
and their Kronecker products
$$eqalign{
A &= Xotimes Y &in {mathbb R}^{mptimes nq} cr
B &= {rm vec}(X)otimes {rm vec}(Y)^T &in {mathbb R}^{mntimes pq}cr
}$$
you want to know how to transform $A implies B$
The elements in these matrices are identical. The only difference is in the matrix dimensions and in the ordering of the elements.
Vectorizing the matrices yields two vectors related by a permutation.
$$eqalign{
P,{rm vec}(A) &= {rm vec}(B) cr
Pa &= b cr
P_{ij},a_j &= b_i cr
}$$
The elements of the permuation are easy to find:
$P_{ij}$ equals one if the $j^{th}$ element of $a$ maps to the $i^{th}$ element of $b$.
The single-entry matrix $E_{mn}$ has all elements equal to zero, except for the $(m,n)$ element which is equal to one. Replacing $(X,Y)$ by single-entry matrices leads to $(A,B)$ becoming single-entry matrices and $(a,b)$ becoming single-entry vectors.
The non-zero elements of $P$ can be found by iterating through every possible single-entry matrix for the input matrices $X ,{rm and}, Y$. Then setting $,P_{ij}=1$, where the non-zero elements of the output vectors $(b,a)$ are observed at $(b_i,a_j)$.
The permutation depends only on the dimensions of the starting matrices. For your derivative expressions the dimensions are $(m,n,p,q)=(4,2,2,3)$ and the transformation is
$${rm vec}Bigg(frac{partial{rm vec}(Y)}{partial{rm vec}(X)}Bigg) = P,,{rm vec}Bigg(frac{partial}{partial X}otimes YBigg) $$
answered Nov 16 at 13:35
greg
7,0701719
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The derivative operation is a distraction, so let's omit it.
Given the matrices
$$eqalign{
X &in {mathbb R}^{mtimes n} cr
Y &in {mathbb R}^{ptimes q} cr
}$$
and their Kronecker products
$$eqalign{
A &= Xotimes Y &in {mathbb R}^{mptimes nq} cr
B &= {rm vec}(X)otimes {rm vec}(Y)^T &in {mathbb R}^{mntimes pq}cr
}$$
you want to know how to transform $A implies B$
The elements in these matrices are identical. The only difference is in the matrix dimensions and in the ordering of the elements.
Vectorizing the matrices yields two vectors related by a permutation.
$$eqalign{
P,{rm vec}(A) &= {rm vec}(B) cr
Pa &= b cr
P_{ij},a_j &= b_i cr
}$$
The elements of the permuation are easy to find:
$P_{ij}$ equals one if the $j^{th}$ element of $a$ maps to the $i^{th}$ element of $b$.
The single-entry matrix $E_{mn}$ has all elements equal to zero, except for the $(m,n)$ element which is equal to one. Replacing $(X,Y)$ by single-entry matrices leads to $(A,B)$ becoming single-entry matrices and $(a,b)$ becoming single-entry vectors.
The non-zero elements of $P$ can be found by iterating through every possible single-entry matrix for the input matrices $X ,{rm and}, Y$. Then setting $,P_{ij}=1$, where the non-zero elements of the output vectors $(b,a)$ are observed at $(b_i,a_j)$.
The permutation depends only on the dimensions of the starting matrices. For your derivative expressions the dimensions are $(m,n,p,q)=(4,2,2,3)$ and the transformation is
$${rm vec}Bigg(frac{partial{rm vec}(Y)}{partial{rm vec}(X)}Bigg) = P,,{rm vec}Bigg(frac{partial}{partial X}otimes YBigg) $$
answered Nov 16 at 13:35
greg
7,0701719
add a comment |
up vote
0
down vote
The derivative operation is a distraction, so let's omit it.
Given the matrices
$$eqalign{
X &in {mathbb R}^{mtimes n} cr
Y &in {mathbb R}^{ptimes q} cr
}$$
and their Kronecker products
$$eqalign{
A &= Xotimes Y &in {mathbb R}^{mptimes nq} cr
B &= {rm vec}(X)otimes {rm vec}(Y)^T &in {mathbb R}^{mntimes pq}cr
}$$
you want to know how to transform $A implies B$
The elements in these matrices are identical. The only difference is in the matrix dimensions and in the ordering of the elements.
Vectorizing the matrices yields two vectors related by a permutation.
$$eqalign{
P,{rm vec}(A) &= {rm vec}(B) cr
Pa &= b cr
P_{ij},a_j &= b_i cr
}$$
The elements of the permuation are easy to find:
$P_{ij}$ equals one if the $j^{th}$ element of $a$ maps to the $i^{th}$ element of $b$.
The single-entry matrix $E_{mn}$ has all elements equal to zero, except for the $(m,n)$ element which is equal to one. Replacing $(X,Y)$ by single-entry matrices leads to $(A,B)$ becoming single-entry matrices and $(a,b)$ becoming single-entry vectors.
The non-zero elements of $P$ can be found by iterating through every possible single-entry matrix for the input matrices $X ,{rm and}, Y$. Then setting $,P_{ij}=1$, where the non-zero elements of the output vectors $(b,a)$ are observed at $(b_i,a_j)$.
The permutation depends only on the dimensions of the starting matrices. For your derivative expressions the dimensions are $(m,n,p,q)=(4,2,2,3)$ and the transformation is
$${rm vec}Bigg(frac{partial{rm vec}(Y)}{partial{rm vec}(X)}Bigg) = P,,{rm vec}Bigg(frac{partial}{partial X}otimes YBigg) $$
answered Nov 16 at 13:35
greg
7,0701719
add a comment |
up vote
0
down vote
up vote
0
down vote
The derivative operation is a distraction, so let's omit it.
Given the matrices
$$eqalign{
X &in {mathbb R}^{mtimes n} cr
Y &in {mathbb R}^{ptimes q} cr
}$$
and their Kronecker products
$$eqalign{
A &= Xotimes Y &in {mathbb R}^{mptimes nq} cr
B &= {rm vec}(X)otimes {rm vec}(Y)^T &in {mathbb R}^{mntimes pq}cr
}$$
you want to know how to transform $A implies B$
The elements in these matrices are identical. The only difference is in the matrix dimensions and in the ordering of the elements.
Vectorizing the matrices yields two vectors related by a permutation.
$$eqalign{
P,{rm vec}(A) &= {rm vec}(B) cr
Pa &= b cr
P_{ij},a_j &= b_i cr
}$$
The elements of the permuation are easy to find:
$P_{ij}$ equals one if the $j^{th}$ element of $a$ maps to the $i^{th}$ element of $b$.
The single-entry matrix $E_{mn}$ has all elements equal to zero, except for the $(m,n)$ element which is equal to one. Replacing $(X,Y)$ by single-entry matrices leads to $(A,B)$ becoming single-entry matrices and $(a,b)$ becoming single-entry vectors.
The non-zero elements of $P$ can be found by iterating through every possible single-entry matrix for the input matrices $X ,{rm and}, Y$. Then setting $,P_{ij}=1$, where the non-zero elements of the output vectors $(b,a)$ are observed at $(b_i,a_j)$.
The permutation depends only on the dimensions of the starting matrices. For your derivative expressions the dimensions are $(m,n,p,q)=(4,2,2,3)$ and the transformation is
$${rm vec}Bigg(frac{partial{rm vec}(Y)}{partial{rm vec}(X)}Bigg) = P,,{rm vec}Bigg(frac{partial}{partial X}otimes YBigg) $$
answered Nov 16 at 13:35
greg
7,0701719
The derivative operation is a distraction, so let's omit it.
Given the matrices
$$eqalign{
X &in {mathbb R}^{mtimes n} cr
Y &in {mathbb R}^{ptimes q} cr
}$$
and their Kronecker products
$$eqalign{
A &= Xotimes Y &in {mathbb R}^{mptimes nq} cr
B &= {rm vec}(X)otimes {rm vec}(Y)^T &in {mathbb R}^{mntimes pq}cr
}$$
you want to know how to transform $A implies B$
The elements in these matrices are identical. The only difference is in the matrix dimensions and in the ordering of the elements.
Vectorizing the matrices yields two vectors related by a permutation.
$$eqalign{
P,{rm vec}(A) &= {rm vec}(B) cr
Pa &= b cr
P_{ij},a_j &= b_i cr
}$$
The elements of the permuation are easy to find:
$P_{ij}$ equals one if the $j^{th}$ element of $a$ maps to the $i^{th}$ element of $b$.
The single-entry matrix $E_{mn}$ has all elements equal to zero, except for the $(m,n)$ element which is equal to one. Replacing $(X,Y)$ by single-entry matrices leads to $(A,B)$ becoming single-entry matrices and $(a,b)$ becoming single-entry vectors.
The non-zero elements of $P$ can be found by iterating through every possible single-entry matrix for the input matrices $X ,{rm and}, Y$. Then setting $,P_{ij}=1$, where the non-zero elements of the output vectors $(b,a)$ are observed at $(b_i,a_j)$.
The permutation depends only on the dimensions of the starting matrices. For your derivative expressions the dimensions are $(m,n,p,q)=(4,2,2,3)$ and the transformation is
$${rm vec}Bigg(frac{partial{rm vec}(Y)}{partial{rm vec}(X)}Bigg) = P,,{rm vec}Bigg(frac{partial}{partial X}otimes YBigg) $$
answered Nov 16 at 13:35
greg
7,0701719
edited yesterday
answered Nov 16 at 13:35
greg
7,0701719
answered Nov 16 at 13:35
greg
7,0701719
answered Nov 16 at 13:35
greg
7,0701719
7,0701719
add a comment |
add a comment |
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