Basic property of kronecker delta $( A otimes B)(A^-1 otimes B^-1)$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to prove $e^A oplus B = e^A otimes e^B$ where $A$ and $B$ are matrices? (Kronecker operations)Notation for the direct sum and Kronecker sum of matricesNonsingularity of submatricesMatrices with the sign pattern of the Kronecker sumTensor Product: Use of Universal Mapping PropertyAre there non-square matrices that are both left and right invertible?Is $mathcalVotimesmathcalW$ defined as a double dual?Matrices Commuting with a Kronecker SumMatrix algebras are a tensor products?Why is the Kronecker sum defined for square matrices?
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Basic property of kronecker delta $( A otimes B)(A^-1 otimes B^-1)$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to prove $e^A oplus B = e^A otimes e^B$ where $A$ and $B$ are matrices? (Kronecker operations)Notation for the direct sum and Kronecker sum of matricesNonsingularity of submatricesMatrices with the sign pattern of the Kronecker sumTensor Product: Use of Universal Mapping PropertyAre there non-square matrices that are both left and right invertible?Is $mathcalVotimesmathcalW$ defined as a double dual?Matrices Commuting with a Kronecker SumMatrix algebras are a tensor products?Why is the Kronecker sum defined for square matrices?
$begingroup$
Given non singular matrices $A_n times n,B_m times m$
$$
( A otimes B)(A^-1 otimes B^-1) = (AA^-1) otimes (BB^-1) = I_n otimes I_m = I_(nm times nm )
$$
I was just reading through mathematical primer for social statistics by John Fox
and saw this on page 17, it wasn't clear to me why this is true though.
linear-algebra tensor-products kronecker-product
asked Mar 25 at 21:15
baxxbaxx
413311
$endgroup$
add a comment |
$begingroup$
Given non singular matrices $A_n times n,B_m times m$
$$
( A otimes B)(A^-1 otimes B^-1) = (AA^-1) otimes (BB^-1) = I_n otimes I_m = I_(nm times nm )
$$
I was just reading through mathematical primer for social statistics by John Fox
and saw this on page 17, it wasn't clear to me why this is true though.
linear-algebra tensor-products kronecker-product
asked Mar 25 at 21:15
baxxbaxx
413311
$endgroup$
$begingroup$
Which part isn't clear? It all follows from how you multiply tensor products of matrices.
$endgroup$
– Adam Latosiński
Mar 25 at 21:25
$begingroup$
It's not explained in the text - i tried computing it long hand with an example, perhaps i messed up a computation because it didn't work out.
$endgroup$
– baxx
Mar 25 at 21:41
add a comment |
$begingroup$
Given non singular matrices $A_n times n,B_m times m$
$$
( A otimes B)(A^-1 otimes B^-1) = (AA^-1) otimes (BB^-1) = I_n otimes I_m = I_(nm times nm )
$$
I was just reading through mathematical primer for social statistics by John Fox
and saw this on page 17, it wasn't clear to me why this is true though.
linear-algebra tensor-products kronecker-product
asked Mar 25 at 21:15
baxxbaxx
413311
$endgroup$
Given non singular matrices $A_n times n,B_m times m$
$$
( A otimes B)(A^-1 otimes B^-1) = (AA^-1) otimes (BB^-1) = I_n otimes I_m = I_(nm times nm )
$$
I was just reading through mathematical primer for social statistics by John Fox
and saw this on page 17, it wasn't clear to me why this is true though.
linear-algebra tensor-products kronecker-product
linear-algebra tensor-products kronecker-product
asked Mar 25 at 21:15
baxxbaxx
413311
asked Mar 25 at 21:15
baxxbaxx
413311
asked Mar 25 at 21:15
baxxbaxx
413311
asked Mar 25 at 21:15
baxxbaxx
413311
asked Mar 25 at 21:15
baxxbaxx
413311
413311
$begingroup$
Which part isn't clear? It all follows from how you multiply tensor products of matrices.
$endgroup$
– Adam Latosiński
Mar 25 at 21:25
$begingroup$
It's not explained in the text - i tried computing it long hand with an example, perhaps i messed up a computation because it didn't work out.
$endgroup$
– baxx
Mar 25 at 21:41
add a comment |
$begingroup$
Which part isn't clear? It all follows from how you multiply tensor products of matrices.
$endgroup$
– Adam Latosiński
Mar 25 at 21:25
$begingroup$
It's not explained in the text - i tried computing it long hand with an example, perhaps i messed up a computation because it didn't work out.
$endgroup$
– baxx
Mar 25 at 21:41
$begingroup$
Which part isn't clear? It all follows from how you multiply tensor products of matrices.
$endgroup$
– Adam Latosiński
Mar 25 at 21:25
$begingroup$
Which part isn't clear? It all follows from how you multiply tensor products of matrices.
$endgroup$
– Adam Latosiński
Mar 25 at 21:25
$begingroup$
It's not explained in the text - i tried computing it long hand with an example, perhaps i messed up a computation because it didn't work out.
$endgroup$
– baxx
Mar 25 at 21:41
$begingroup$
It's not explained in the text - i tried computing it long hand with an example, perhaps i messed up a computation because it didn't work out.
$endgroup$
– baxx
Mar 25 at 21:41
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The basic idea is that
$$Aotimes B =beginbmatrix a_11B & a_12B & a_13B dots \ a_21B & a_22B & a_23B dots \ vdots & vdots & vdots endbmatrix $$
$$A^-1otimes B^-1 =beginbmatrix a_11'B^-1 & a_12'B^-1 & a_13'B^-1 dots \ a_21'B^-1 & a_22'B^-1 & a_23'B^-1 dots \ vdots & vdots & vdots endbmatrix $$
$$(Aotimes B)(A^-1otimes B^-1) =beginbmatrix c_11BB^-1 & c_12BB^-1 & c_13BB^-1 dots \ c_21BB^-1 & c_22BB^-1 & c_23BB^-1 dots \ vdots & vdots & vdots endbmatrix = beginbmatrix c_11I & c_12I & c_13I dots \ c_21I & c_22I & c_23I dots \ vdots & vdots & vdots endbmatrix$$
where $c_ij$ is a an element of $AA^-1$, so $c_ij = delta_ij$ and the equation holds. This is extremely schematic of course, just to give you some "visuals". You should calculate $Aotimes B$, $A^-1otimes B^-1$ directly by definition of Kronecker product and then multiply them to see how indices behave.
answered Mar 25 at 21:45
AromaTheLoopAromaTheLoop
616
$endgroup$
add a comment |
$begingroup$
More generally, if $A,B,C,D$ are matrices such that the products $AC$ and $BD$ are defined, then $(Aotimes B)(Cotimes D)=ACotimes BD$. This follows from the definition of Kronecker product.
Another relevant property here is $I_motimes I_n=I_mn$.
answered Mar 25 at 21:53
chhrochhro
1,452311
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
The basic idea is that
$$Aotimes B =beginbmatrix a_11B & a_12B & a_13B dots \ a_21B & a_22B & a_23B dots \ vdots & vdots & vdots endbmatrix $$
$$A^-1otimes B^-1 =beginbmatrix a_11'B^-1 & a_12'B^-1 & a_13'B^-1 dots \ a_21'B^-1 & a_22'B^-1 & a_23'B^-1 dots \ vdots & vdots & vdots endbmatrix $$
$$(Aotimes B)(A^-1otimes B^-1) =beginbmatrix c_11BB^-1 & c_12BB^-1 & c_13BB^-1 dots \ c_21BB^-1 & c_22BB^-1 & c_23BB^-1 dots \ vdots & vdots & vdots endbmatrix = beginbmatrix c_11I & c_12I & c_13I dots \ c_21I & c_22I & c_23I dots \ vdots & vdots & vdots endbmatrix$$
where $c_ij$ is a an element of $AA^-1$, so $c_ij = delta_ij$ and the equation holds. This is extremely schematic of course, just to give you some "visuals". You should calculate $Aotimes B$, $A^-1otimes B^-1$ directly by definition of Kronecker product and then multiply them to see how indices behave.
answered Mar 25 at 21:45
AromaTheLoopAromaTheLoop
616
$endgroup$
add a comment |
$begingroup$
The basic idea is that
$$Aotimes B =beginbmatrix a_11B & a_12B & a_13B dots \ a_21B & a_22B & a_23B dots \ vdots & vdots & vdots endbmatrix $$
$$A^-1otimes B^-1 =beginbmatrix a_11'B^-1 & a_12'B^-1 & a_13'B^-1 dots \ a_21'B^-1 & a_22'B^-1 & a_23'B^-1 dots \ vdots & vdots & vdots endbmatrix $$
$$(Aotimes B)(A^-1otimes B^-1) =beginbmatrix c_11BB^-1 & c_12BB^-1 & c_13BB^-1 dots \ c_21BB^-1 & c_22BB^-1 & c_23BB^-1 dots \ vdots & vdots & vdots endbmatrix = beginbmatrix c_11I & c_12I & c_13I dots \ c_21I & c_22I & c_23I dots \ vdots & vdots & vdots endbmatrix$$
where $c_ij$ is a an element of $AA^-1$, so $c_ij = delta_ij$ and the equation holds. This is extremely schematic of course, just to give you some "visuals". You should calculate $Aotimes B$, $A^-1otimes B^-1$ directly by definition of Kronecker product and then multiply them to see how indices behave.
answered Mar 25 at 21:45
AromaTheLoopAromaTheLoop
616
$endgroup$
add a comment |
$begingroup$
The basic idea is that
$$Aotimes B =beginbmatrix a_11B & a_12B & a_13B dots \ a_21B & a_22B & a_23B dots \ vdots & vdots & vdots endbmatrix $$
$$A^-1otimes B^-1 =beginbmatrix a_11'B^-1 & a_12'B^-1 & a_13'B^-1 dots \ a_21'B^-1 & a_22'B^-1 & a_23'B^-1 dots \ vdots & vdots & vdots endbmatrix $$
$$(Aotimes B)(A^-1otimes B^-1) =beginbmatrix c_11BB^-1 & c_12BB^-1 & c_13BB^-1 dots \ c_21BB^-1 & c_22BB^-1 & c_23BB^-1 dots \ vdots & vdots & vdots endbmatrix = beginbmatrix c_11I & c_12I & c_13I dots \ c_21I & c_22I & c_23I dots \ vdots & vdots & vdots endbmatrix$$
where $c_ij$ is a an element of $AA^-1$, so $c_ij = delta_ij$ and the equation holds. This is extremely schematic of course, just to give you some "visuals". You should calculate $Aotimes B$, $A^-1otimes B^-1$ directly by definition of Kronecker product and then multiply them to see how indices behave.
answered Mar 25 at 21:45
AromaTheLoopAromaTheLoop
616
$endgroup$
The basic idea is that
$$Aotimes B =beginbmatrix a_11B & a_12B & a_13B dots \ a_21B & a_22B & a_23B dots \ vdots & vdots & vdots endbmatrix $$
$$A^-1otimes B^-1 =beginbmatrix a_11'B^-1 & a_12'B^-1 & a_13'B^-1 dots \ a_21'B^-1 & a_22'B^-1 & a_23'B^-1 dots \ vdots & vdots & vdots endbmatrix $$
$$(Aotimes B)(A^-1otimes B^-1) =beginbmatrix c_11BB^-1 & c_12BB^-1 & c_13BB^-1 dots \ c_21BB^-1 & c_22BB^-1 & c_23BB^-1 dots \ vdots & vdots & vdots endbmatrix = beginbmatrix c_11I & c_12I & c_13I dots \ c_21I & c_22I & c_23I dots \ vdots & vdots & vdots endbmatrix$$
where $c_ij$ is a an element of $AA^-1$, so $c_ij = delta_ij$ and the equation holds. This is extremely schematic of course, just to give you some "visuals". You should calculate $Aotimes B$, $A^-1otimes B^-1$ directly by definition of Kronecker product and then multiply them to see how indices behave.
answered Mar 25 at 21:45
AromaTheLoopAromaTheLoop
616
answered Mar 25 at 21:45
AromaTheLoopAromaTheLoop
616
answered Mar 25 at 21:45
AromaTheLoopAromaTheLoop
616
answered Mar 25 at 21:45
AromaTheLoopAromaTheLoop
616
616
add a comment |
add a comment |
$begingroup$
More generally, if $A,B,C,D$ are matrices such that the products $AC$ and $BD$ are defined, then $(Aotimes B)(Cotimes D)=ACotimes BD$. This follows from the definition of Kronecker product.
Another relevant property here is $I_motimes I_n=I_mn$.
answered Mar 25 at 21:53
chhrochhro
1,452311
$endgroup$
add a comment |
$begingroup$
More generally, if $A,B,C,D$ are matrices such that the products $AC$ and $BD$ are defined, then $(Aotimes B)(Cotimes D)=ACotimes BD$. This follows from the definition of Kronecker product.
Another relevant property here is $I_motimes I_n=I_mn$.
answered Mar 25 at 21:53
chhrochhro
1,452311
$endgroup$
add a comment |
$begingroup$
More generally, if $A,B,C,D$ are matrices such that the products $AC$ and $BD$ are defined, then $(Aotimes B)(Cotimes D)=ACotimes BD$. This follows from the definition of Kronecker product.
Another relevant property here is $I_motimes I_n=I_mn$.
answered Mar 25 at 21:53
chhrochhro
1,452311
$endgroup$
More generally, if $A,B,C,D$ are matrices such that the products $AC$ and $BD$ are defined, then $(Aotimes B)(Cotimes D)=ACotimes BD$. This follows from the definition of Kronecker product.
Another relevant property here is $I_motimes I_n=I_mn$.
answered Mar 25 at 21:53
chhrochhro
1,452311
answered Mar 25 at 21:53
chhrochhro
1,452311
answered Mar 25 at 21:53
chhrochhro
1,452311
answered Mar 25 at 21:53
chhrochhro
1,452311
1,452311
add a comment |
add a comment |
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$begingroup$
Which part isn't clear? It all follows from how you multiply tensor products of matrices.
$endgroup$
– Adam Latosiński
Mar 25 at 21:25
$begingroup$
It's not explained in the text - i tried computing it long hand with an example, perhaps i messed up a computation because it didn't work out.
$endgroup$
– baxx
Mar 25 at 21:41