Mixing
Kronecker product of identity and matrix product
$begingroup$
How is the following property true? Let $I$ be the identity matrix and $A$, $B$ be appropriately sized real matrices. Then
$$I otimes left(left( I otimes Aright) B right) = left( I otimes I otimes A right)left( I otimes Bright).$$
I think I'm missing something basic about the Kronecker product.
kronecker-product
asked Jan 28 at 8:24
G. GareG. Gare
31013
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add a comment |
$begingroup$
How is the following property true? Let $I$ be the identity matrix and $A$, $B$ be appropriately sized real matrices. Then
$$I otimes left(left( I otimes Aright) B right) = left( I otimes I otimes A right)left( I otimes Bright).$$
I think I'm missing something basic about the Kronecker product.
kronecker-product
asked Jan 28 at 8:24
G. GareG. Gare
31013
$endgroup$
add a comment |
$begingroup$
How is the following property true? Let $I$ be the identity matrix and $A$, $B$ be appropriately sized real matrices. Then
$$I otimes left(left( I otimes Aright) B right) = left( I otimes I otimes A right)left( I otimes Bright).$$
I think I'm missing something basic about the Kronecker product.
kronecker-product
asked Jan 28 at 8:24
G. GareG. Gare
31013
$endgroup$
How is the following property true? Let $I$ be the identity matrix and $A$, $B$ be appropriately sized real matrices. Then
$$I otimes left(left( I otimes Aright) B right) = left( I otimes I otimes A right)left( I otimes Bright).$$
I think I'm missing something basic about the Kronecker product.
kronecker-product
kronecker-product
asked Jan 28 at 8:24
G. GareG. Gare
31013
asked Jan 28 at 8:24
G. GareG. Gare
31013
asked Jan 28 at 8:24
G. GareG. Gare
31013
asked Jan 28 at 8:24
G. GareG. Gare
31013
asked Jan 28 at 8:24
G. GareG. Gare
31013
31013
add a comment |
add a comment |
1 Answer
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The kronecker product is associative, so
$(Aotimes B)otimes C=Aotimes(Botimes C)$ and use that $(AB)otimes(CD)=(Aotimes C)(Botimes D)$. Now, you can just expand.
answered Jan 28 at 8:28
JamesJames
965318
$endgroup$
$begingroup$
Yes. I didn't think that $I = I^2$. Thanks!
$endgroup$
– G. Gare
Jan 28 at 8:37
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The kronecker product is associative, so
$(Aotimes B)otimes C=Aotimes(Botimes C)$ and use that $(AB)otimes(CD)=(Aotimes C)(Botimes D)$. Now, you can just expand.
answered Jan 28 at 8:28
JamesJames
965318
$endgroup$
$begingroup$
Yes. I didn't think that $I = I^2$. Thanks!
$endgroup$
– G. Gare
Jan 28 at 8:37
add a comment |
$begingroup$
The kronecker product is associative, so
$(Aotimes B)otimes C=Aotimes(Botimes C)$ and use that $(AB)otimes(CD)=(Aotimes C)(Botimes D)$. Now, you can just expand.
answered Jan 28 at 8:28
JamesJames
965318
$endgroup$
$begingroup$
Yes. I didn't think that $I = I^2$. Thanks!
$endgroup$
– G. Gare
Jan 28 at 8:37
add a comment |
$begingroup$
The kronecker product is associative, so
$(Aotimes B)otimes C=Aotimes(Botimes C)$ and use that $(AB)otimes(CD)=(Aotimes C)(Botimes D)$. Now, you can just expand.
answered Jan 28 at 8:28
JamesJames
965318
$endgroup$
The kronecker product is associative, so
$(Aotimes B)otimes C=Aotimes(Botimes C)$ and use that $(AB)otimes(CD)=(Aotimes C)(Botimes D)$. Now, you can just expand.
answered Jan 28 at 8:28
JamesJames
965318
answered Jan 28 at 8:28
JamesJames
965318
answered Jan 28 at 8:28
JamesJames
965318
answered Jan 28 at 8:28
JamesJames
965318
965318
$begingroup$
Yes. I didn't think that $I = I^2$. Thanks!
$endgroup$
– G. Gare
Jan 28 at 8:37
add a comment |
$begingroup$
Yes. I didn't think that $I = I^2$. Thanks!
$endgroup$
– G. Gare
Jan 28 at 8:37
$begingroup$
Yes. I didn't think that $I = I^2$. Thanks!
$endgroup$
– G. Gare
Jan 28 at 8:37
$begingroup$
Yes. I didn't think that $I = I^2$. Thanks!
$endgroup$
– G. Gare
Jan 28 at 8:37
add a comment |
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