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0

$begingroup$

I have on my hands a linear system of equations of the following form
$$
sum_j=1^Ksum_q=1^N A_ijpq x_jq = b_ip quad(i=1dots K,p=1dots N)
$$
in which the $x_jq$ are unknown and the rest are given data.
I would like to recast this in the form $tilde Atilde x=tilde b$, where $tilde A$ is a $KN$-by-$KN$ matrix and $tilde x$ and $tilde b$ are each vectors of length $KN$. If I fix $K$ and $N$ to concrete values, I can work out by hand what $tilde A$, $tilde x$, and $tilde b$ should be. I am having trouble seeing the generalization for arbitrary $K$ and $N$. Surely there is one, though. What is it?

linear-algebra matrices block-matrices

share|cite|improve this question

edited Mar 28 at 17:04

Rodrigo de Azevedo

13.1k41960

asked Mar 21 at 15:05

EndulumEndulum

1566

$endgroup$

add a comment | 

0

$begingroup$

I have on my hands a linear system of equations of the following form
$$
sum_j=1^Ksum_q=1^N A_ijpq x_jq = b_ip quad(i=1dots K,p=1dots N)
$$
in which the $x_jq$ are unknown and the rest are given data.
I would like to recast this in the form $tilde Atilde x=tilde b$, where $tilde A$ is a $KN$-by-$KN$ matrix and $tilde x$ and $tilde b$ are each vectors of length $KN$. If I fix $K$ and $N$ to concrete values, I can work out by hand what $tilde A$, $tilde x$, and $tilde b$ should be. I am having trouble seeing the generalization for arbitrary $K$ and $N$. Surely there is one, though. What is it?

linear-algebra matrices block-matrices

share|cite|improve this question

edited Mar 28 at 17:04

Rodrigo de Azevedo

13.1k41960

asked Mar 21 at 15:05

EndulumEndulum

1566

$endgroup$

add a comment | 

$begingroup$

I have on my hands a linear system of equations of the following form
$$
sum_j=1^Ksum_q=1^N A_ijpq x_jq = b_ip quad(i=1dots K,p=1dots N)
$$
in which the $x_jq$ are unknown and the rest are given data.
I would like to recast this in the form $tilde Atilde x=tilde b$, where $tilde A$ is a $KN$-by-$KN$ matrix and $tilde x$ and $tilde b$ are each vectors of length $KN$. If I fix $K$ and $N$ to concrete values, I can work out by hand what $tilde A$, $tilde x$, and $tilde b$ should be. I am having trouble seeing the generalization for arbitrary $K$ and $N$. Surely there is one, though. What is it?

linear-algebra matrices block-matrices

share|cite|improve this question

edited Mar 28 at 17:04

Rodrigo de Azevedo

13.1k41960

asked Mar 21 at 15:05

EndulumEndulum

1566

$endgroup$

share|cite|improve this question

edited Mar 28 at 17:04

Rodrigo de Azevedo

13.1k41960

asked Mar 21 at 15:05

EndulumEndulum

1566

share|cite|improve this question

edited Mar 28 at 17:04

Rodrigo de Azevedo

13.1k41960

asked Mar 21 at 15:05

EndulumEndulum

1566

edited Mar 28 at 17:04

Rodrigo de Azevedo

13.1k41960

edited Mar 28 at 17:04

Rodrigo de Azevedo

13.1k41960

asked Mar 21 at 15:05

EndulumEndulum

1566

asked Mar 21 at 15:05

EndulumEndulum

1566

1 Answer
1

active

oldest

votes

0

$begingroup$

This turns out to be quite easy to do if one introduces new "super-indices" that run from $1dots KN$ according to some ordering convention (i.e., row- versus column-major). For instance, defining
$$
(ip) = i + K(p-1) \
(jq) = j + K(q-1)
$$
allows the tensor equation in question to be recast from
$$
sum_j=1^Ksum_q=1^N A_ijpqx_jq = b_ip qquad (i=1dots K, p=1dots N)
$$
to the matrix equation
$$
sum_(jq)=1^KN tilde A_(ip),(jq) tilde x_(jq) = tilde b_(ip) qquad ((jq)=1..KN)
$$
with the correspondence $tilde A_(i,p),(j,q) = A_ijpq$, etc.

share|cite|improve this answer

answered Mar 28 at 17:02

EndulumEndulum

1566

$endgroup$

add a comment | 

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0

$begingroup$

This turns out to be quite easy to do if one introduces new "super-indices" that run from $1dots KN$ according to some ordering convention (i.e., row- versus column-major). For instance, defining
$$
(ip) = i + K(p-1) \
(jq) = j + K(q-1)
$$
allows the tensor equation in question to be recast from
$$
sum_j=1^Ksum_q=1^N A_ijpqx_jq = b_ip qquad (i=1dots K, p=1dots N)
$$
to the matrix equation
$$
sum_(jq)=1^KN tilde A_(ip),(jq) tilde x_(jq) = tilde b_(ip) qquad ((jq)=1..KN)
$$
with the correspondence $tilde A_(i,p),(j,q) = A_ijpq$, etc.

share|cite|improve this answer

answered Mar 28 at 17:02

EndulumEndulum

1566

$endgroup$

add a comment | 

0

$begingroup$

This turns out to be quite easy to do if one introduces new "super-indices" that run from $1dots KN$ according to some ordering convention (i.e., row- versus column-major). For instance, defining
$$
(ip) = i + K(p-1) \
(jq) = j + K(q-1)
$$
allows the tensor equation in question to be recast from
$$
sum_j=1^Ksum_q=1^N A_ijpqx_jq = b_ip qquad (i=1dots K, p=1dots N)
$$
to the matrix equation
$$
sum_(jq)=1^KN tilde A_(ip),(jq) tilde x_(jq) = tilde b_(ip) qquad ((jq)=1..KN)
$$
with the correspondence $tilde A_(i,p),(j,q) = A_ijpq$, etc.

share|cite|improve this answer

answered Mar 28 at 17:02

EndulumEndulum

1566

$endgroup$

add a comment | 

$begingroup$

This turns out to be quite easy to do if one introduces new "super-indices" that run from $1dots KN$ according to some ordering convention (i.e., row- versus column-major). For instance, defining
$$
(ip) = i + K(p-1) \
(jq) = j + K(q-1)
$$
allows the tensor equation in question to be recast from
$$
sum_j=1^Ksum_q=1^N A_ijpqx_jq = b_ip qquad (i=1dots K, p=1dots N)
$$
to the matrix equation
$$
sum_(jq)=1^KN tilde A_(ip),(jq) tilde x_(jq) = tilde b_(ip) qquad ((jq)=1..KN)
$$
with the correspondence $tilde A_(i,p),(j,q) = A_ijpq$, etc.

share|cite|improve this answer

answered Mar 28 at 17:02

EndulumEndulum

1566

$endgroup$

share|cite|improve this answer

answered Mar 28 at 17:02

EndulumEndulum

1566

answered Mar 28 at 17:02

EndulumEndulum

1566

answered Mar 28 at 17:02

EndulumEndulum

1566