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How to construct the matrix representation of a bilinear transformation?

0

$begingroup$

Suppose $fcolon mathbbR^mtimesmathbbR^mtomathbbR^n$ is a bilinear transformation. How do I define and construct the matrix representation of $f$ with respect to the canonical bases of the vector spaces involved? How do I represent in matrix form the action of $f$ on the pair $(u,v)inmathbbR^mtimesmathbbR^m$?

This is not a homework. I'm self studying this topic and can't find this construction in the books I have at my disposal.

linear-algebra multilinear-algebra

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asked Apr 1 at 13:56

booNlatoTbooNlatoT

164

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add a comment | 

0

$begingroup$

Suppose $fcolon mathbbR^mtimesmathbbR^mtomathbbR^n$ is a bilinear transformation. How do I define and construct the matrix representation of $f$ with respect to the canonical bases of the vector spaces involved? How do I represent in matrix form the action of $f$ on the pair $(u,v)inmathbbR^mtimesmathbbR^m$?

This is not a homework. I'm self studying this topic and can't find this construction in the books I have at my disposal.

linear-algebra multilinear-algebra

share|cite|improve this question

asked Apr 1 at 13:56

booNlatoTbooNlatoT

164

$endgroup$

add a comment | 

$begingroup$

Suppose $fcolon mathbbR^mtimesmathbbR^mtomathbbR^n$ is a bilinear transformation. How do I define and construct the matrix representation of $f$ with respect to the canonical bases of the vector spaces involved? How do I represent in matrix form the action of $f$ on the pair $(u,v)inmathbbR^mtimesmathbbR^m$?

This is not a homework. I'm self studying this topic and can't find this construction in the books I have at my disposal.

linear-algebra multilinear-algebra

share|cite|improve this question

asked Apr 1 at 13:56

booNlatoTbooNlatoT

164

$endgroup$

share|cite|improve this question

asked Apr 1 at 13:56

booNlatoTbooNlatoT

164

share|cite|improve this question

asked Apr 1 at 13:56

booNlatoTbooNlatoT

164

asked Apr 1 at 13:56

booNlatoTbooNlatoT

164

asked Apr 1 at 13:56

booNlatoTbooNlatoT

164

2 Answers
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0

$begingroup$

I would say you can't! You will need three indices such an array!

If $n=1$, you can use Gramian matrices. For other $n$, the only thing I can think of is creating one of those for each of the $n$ components.

You could of course just as well make a "canonical application" $c:mathbbR^2m tomathbbR^mtimesmathbbR^m$ and work with $ccdot f$ isntead but I don't think that's what you are looking for

share|cite|improve this answer

edited Apr 1 at 14:16

MPW

31.2k12157

answered Apr 1 at 14:05

DavidDavid

2377

$endgroup$

add a comment | 

0

$begingroup$

When $f:Bbb R^ntimesBbb R^ntoBbb R$ the answer is
$$[f]=[f(e_i,e_j)],$$
where the $e_i$ are the canonical base vectors.

Now if $f:Bbb R^ntimesBbb R^ntoBbb R^2$ then
$f(v,w)=(f_1(v,w),f_2(v,w))$ with both $f_1,f_2$ bilinear so a pair of matrices
can be associate as:
$$[f_1]=[f(e_i,e_j)]quad mboxandquad [f_2]=[f_2(e_i,e_j)]$$
Now you can figure it up what is next.

share|cite|improve this answer

edited Apr 2 at 21:20

answered Apr 2 at 21:08

janmarqzjanmarqz

6,29241630

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add a comment | 

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0

$begingroup$

I would say you can't! You will need three indices such an array!

If $n=1$, you can use Gramian matrices. For other $n$, the only thing I can think of is creating one of those for each of the $n$ components.

You could of course just as well make a "canonical application" $c:mathbbR^2m tomathbbR^mtimesmathbbR^m$ and work with $ccdot f$ isntead but I don't think that's what you are looking for

share|cite|improve this answer

edited Apr 1 at 14:16

MPW

31.2k12157

answered Apr 1 at 14:05

DavidDavid

2377

$endgroup$

add a comment | 

0

$begingroup$

I would say you can't! You will need three indices such an array!

If $n=1$, you can use Gramian matrices. For other $n$, the only thing I can think of is creating one of those for each of the $n$ components.

You could of course just as well make a "canonical application" $c:mathbbR^2m tomathbbR^mtimesmathbbR^m$ and work with $ccdot f$ isntead but I don't think that's what you are looking for

share|cite|improve this answer

edited Apr 1 at 14:16

MPW

31.2k12157

answered Apr 1 at 14:05

DavidDavid

2377

$endgroup$

add a comment | 

$begingroup$

I would say you can't! You will need three indices such an array!

If $n=1$, you can use Gramian matrices. For other $n$, the only thing I can think of is creating one of those for each of the $n$ components.

You could of course just as well make a "canonical application" $c:mathbbR^2m tomathbbR^mtimesmathbbR^m$ and work with $ccdot f$ isntead but I don't think that's what you are looking for

share|cite|improve this answer

edited Apr 1 at 14:16

MPW

31.2k12157

answered Apr 1 at 14:05

DavidDavid

2377

$endgroup$

share|cite|improve this answer

edited Apr 1 at 14:16

MPW

31.2k12157

answered Apr 1 at 14:05

DavidDavid

2377

edited Apr 1 at 14:16

MPW

31.2k12157

edited Apr 1 at 14:16

MPW

31.2k12157

answered Apr 1 at 14:05

DavidDavid

2377

answered Apr 1 at 14:05

DavidDavid

2377

0

$begingroup$

When $f:Bbb R^ntimesBbb R^ntoBbb R$ the answer is
$$[f]=[f(e_i,e_j)],$$
where the $e_i$ are the canonical base vectors.

Now if $f:Bbb R^ntimesBbb R^ntoBbb R^2$ then
$f(v,w)=(f_1(v,w),f_2(v,w))$ with both $f_1,f_2$ bilinear so a pair of matrices
can be associate as:
$$[f_1]=[f(e_i,e_j)]quad mboxandquad [f_2]=[f_2(e_i,e_j)]$$
Now you can figure it up what is next.

share|cite|improve this answer

edited Apr 2 at 21:20

answered Apr 2 at 21:08

janmarqzjanmarqz

6,29241630

$endgroup$

add a comment | 

0

$begingroup$

When $f:Bbb R^ntimesBbb R^ntoBbb R$ the answer is
$$[f]=[f(e_i,e_j)],$$
where the $e_i$ are the canonical base vectors.

Now if $f:Bbb R^ntimesBbb R^ntoBbb R^2$ then
$f(v,w)=(f_1(v,w),f_2(v,w))$ with both $f_1,f_2$ bilinear so a pair of matrices
can be associate as:
$$[f_1]=[f(e_i,e_j)]quad mboxandquad [f_2]=[f_2(e_i,e_j)]$$
Now you can figure it up what is next.

share|cite|improve this answer

edited Apr 2 at 21:20

answered Apr 2 at 21:08

janmarqzjanmarqz

6,29241630

$endgroup$

add a comment | 

$begingroup$

When $f:Bbb R^ntimesBbb R^ntoBbb R$ the answer is
$$[f]=[f(e_i,e_j)],$$
where the $e_i$ are the canonical base vectors.

Now if $f:Bbb R^ntimesBbb R^ntoBbb R^2$ then
$f(v,w)=(f_1(v,w),f_2(v,w))$ with both $f_1,f_2$ bilinear so a pair of matrices
can be associate as:
$$[f_1]=[f(e_i,e_j)]quad mboxandquad [f_2]=[f_2(e_i,e_j)]$$
Now you can figure it up what is next.

share|cite|improve this answer

edited Apr 2 at 21:20

answered Apr 2 at 21:08

janmarqzjanmarqz

6,29241630

$endgroup$

share|cite|improve this answer

edited Apr 2 at 21:20

answered Apr 2 at 21:08

janmarqzjanmarqz

6,29241630

answered Apr 2 at 21:08

janmarqzjanmarqz

6,29241630

answered Apr 2 at 21:08

janmarqzjanmarqz

6,29241630