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Relation between rank of a linear transformation and pullback ($r$-linear forms.)$
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Problem. The rank of the linear transformation $A: S to U$ is the largest integer $r$ such that $A^*: mathcalA_r(S) to mathcalA_r(U)$ is not equal to zero.
Notations.
$mathcalA_r(S)$ denotes the vector space of $r$-linear forms over $S$.
$A^*$ is defined by
$$(A^*omega)(v_1,...,v_r) = omega(Acdot v_1,...,Acdot v_r),$$
for $omega in mathcalA_r(U)$ and $v_i in S$. The transformation $A^*$ is the induced by $A$ in the forms of
degree $r$ and the form $A^*omega$ is the pull-back of the form
$omega$ to space $S$.
Idea. Let $r$ the largest integer such that $A^*$ is not zero and rank$(A) = m$. Given a set $C = lbrace v_1,...,v_k mid v_i in Srbrace$ with $k > m$ so, $C$ is LD over $S$. Thus there is $i neq j$ such that $v_i = alpha v_j$ where $alpha in mathbbR$. Therefore
$$omega(v_1,...,v_i,...,v_j,...,v_k) = omega(v_1,...,alpha v_j,...,v_j,...,v_k) = 0,$$
then $r leq m$.
Now, if $tildeC = lbrace v_1,...,v_k mid v_i in S rbrace$ with $k leq m$ we can suppose $tildeC$ LI, since rank$(S) = m$. Thus,
$$omega(v_1,...,v_k) neq 0$$
since $v_i$ are LI. Then $omega neq 0$.
Is this correct?
real-analysis linear-algebra
asked Jul 16 at 19:29
Lucas Corrêa
1,204319
add a comment |Â
up vote
1
down vote
favorite
Problem. The rank of the linear transformation $A: S to U$ is the largest integer $r$ such that $A^*: mathcalA_r(S) to mathcalA_r(U)$ is not equal to zero.
Notations.
$mathcalA_r(S)$ denotes the vector space of $r$-linear forms over $S$.
$A^*$ is defined by
$$(A^*omega)(v_1,...,v_r) = omega(Acdot v_1,...,Acdot v_r),$$
for $omega in mathcalA_r(U)$ and $v_i in S$. The transformation $A^*$ is the induced by $A$ in the forms of
degree $r$ and the form $A^*omega$ is the pull-back of the form
$omega$ to space $S$.
Idea. Let $r$ the largest integer such that $A^*$ is not zero and rank$(A) = m$. Given a set $C = lbrace v_1,...,v_k mid v_i in Srbrace$ with $k > m$ so, $C$ is LD over $S$. Thus there is $i neq j$ such that $v_i = alpha v_j$ where $alpha in mathbbR$. Therefore
$$omega(v_1,...,v_i,...,v_j,...,v_k) = omega(v_1,...,alpha v_j,...,v_j,...,v_k) = 0,$$
then $r leq m$.
Now, if $tildeC = lbrace v_1,...,v_k mid v_i in S rbrace$ with $k leq m$ we can suppose $tildeC$ LI, since rank$(S) = m$. Thus,
$$omega(v_1,...,v_k) neq 0$$
since $v_i$ are LI. Then $omega neq 0$.
Is this correct?
real-analysis linear-algebra
asked Jul 16 at 19:29
Lucas Corrêa
1,204319
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Problem. The rank of the linear transformation $A: S to U$ is the largest integer $r$ such that $A^*: mathcalA_r(S) to mathcalA_r(U)$ is not equal to zero.
Notations.
$mathcalA_r(S)$ denotes the vector space of $r$-linear forms over $S$.
$A^*$ is defined by
$$(A^*omega)(v_1,...,v_r) = omega(Acdot v_1,...,Acdot v_r),$$
for $omega in mathcalA_r(U)$ and $v_i in S$. The transformation $A^*$ is the induced by $A$ in the forms of
degree $r$ and the form $A^*omega$ is the pull-back of the form
$omega$ to space $S$.
Idea. Let $r$ the largest integer such that $A^*$ is not zero and rank$(A) = m$. Given a set $C = lbrace v_1,...,v_k mid v_i in Srbrace$ with $k > m$ so, $C$ is LD over $S$. Thus there is $i neq j$ such that $v_i = alpha v_j$ where $alpha in mathbbR$. Therefore
$$omega(v_1,...,v_i,...,v_j,...,v_k) = omega(v_1,...,alpha v_j,...,v_j,...,v_k) = 0,$$
then $r leq m$.
Now, if $tildeC = lbrace v_1,...,v_k mid v_i in S rbrace$ with $k leq m$ we can suppose $tildeC$ LI, since rank$(S) = m$. Thus,
$$omega(v_1,...,v_k) neq 0$$
since $v_i$ are LI. Then $omega neq 0$.
Is this correct?
real-analysis linear-algebra
asked Jul 16 at 19:29
Lucas Corrêa
1,204319
Problem. The rank of the linear transformation $A: S to U$ is the largest integer $r$ such that $A^*: mathcalA_r(S) to mathcalA_r(U)$ is not equal to zero.
Notations.
$mathcalA_r(S)$ denotes the vector space of $r$-linear forms over $S$.
$A^*$ is defined by
$$(A^*omega)(v_1,...,v_r) = omega(Acdot v_1,...,Acdot v_r),$$
for $omega in mathcalA_r(U)$ and $v_i in S$. The transformation $A^*$ is the induced by $A$ in the forms of
degree $r$ and the form $A^*omega$ is the pull-back of the form
$omega$ to space $S$.
Idea. Let $r$ the largest integer such that $A^*$ is not zero and rank$(A) = m$. Given a set $C = lbrace v_1,...,v_k mid v_i in Srbrace$ with $k > m$ so, $C$ is LD over $S$. Thus there is $i neq j$ such that $v_i = alpha v_j$ where $alpha in mathbbR$. Therefore
$$omega(v_1,...,v_i,...,v_j,...,v_k) = omega(v_1,...,alpha v_j,...,v_j,...,v_k) = 0,$$
then $r leq m$.
Now, if $tildeC = lbrace v_1,...,v_k mid v_i in S rbrace$ with $k leq m$ we can suppose $tildeC$ LI, since rank$(S) = m$. Thus,
$$omega(v_1,...,v_k) neq 0$$
since $v_i$ are LI. Then $omega neq 0$.
Is this correct?
real-analysis linear-algebra
asked Jul 16 at 19:29
Lucas Corrêa
1,204319
edited Jul 17 at 1:28
asked Jul 16 at 19:29
Lucas Corrêa
1,204319
asked Jul 16 at 19:29
Lucas Corrêa
1,204319
asked Jul 16 at 19:29
Lucas Corrêa
1,204319
1,204319
add a comment |Â
add a comment |Â
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