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Matrix-based proof of transformation rule for a wedge product of covectors by wedge product of covectors

Transformation Rule for a Wedge Product of CovectorsDescribe $f$ in terms of the tensor products of $alpha^i$ and $alpha^j$. Is this inner product? What are its coefficients?Derivations of bases of $A_k(V)$ and $L_k(V)$: what's the difference?An Expression for the Wedge ProductOrthogonal direct sum decomposition and projectionsSubspace for a matrix representation of a linear transformationAntisymmetry of exterior product for alternating multilinear formsIs the exterior wedge product of differential forms injective?Formula for wedge productIs this a valid way to think of decomposable $k$-forms?The volume element for a subspace of $mathbb R^N$Prove that Gramian matrix is Invertible iff $(v_1,…,v_k) $ is linearly independentDerivations of bases of $A_k(V)$ and $L_k(V)$: what's the difference?

0

$begingroup$

I am supposed to show

$$beta^1 ∧ ··· ∧ beta^k = (det A) gamma^1 ∧ ··· ∧ gamma^k$$

for covectors $beta^1, ···, beta^k, gamma^1, ···, gamma^k$ of a
vector space finite dimensional real vector space $V$ of dimension $n$
such that $vecbeta=Avecgamma$, where

$$vecbeta = [beta^1 ... beta^k]^T, vecgamma = [gamma^1 ...
> gamma^k]^T, A = [a_1^T ... a_k^T]^T, a_i^T = [a^i_1 ... a^i_k]$$

The solutions I have found (like this one on stackexchange and two more that are not on stackexchange) are not matrix-based and do not seem to utilize the following fact about the wedge product of covectors

$$(beta^1 ∧ ··· ∧ beta^k) (v_1, ..., v_k) = det[beta^i(v_j)]$$

• So I'm composing a matrix-based proof now. Please verify.

Let $v_1, ..., v_k in V$. Then $$LHS = (beta^1 ∧ ··· ∧ beta^k) (v_1, ..., v_k) = det[beta^i(v_j)] = det[vecbeta(v_1) ... vecbeta(v_k)],$$ where $vecbeta(v_j) = [beta^1(v_j) ... beta^k(v_j)]^T$.

Now,

$$[vecbeta(v_1) ... vecbeta(v_k)] = [Avecgamma(v_1) ... Avecgamma(v_k)] = A [vecgamma(v_1) ... vecgamma(v_k)] = A [gamma^i(v_j)]$$

Hence,

$$[vecbeta(v_1) ... vecbeta(v_k)] = A [gamma^i(v_j)]$$

$$implies LHS = det[vecbeta(v_1) ... vecbeta(v_k)] = det (A [gamma^i(v_j)])= det (A) det([gamma^i(v_j)])$$

$$ = det(A) (gamma^1 ∧ ··· ∧ gamma^k)(v_1, ..., v_k) = RHS$$

---

• My book is An Introduction to Manifolds by Loring W. Tu. This is Exercise 3.7 and is called "Transformation rule for a wedge product of covectors". The above fact is Proposition 3.27 called "wedge product of 1-covectors" (covector is defined as 1-covector).




• Motivation for a matrix-based proof: I believe there's a way to do a matrix-based proof for another exercise, Exercise 3.8, based on an analogous result for Proposition 3.27 and Exercise 3.1 on inner product (specifically the generalization of Exercise 3.1 given in Exercise 3.3). If the above proof is unsuccessful, then I might not continue to attempt a matrix-based proof for Exercise 3.8. If the above proof is successful, then I might consider posting my attempt of a matrix-based proof for Exercise 3.8 in another question.

linear-algebra differential-geometry vector-spaces inner-product-space tensor-products

share|cite|improve this question

edited Mar 11 at 11:33

Selene Auckland

asked Feb 28 at 11:36

Selene AucklandSelene Auckland

6911

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  are you sure $nneq k$? If we scale $A$ by a constant $q$, the left-hand side of the equation should scale by $q^k$, yet the right hand side should scale by $det (q*A) / (det A) = q^n$.
  $endgroup$
  – user3257842
  Feb 28 at 14:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user3257842 $A$ is $k times k$ and not $n times n$, so how did you get $det(q * A) = q^n det (A)$ please?
  $endgroup$
  – Selene Auckland
  Mar 2 at 7:46
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I got confused when I saw $vecbeta=Avecgamma$, and took $A$ as an operator that acts on the elements of $V$. However, looking at the definitions, it is more appropriate to say $A$ acts on indices, defining a specific way to obtain the $beta^q, qin1..k$ vectors as index-dependent linear combinations of the $gamma^q, qin1..k$ vectors.
  $endgroup$
  – user3257842
  Mar 3 at 21:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user3257842 So, proof is correct?
  $endgroup$
  – Selene Auckland
  Mar 4 at 4:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user3257842 Thanks, but what do you mean? Is my proof correct?
  $endgroup$
  – Selene Auckland
  Mar 7 at 9:41
  

  
  
  
  

add a comment | 

0

$begingroup$

I am supposed to show

$$beta^1 ∧ ··· ∧ beta^k = (det A) gamma^1 ∧ ··· ∧ gamma^k$$

for covectors $beta^1, ···, beta^k, gamma^1, ···, gamma^k$ of a
vector space finite dimensional real vector space $V$ of dimension $n$
such that $vecbeta=Avecgamma$, where

$$vecbeta = [beta^1 ... beta^k]^T, vecgamma = [gamma^1 ...
> gamma^k]^T, A = [a_1^T ... a_k^T]^T, a_i^T = [a^i_1 ... a^i_k]$$

The solutions I have found (like this one on stackexchange and two more that are not on stackexchange) are not matrix-based and do not seem to utilize the following fact about the wedge product of covectors

$$(beta^1 ∧ ··· ∧ beta^k) (v_1, ..., v_k) = det[beta^i(v_j)]$$

• So I'm composing a matrix-based proof now. Please verify.

Let $v_1, ..., v_k in V$. Then $$LHS = (beta^1 ∧ ··· ∧ beta^k) (v_1, ..., v_k) = det[beta^i(v_j)] = det[vecbeta(v_1) ... vecbeta(v_k)],$$ where $vecbeta(v_j) = [beta^1(v_j) ... beta^k(v_j)]^T$.

Now,

$$[vecbeta(v_1) ... vecbeta(v_k)] = [Avecgamma(v_1) ... Avecgamma(v_k)] = A [vecgamma(v_1) ... vecgamma(v_k)] = A [gamma^i(v_j)]$$

Hence,

$$[vecbeta(v_1) ... vecbeta(v_k)] = A [gamma^i(v_j)]$$

$$implies LHS = det[vecbeta(v_1) ... vecbeta(v_k)] = det (A [gamma^i(v_j)])= det (A) det([gamma^i(v_j)])$$

$$ = det(A) (gamma^1 ∧ ··· ∧ gamma^k)(v_1, ..., v_k) = RHS$$

---

• My book is An Introduction to Manifolds by Loring W. Tu. This is Exercise 3.7 and is called "Transformation rule for a wedge product of covectors". The above fact is Proposition 3.27 called "wedge product of 1-covectors" (covector is defined as 1-covector).




• Motivation for a matrix-based proof: I believe there's a way to do a matrix-based proof for another exercise, Exercise 3.8, based on an analogous result for Proposition 3.27 and Exercise 3.1 on inner product (specifically the generalization of Exercise 3.1 given in Exercise 3.3). If the above proof is unsuccessful, then I might not continue to attempt a matrix-based proof for Exercise 3.8. If the above proof is successful, then I might consider posting my attempt of a matrix-based proof for Exercise 3.8 in another question.

linear-algebra differential-geometry vector-spaces inner-product-space tensor-products

share|cite|improve this question

edited Mar 11 at 11:33

Selene Auckland

asked Feb 28 at 11:36

Selene AucklandSelene Auckland

6911

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  are you sure $nneq k$? If we scale $A$ by a constant $q$, the left-hand side of the equation should scale by $q^k$, yet the right hand side should scale by $det (q*A) / (det A) = q^n$.
  $endgroup$
  – user3257842
  Feb 28 at 14:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user3257842 $A$ is $k times k$ and not $n times n$, so how did you get $det(q * A) = q^n det (A)$ please?
  $endgroup$
  – Selene Auckland
  Mar 2 at 7:46
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I got confused when I saw $vecbeta=Avecgamma$, and took $A$ as an operator that acts on the elements of $V$. However, looking at the definitions, it is more appropriate to say $A$ acts on indices, defining a specific way to obtain the $beta^q, qin1..k$ vectors as index-dependent linear combinations of the $gamma^q, qin1..k$ vectors.
  $endgroup$
  – user3257842
  Mar 3 at 21:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user3257842 So, proof is correct?
  $endgroup$
  – Selene Auckland
  Mar 4 at 4:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user3257842 Thanks, but what do you mean? Is my proof correct?
  $endgroup$
  – Selene Auckland
  Mar 7 at 9:41
  

  
  
  
  

add a comment | 

$begingroup$

I am supposed to show

$$beta^1 ∧ ··· ∧ beta^k = (det A) gamma^1 ∧ ··· ∧ gamma^k$$

for covectors $beta^1, ···, beta^k, gamma^1, ···, gamma^k$ of a
vector space finite dimensional real vector space $V$ of dimension $n$
such that $vecbeta=Avecgamma$, where

$$vecbeta = [beta^1 ... beta^k]^T, vecgamma = [gamma^1 ...
> gamma^k]^T, A = [a_1^T ... a_k^T]^T, a_i^T = [a^i_1 ... a^i_k]$$

The solutions I have found (like this one on stackexchange and two more that are not on stackexchange) are not matrix-based and do not seem to utilize the following fact about the wedge product of covectors

$$(beta^1 ∧ ··· ∧ beta^k) (v_1, ..., v_k) = det[beta^i(v_j)]$$

• So I'm composing a matrix-based proof now. Please verify.

Let $v_1, ..., v_k in V$. Then $$LHS = (beta^1 ∧ ··· ∧ beta^k) (v_1, ..., v_k) = det[beta^i(v_j)] = det[vecbeta(v_1) ... vecbeta(v_k)],$$ where $vecbeta(v_j) = [beta^1(v_j) ... beta^k(v_j)]^T$.

Now,

$$[vecbeta(v_1) ... vecbeta(v_k)] = [Avecgamma(v_1) ... Avecgamma(v_k)] = A [vecgamma(v_1) ... vecgamma(v_k)] = A [gamma^i(v_j)]$$

Hence,

$$[vecbeta(v_1) ... vecbeta(v_k)] = A [gamma^i(v_j)]$$

$$implies LHS = det[vecbeta(v_1) ... vecbeta(v_k)] = det (A [gamma^i(v_j)])= det (A) det([gamma^i(v_j)])$$

$$ = det(A) (gamma^1 ∧ ··· ∧ gamma^k)(v_1, ..., v_k) = RHS$$

---

• My book is An Introduction to Manifolds by Loring W. Tu. This is Exercise 3.7 and is called "Transformation rule for a wedge product of covectors". The above fact is Proposition 3.27 called "wedge product of 1-covectors" (covector is defined as 1-covector).




• Motivation for a matrix-based proof: I believe there's a way to do a matrix-based proof for another exercise, Exercise 3.8, based on an analogous result for Proposition 3.27 and Exercise 3.1 on inner product (specifically the generalization of Exercise 3.1 given in Exercise 3.3). If the above proof is unsuccessful, then I might not continue to attempt a matrix-based proof for Exercise 3.8. If the above proof is successful, then I might consider posting my attempt of a matrix-based proof for Exercise 3.8 in another question.

linear-algebra differential-geometry vector-spaces inner-product-space tensor-products

share|cite|improve this question

edited Mar 11 at 11:33

Selene Auckland

asked Feb 28 at 11:36

Selene AucklandSelene Auckland

6911

$endgroup$

share|cite|improve this question

edited Mar 11 at 11:33

Selene Auckland

asked Feb 28 at 11:36

Selene AucklandSelene Auckland

6911

share|cite|improve this question

edited Mar 11 at 11:33

Selene Auckland

asked Feb 28 at 11:36

Selene AucklandSelene Auckland

6911

asked Feb 28 at 11:36

Selene AucklandSelene Auckland

6911

asked Feb 28 at 11:36

Selene AucklandSelene Auckland

6911