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 $\vec{\beta}=A\vec{\gamma}$, where

$$\vec{\beta} = [\beta^1 ... \beta^k]^T, \vec{\gamma} = [\gamma^1 ... > \gamma^k]^T, A = [a_1^T ... a_k^T]^T, a_i^T = [a^i_1 ... a^i_k]$$

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 $$LHS = (\beta^1 ∧ ··· ∧ \beta^k) (v_1, ..., v_k) = \det[\beta^{i}(v_j)] = \det[\vec{\beta}(v_1) ... \vec{\beta}(v_k)],$$ where $\vec{\beta}(v_j) = [\beta^1(v_j) ... \beta^k(v_j)]^T$.


$$[\vec{\beta}(v_1) ... \vec{\beta}(v_k)] = [A\vec{\gamma}(v_1) ... A\vec{\gamma}(v_k)] = A [\vec{\gamma}(v_1) ... \vec{\gamma}(v_k)] = A [\gamma^i(v_j)]$$


$$[\vec{\beta}(v_1) ... \vec{\beta}(v_k)] = A [\gamma^i(v_j)]$$

$$\implies LHS = \det[\vec{\beta}(v_1) ... \vec{\beta}(v_k)] = \det (A [\gamma^i(v_j)])= \det (A) \det([\gamma^i(v_j)])$$

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

  • 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.

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    $\begingroup$ are you sure $n\neq k$? 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
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    $\begingroup$ @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 $\vec{\beta}=A\vec{\gamma}$, 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 $\beta^q, q\in\{1..k\}$ vectors as index-dependent linear combinations of the $\gamma^q, q\in\{1..k\}$ 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

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