Using exterior product to prove formula for expansion of a determinant according to its first k rows:

Question

I am self studying the lecture notes of math52H standford by Eliasberg, here is a exercise I have some trouble with doing it.

$\dim(V) = n$, and

$x_1,x_2 ...x_n$ are the coordinates of dual space $V^{\ast} $

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Answer 1

Let $(e_1, \dots, e_n)$ be the standard basis of $\mathbb{R}^n$ (so that $(x_1, \dots, x_n)$ is the dual basis of $(e_1, \dots, e_n)$). Evaluating both sides on $e_1, \dots, e_n$ we get

$$ \det(A) = \det(A) (x_1 \wedge \dots \wedge x_n)(e_1, \dots, e_n) = A^{*}(x_1 \wedge \dots \wedge x_n)(e_1, \dots, e_n) = \left( A^{*}(x_1 \wedge \dots \wedge x_k) \wedge A^{*}(x_{k+1} \wedge \dots \wedge x_n) \right)(e_1 , \dots, e_n) \\= \sum_{i_1 < \dots < i_k, i_{k+1} < \dots < i_n} (-1)^{\text{inv}(i_1,\dots,i_n)} A^{*}(x_1 \wedge \dots \wedge x_k)(e_{i_1}, \dots, e_{i_k}) A^{*}(x_{k+1} \wedge \dots \wedge x_n)(e_{i_{k+1}}, \dots, e_{i_n}).$$

Now mimic the proof of proposition 2.17 to calculate the terms $A^{*}(x_1 \wedge \dots \wedge x_k)(e_{i_1}, \dots, e_{i_k})$ and $A^{*}(x_{k+1} \wedge \dots \wedge x_n)(e_{i_{k+1}}, \dots, e_{i_n})$ which will give you the required equality.