Proof of Theorem 7 (Chapter 5) in Hoffman and Kunze's *Linear Algebra* is unclear

Question

Let $V$ be a free module of rank $n$ over a commutative ring $K$ with identity. We denote the space of all $r$-linear forms on $V$ by $M^r(V)$ and the space of all alternating $r$-linear forms by $\Lambda^r(V)$. For $L \in M^r(V)$ and any permutation $\sigma$ of $\{1,\dots,r\}$, we obtain another $r$-linear function $L_\sigma$ by defining $$L_\sigma(\alpha_1,\dots,\alpha_r) = L(\alpha_{\sigma 1},\dots,\alpha_{\sigma r})$$ for all $(\alpha_1,\dots,\alpha_r) \in V^r$. For each $L \in M^r(V)$, we define the alternating $r$-linear function $\pi_r L$ by $$\pi_r L = \sum_\sigma (\operatorname{sgn} \sigma) L_\sigma$$ where the sum is over all permutations $\sigma$ of $\{1,\dots,r\}$.

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Now, Theorem 7 of Chapter 5 in Hoffman and Kunze's Linear Algebra states the following:

Theorem $7$. Let $K$ be a commutative ring with identity and let $V$ be a free $K$-module of rank $n$. If $r > n$, then $\Lambda^r(V) = \{0\}$. If $1 \leq r \leq n$, then $\Lambda^r(V)$ is a free $K$-module of rank $\binom{n}{r}$.

Proof. Suppose $\{ \beta_1,\dots,\beta_n \}$ is an ordered basis for $V$ with dual basis $\{ f_1,\dots,f_n\}$. If $L \in M^r(V)$, then $$L = \sum_H L(\beta_{h_1},\dots,\beta_{h_r})\ f_{h_1}\! \otimes \dots \otimes f_{h_r} \tag{5-37}$$ where the sum extends over all $r$-tuples $H = (h_1,\dots,h_r)$ of integers between $1$ and $n$. If $L \in \Lambda^r(V)$, this sum need be extended only over the $r$-tuples $H$ for which $h_1,\dots,h_r$ are distinct because if $L$ is alternating then $$L(\beta_{h_1},\dots,\beta_{h_r}) = 0$$ whenever two subscripts $h_i$ are the same. If $r > n$ then in each $r$-tuple some integer must be repeated. Thus $\Lambda^r(V) = \{0\}$ if $r > n$.

Now, suppose $1 \leq r \leq n$. We define an $r$-shuffle of $\{ 1,\dots, n\}$ to be an $r$-tuple $J = (j_1,\dots,j_r)$ such that $1 \leq j_1 < \dots < j_r \leq n$. There are $$\binom{n}{r} = \frac{n!}{r!(n-r)!}$$ such shuffles. Suppose we fix an $r$-shuffle $J$. Let $L_J$ be the sum of all the terms in $(5\text{-}37)$ for which the indexing $r$-tuple $H$ is a permutation of the $r$-shuffle $J$. If $\sigma$ is a permutation of $\{1,\dots,r\}$, then $$L(\beta_{j_{\sigma 1}},\dots,\beta_{j_{\sigma r}}) = (\operatorname{sgn} \sigma) L(\beta_{j_1},\dots,\beta_{j_r}).$$ Thus, $$L_J = L(\beta_{j_1},\dots,\beta_{j_r}) D_J \tag{5-38}$$ where $$\begin{align} D_J &= \sum_\sigma (\operatorname{sgn} \sigma)\ f_{j_{\sigma 1}}\! \otimes \dots \otimes f_{j_{\sigma r}} \tag{5-39}\\ &= \pi_r(f_{j_1}\! \otimes \dots \otimes f_{j_r}). \end{align}$$ We see from $(5\text{-}39)$ that each $D_J$ is alternating and that $$L = \sum_{\text{shuffles $J$}} L(\beta_{j_1},\dots,\beta_{j_r}) D_J \tag{5-40}$$ for every $L$ in $\Lambda^r(V)$. The assertion is that the $\binom{n}{r}$ forms $D_J$ constitute a basis for $\Lambda^r(V)$. We have seen that they span $\Lambda^r(V)$. It is easy to see that they are independent. Hence, proved.

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My doubt is how in Equation $(5\text{-}39)$ we can go from the first line to the second line. It does not seem to follow directly from the definition of $\pi_r L$, because $$\pi_r(f_{j_1}\! \otimes \dots \otimes f_{j_r}) := \sum_\sigma (\operatorname{sgn} \sigma) (f_{j_1}\! \otimes \dots \otimes f_{j_r} )_\sigma \stackrel{?}{=} \sum_{\sigma} (\operatorname{sgn} \sigma)\ f_{j_{\sigma 1}}\! \otimes \dots \otimes f_{j_{\sigma r}}.$$

If someone can give me a step-by-step proof of the equality it would be really helpful.

Answer 1

The key observation is the following identity which we state as a lemma.

Lemma: Let $1\leq r \leq n$. Consider the $r$-tuple $(j_1,\dots,j_r)$, $1 \leq j_1,\dots,j_r \leq n$. Let $\sigma$ be a permutation of $\{1,\dots,r\}$. Then, we have $$f_{j_{\sigma 1}}\! \otimes \dots \otimes f_{j_{\sigma r}} = (f_{j_1}\! \otimes \dots \otimes f_{j_r})_{\sigma^{-1}}.$$

Proof: For $(\alpha_1,\dots,\alpha_r) \in V^r$, we have $$ \begin{align} (f_{j_{\sigma 1}}\! \otimes \dots \otimes f_{j_{\sigma r}})(\alpha_1,\dots,\alpha_r) &= f_{j_{\sigma 1}}(\alpha_1) \cdots f_{j_{\sigma r}}(\alpha_r) \\ &= f_{j_1}(\alpha_{\sigma^{-1} 1}) \cdots f_{j_r}(\alpha_{\sigma^{-1} r}) &&\text{(by rearranging terms)} \\ &= (f_{j_1}\! \otimes \dots \otimes f_{j_r})(\alpha_{\sigma^{-1} 1},\dots,\alpha_{\sigma^{-1} r}) \\ &= (f_{j_1}\! \otimes \dots \otimes f_{j_r})_{\sigma^{-1}}(\alpha_1,\dots,\alpha_r). \end{align} $$ Since $(\alpha_1,\dots,\alpha_r)$ was an arbitrary element of $V^r$, we get $$f_{j_{\sigma 1}}\! \otimes \dots \otimes f_{j_{\sigma r}} = (f_{j_1}\! \otimes \dots \otimes f_{j_r})_{\sigma^{-1}}.$$ Hence, proved.

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Using this lemma, we show that equation $(5\text{-}39)$ is correct as follows. $$ \begin{align} \pi_r(f_{j_1}\! \otimes \dots \otimes f_{j_r}) &:= \sum_{\sigma} (\operatorname{sgn} \sigma) (f_{j_1}\! \otimes \dots \otimes f_{j_r})_\sigma\\ &= \sum_\sigma (\operatorname{sgn} \sigma)\ f_{j_{\sigma^{-1} 1}}\! \otimes \dots \otimes f_{j_{\sigma^{-1} r}} &&(\text{by above lemma})\\ &= \sum_\sigma (\operatorname{sgn} \sigma^{-1})\ f_{j_{\sigma^{-1} 1}}\!\otimes \dots \otimes f_{j_{\sigma^{-1} r}} &&(\because \operatorname{sgn} \sigma = \operatorname{sgn} \sigma^{-1}). \end{align} $$ Now, as $\sigma$ runs (once) over all permutations of $\{ 1,\dots,r \}$, so does $\sigma^{-1}$. Therefore, $$ \pi_r(f_{j_1}\! \otimes \dots \otimes f_{j_r}) = \sum_{\tau}(\operatorname{sgn} \tau)\ f_{j_{\tau 1}}\! \otimes \dots \otimes f_{j_{\tau r}} $$ where the sum is taken over all permutations $\tau$ of $\{ 1,\dots,r \}$. But, the right-hand side is precisely $D_J$. Hence, proved.

Answer 2

The proof is given in the preceding Example 12 in the text (see page 169).