If $f:(\mathbb{K}^n)^n \rightarrow \mathbb{K}$ is multilinear and alternating, prove: $f(T(u_1),T(u_2),...,T(u_n)=\det(A)f(u_1,...,u_n)$

Question

Let $V$ a $\mathbb{K}$-vector space of finite dimension $n$. Let $B=\{v_1,...,v_n\}$ a basis of $V$. Let $A=[a_{ij}]$ a matrix in ${\rm Mat}_n(\mathbb{K})$. Consider the linear transformation $T:V\rightarrow V$ constructed by $A$ defined by elements of $B$ as $$T(v_j)=a_{1j}v_1+...+a_{nj}v_n.$$

If $f:(\mathbb{K}^n)^n \rightarrow \mathbb{K}$ is multilinear and alternating, prove that $$f(T(u_1),T(u_2),...,T(u_n))=\det(A)f(u_1,...,u_n)$$ for all $u_1,...,u_n\in V$

My work:

As $f$ is multilinear and alternating then $f=f(v_1,...,v_n)\det_B$ Then $f(T(u_1),...,T(u_n))=f(v_1,...v_n)\det_B(T(u_1),...,T(u_n))$

Here I'm stuck. Can someone help me?

Answer 1

Since $f$ is multilinear and alternating, we can assume without loss of generality that $u_i = v_i$ for all $i$. Then brute force your way through:

$$\begin{align} f(T(v_1),\ldots, T(v_n)) &= f\left(\sum_{i_1=1}^n a_{i_11}v_{i_1},\ldots,\sum_{i_n=1}^na_{i_nn}v_{i_n} \right) \\ &= \sum_{i_1,\ldots,i_n=1}^n a_{i_11}\ldots a_{i_nn}f(v_{i_1},\ldots,v_{i_n}) \\ &\stackrel{(1)}{=} \sum_{\sigma \in S_n} a_{\sigma(1)1}\ldots a_{\sigma(n)n} f(v_{\sigma(1)},\ldots, v_{\sigma(n)}) \\ &= \sum_{\sigma \in S_n} {\rm sgn}(\sigma) a_{\sigma(1)1}\ldots a_{\sigma(n)n}f(v_1,\ldots,v_n) \\ &= \left(\sum_{\sigma \in S_n} {\rm sgn}(\sigma) a_{\sigma(1)1}\ldots a_{\sigma(n)n}\right)f(v_1,\ldots,v_n) \\ &\stackrel{(2)}{=} \left(\sum_{\sigma \in S_n} {\rm sgn}(\sigma) a_{1\sigma(1)}\ldots a_{n\sigma(n)}\right)f(v_1,\ldots,v_n) \\ &\stackrel{(3)}{=} \det(A) f(v_1,\ldots,v_n)\end{align}$$where:

• in $(1)$ we use that all the terms where some index repeats are zero, and we consider $\sigma \in S_n$ the permutation such that $i_1 = \sigma(1)$, ..., $i_n = \sigma(n)$;

• in $(2)$ we make a change of variable $\tau = \sigma^{-1}$, note that ${\rm sgn}(\tau) = {\rm sgn}(\sigma)$ and rename $\tau \to \sigma$ in the same step. Be very careful, we're not saying that $a_{\sigma(i)i} = a_{i\sigma(i)}$ for all $i$, but that $a_{\sigma(1)1}\ldots a_{\sigma(n)n} = a_{1\sigma(1)}\ldots a_{n\sigma(n)}$;

• $(3)$ is the definition of the determinant.

Answer 2

As you noticed, you have

$$f(T(u_1),...,T(u_n))=f(v_1,...v_n)det_B(T(u_1),...,T(u_n)).$$

You also have $$f(u_1,...,u_n)=f(v_1,...v_n)det_B(u_1,...,u_n).$$

If $\{u_1,\dots,u_n\}$ is a dependent family then the requested equality is trivial. Otherwise, $B^\prime =(u_1,\dots,u_n)$ is a basis of $V=\mathbb K^n$ and $det_B(u_1,...,u_n) = det (M_B^{B^\prime})\neq 0$. Hence $$f(T(u_1), \dots , T(u_n)) = det (M_{B^\prime}^B) det_B(T(u_1),...,T(u_n))f(u_1,\dots,u_n),$$ as $\left(M_B^{B^\prime}\right)^{-1}=M_{B^\prime}^B$ and $det(B^{-1})= 1/det(B)$ for any inversible matrix.

Finally, you have $Mat_B(T(u_1), \dots, T(u_n))=A M_B^{B^\prime}$, therefore $$det_B(T(u_1),...,T(u_n))=det(A)det(M_B^{B^\prime}).$$

You can conclude as $M_B^{B^\prime}M_{B^\prime}^B=I_n$, the identity matrix.