Let $V$ be a vector space of dimension $n$ with basis $\{v_1,\cdots,v_n\}$.

Let $\phi$ be an n-alternating multilinear map and $A:V\rightarrow V$ is any map (matrix form) then we have to prove that

$\phi(Av_1,\cdots,Av_n)=\det(A) \phi(v_1,v_2,\cdots,v_n)$

I could do this for $n=2$ but could not generalize it..



As $\phi$ is alternating $\phi(v_1,v_1)=\phi(v_2,v_2)=0$ and $\phi(v_2,v_1)=-\phi(v_1,v_2)$. So, we have


Now, i am having difficulty in generalizing this..

$$\det(A)\phi(v_1,v_2,\cdots,v_n)=\sum_{\sigma}sgn(\sigma)\Pi_{i} a_{i\sigma(i)}\phi(v_1,v_2,\cdots,v_n)$$

As $\phi$ is alternating we have $sgn(\sigma)\phi(v_1,v_2,\cdots,v_n)=\phi(v_{\sigma(1)},v_{\sigma(2)},\cdots,v_{\sigma(n)})$

So, we have $$\det(A)\phi(v_1,v_2,\cdots,v_n)=\sum_{\sigma}\Pi_{i} a_{i\sigma(i)}\phi(v_{\sigma(1)},v_{\sigma(2)},\cdots,v_{\sigma(n)})$$

I am not able to relate this to $\phi(Av_1,\cdots,Av_n)$.

See that $A=(a_{ij})$..

Please provide some hints


It seems as though you have the right idea, but you're getting caught up on the notation. Here's a suitable generalization of your proof: $$ \begin{align} \phi(Av_1,\cdots,Av_n) &= \phi\left(\sum_{j_1=1}^n a_{1j_1} v_{j_1}, \sum_{j_2=1}^n a_{2j_2} v_{j_2}, \dots, \sum_{j_n=1}^n a_{nj_n} v_{j_n}\right) \\ & = \sum_{j_1=1}^n \sum_{j_2=1}^n \cdots \sum_{j_n=1}^n \left(\left[\prod_{i=1}^n a_{i j_i}\right] \phi\left( v_{j_1}, v_{j_2}, \dots, v_{j_n}\right)\right) \\ & = \sum_{(j_1,\dots,j_n) \in \{1,\dots,n\}^n} \left(\left[\prod_{i=1}^n a_{i j_i}\right] \phi\left( v_{j_1}, v_{j_2}, \dots, v_{j_n}\right)\right) \end{align} $$ Now, we note that if any two of the $j_{k}$ are equal, then $\phi\left( v_{j_1}, v_{j_2}, \dots, v_{j_n}\right) = 0$, so that we may remove the corresponding term from the sum. We may simply sum over all tuples $(j_1,\dots,j_n)$ that (may) produce a non-zero term, which means we want all tuples of $j_k$ such that each pair of $j_k$ are mutually distinct. That is, $(j_1,\dots,j_n)$ must be a permutation of $(1,\dots,n)$. That is, the only tuples we want to consider are those for which there is a $\sigma \in S_n$ such that $j_k = \sigma(k)$. We may therefore rewrite the above sum as \begin{align} \phi(Av_1,\cdots,Av_n) &= \sum_{\sigma \in S_n} \left(\left[\prod_{i=1}^n a_{i \sigma(i)}\right] \phi\left( v_{\sigma(1)}, v_{\sigma(2)}, \dots, v_{\sigma(n)}\right)\right) \\ & = \sum_{\sigma \in S_n} \left(\left[\prod_{i=1}^n a_{i \sigma(i)}\right] \operatorname{sgn}(\sigma)\, \phi\left( v_{1}, v_{2}, \dots, v_{n}\right)\right) \\ & = \left(\sum_{\sigma \in S_n} \left[\prod_{i=1}^n a_{i \sigma(i)}\right] \operatorname{sgn}(\sigma)\right) \phi\left( v_{1}, v_{2}, \dots, v_{n}\right) \end{align}

  • $\begingroup$ I was getting confused with the notations.. Now it is clear.. Thank you so much.. I have nothing to ask.. :) $\endgroup$ – user87543 Aug 7 '15 at 17:19
  • $\begingroup$ you're welcome :D $\endgroup$ – Omnomnomnom Aug 7 '15 at 21:00

The map $V^n\to K$, $(v_1,\ldots, v_n)\mapsto \phi(Av_1,\ldots, Av_n)$ is readily checked to be alternating multilinear. As the space of alternating $n$-forms is onedimensional, it must be a multiple $c(A)\phi$ of $\phi$. To determine $c(A)$ it suffices to evaluate $\phi(e_1,\ldots, e_n)$, i.e., $\phi$ applied to the columns of $A$. By alternating multilinear property, adding a column to another column does not change the value of $\phi$, whereas multiplying a column with a constant multiplies the value by the same constant. As these (Gauß) steps can be used to transform $A$ to the identity matrix while extracting $\det(A)$ as factor, we conclude that indeed $c(A)=\det(A)$.

  • $\begingroup$ I do not understand third line... i understand that it has to be constant multiple of $\phi$.. I could not understand after that... $\endgroup$ – user87543 Aug 7 '15 at 17:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy