2
$\begingroup$

Need help by the notation:

For $i \in \{1,...,n\}$

$\displaystyle \det(A)=\sum_{\sigma \in S_n}\operatorname{sgn} (\sigma)\cdot \prod_{k=1}^{n}a_{k,\sigma(k)}=\sum_{j=1}^{n}\sum_{\sigma \in S_n, \atop \sigma(i)=j}\operatorname{sgn}(\sigma)\cdot \prod_{k \in \{1,...,n\}, \atop k \ne i}a_{k,\sigma(k)}\cdot a_{i,j}$

So, if the first equality gives the definition of the determinant, then the notation of the second one is a problem for me at the moment. How should I put $i$-s in it?

I take e.g. an $A \in K^{3 \times 3}$ matrix, so I have $S_3=\{ \operatorname{id} , (1,2), (1,3), (2,3), (1,2,3), (3,2,1)\}$. $\operatorname{sgn}(1,2)=\operatorname{sgn}(1,3)=\operatorname{sgn}(2,3)=-1$. Then I can rewrite the formula:

$\displaystyle \det(A)=\ldots =\sum_{\sigma \in S_3, \atop \sigma(i)=1}\operatorname{sgn}(\sigma)\cdot \prod_{k \in \{1,2,3\}, \atop k \ne i}a_{k,\sigma(k)}\cdot a_{i,1}+\sum_{\sigma \in S_3, \atop \sigma(i)=2}\operatorname{sgn}(\sigma)\cdot \prod_{k \in \{1,2,3\}, \atop k \ne i}a_{k,\sigma(k)}\cdot a_{i,2}+ \sum_{\sigma \in S_3, \atop \sigma(i)=3}\operatorname{sgn}(\sigma)\cdot \prod_{k \in \{1,2,3\}, \atop k \ne i}a_{k,\sigma(k)}\cdot a_{i,3}$

But what about the $i$-s? Should I take all the $i$-s $\in \{1,2,3\}$ or should I pick up only one $i$? Anyway, how do I compute next to get a more simple formula?

As we can write $\sum_{k=1}^{n}a_n$ as $a_1 + a_2 + \ldots + a_n$, I want to do the same thing here. But, anyway, I would be glad to understand what the notations means and how it should be computed with the formula.

Thanks.

UPD: where the second formula comes from at all?

$\endgroup$
1
$\begingroup$

Let $i\in\left\{1,\dots,n\right\}$ be fixed. Define for $j\in\left\{1,\dots,n\right\}$ the set $$E_j:=\left\{\sigma\in\mathcal S_n \mid \sigma\left(i\right)=j\right\}.$$ Since all the elements of $\mathcal S_n$ are bijections, the sets $E_j$ are pairwise disjoint and their union is $\mathcal S_n$. If $\sigma$ belongs to $E_j$, then $$\prod_{k=1}^{n}a_{k,\sigma(k)}= a_{i,j}\cdot\prod_{k \in \{1,...,n\}, \atop k \ne i}a_{k,\sigma(k)}.$$

| cite | improve this answer | |
$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.