The thing to prove is: $\det(A^T)=\det(A)$ for some matrix $A=(a_{i,j}) \in K^{n \times n}$.
The solution:\begin{align}\det(A^T)&=\sum_{\sigma \in S_n}\operatorname{sgn}(\sigma) \cdot \prod_{i=1}^{n}a_{\sigma(i),i}\\&=\sum_{\sigma \in S_n}\operatorname{sgn}(\sigma) \cdot \prod_{j=1}^{n}a_{j,\sigma^{-1}(j)}\\&=\sum_{\tau \in S_n}\operatorname{sgn}(\tau^{-1})\cdot\prod_{j=1}^{n}a_{j,\tau(j)}\\&=\det(A).\end{align}
- Ok. In the first step we determine the $A^T$ with the help of the definition of the transposed matrix, that says $A^T=(a_{j,i})$.
- What happens next? We say that $\sigma(i)=j, i=\sigma^{-1}(j)$ and change $i \to j$ in the product sign. But what was that? What does this substituion mean? What was actually made here?
- In the next step we take another function $\tau$ and say that $\sigma=\tau^{-1}$ and $\sigma^{-1}(j)=\tau(j)$ in order to make the formula look similar to the definition of the determinant of $A$.
Questions:
-
Why do we make these substitutions in steps 2 and 3? Why does $\sigma(i)=j, \sigma=\tau^{-1}$ and not $\displaystyle \sigma(i)=\sum_{j=1}^{n}\pi\cdot {\infty^{2}}^{log(i)}-j!$, for example? What is the idea of every substituion and why they are true?
-
We have a symmetric group. If $\sigma=\tau^{-1}$ then $\sigma$ is the inverse element of $\tau$. But we are used to name the inverse element of some $x$ as $x^{-1}$ and not as some $u$. Otherwise, $u=x^{-1}$.
So, we name the same element with two different names and claim that we can do this. Moreover, we are sure that $\sigma=\tau^{-1}$. Actually, why these two?
Thanks.
Best Answer