I want to prove that $$\det(A)=\det(A^T)$$ and the one step I don't understand (the problem is guiding you thought it is to prove $$P^T_{\sigma} = P_{\sigma^{-1}}$$ where P is a permutation matrix.
(Note to self: dominika).
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http://www.math.upenn.edu/~ekorman/teaching/det.pdf here is the proof. – mesel Mar 07 '14 at 14:40
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it is already asked in here also http://math.stackexchange.com/questions/298583/how-do-i-prove-that-det-a-det-at – mesel Mar 07 '14 at 14:41
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I did not see the comments before posting my answer. – Supriyo Mar 07 '14 at 14:45
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Clearly $P_{\sigma^{-1}} P_\sigma = I$, because if you apply a permutation $\sigma$ to the indices, and then its inverse $\sigma^{-1}$, you get the initial configuration. This relation implies $P_\sigma^{-1} = P_{\sigma^{-1}}$.
Moreover we know that $P_\sigma$ is obtained by applying $\sigma$ to the columns of $I$. Then $P_\sigma^T$ is obtained by applying $\sigma$ to the rows of $I$ and, thus is not difficult to realize that $P_\sigma P_\sigma^T = I$. But this implies $P_\sigma^{-1} = P_{\sigma}^T$.
Thus $P^T_\sigma = P_{\sigma^{-1}}$.
Paglia
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The proof I know does not use the same way.
$$ \det A = \sum_\sigma {\epsilon(\sigma)} \prod_i A_{i,\sigma (i)}\\ \det A^T = \sum_\sigma {\epsilon(\sigma)} \prod_i A_{\sigma (i),i} \\\forall\sigma\ \ \prod_i A_{\sigma(i),i} = \prod_j A_{j,\sigma^{-1}(j)} $$ As $\sigma \sigma^{-1} = {\rm id}, 1 = \epsilon( {\rm id})= \epsilon(\sigma \sigma^{-1} ) =\epsilon(\sigma)\epsilon(\sigma^{-1}) $ so $$\epsilon(\sigma)=\epsilon(\sigma^{-1})$$
Now as $\sigma \to \sigma^{-1}$ is a bijection: $$ \det A^T = \sum_\sigma {\epsilon(\sigma^{-1})}\prod_j A_{j,\sigma^{-1}(j)} = \sum_\tau {\epsilon(\tau)}\prod_j A_{j,\tau(j)} = \det A $$
mookid
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Could you please check again if $ϵ(σ)$ is a multiplicative or additive homomorphism? In the additive version $0=ϵ(id)=ϵ(σ)+ϵ(σ^{-1})$, however, the important conclusion $(-1)^{ϵ(σ^{-1})}=(-1)^{-ϵ(σ)}=(-1)^{ϵ(σ)}$ remains the same. I.e., $sgn(σ)=(-1)^{ϵ(σ)}$ is the multiplicative homomorphism to the additive $ϵ(σ)$. – Lutz Lehmann Mar 07 '14 at 16:04
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Then the question becomes, what is the definition of $ϵ(σ)$ and why not write directly $ϵ(σ)$ instead of $(-1)^{ϵ(σ)}$. By the way, if always $ϵ(σ)=\pm1$, then also always $(-1)^{ϵ(σ)}=-1$. – Lutz Lehmann Mar 07 '14 at 16:09
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Write the definition of $\det(A)$ explicitly.
$$\det (A) = \sum_{\phi} sgn(\phi) a_{1, \phi(1)} a_{2, \phi(2)} \dots a_{n,\phi(n)}$$ $\phi$ are all permutations on $\{1,2,\dots,n\}$
So observe that All terms of $A = (a_{i,j})$ are in the sum in some manner.
$A^t = (a_{j,i})$. So
$$\det (A^t) = \sum_{\phi} sgn(\phi) a_{\phi(1),1} a_{\phi(2),2} \dots a_{\phi(n),n}$$.
Now take arrange the terms of the above sum in a different order. See $\phi(1), \phi(2), \dots \phi(n)$ are $1,2,\dots n$ in some order. So let me write the above sum as
$$\det (A^t) = \sum_{\xi} sgn(\xi) a_{\xi(1),1} a_{\xi(2),2} \dots a_{\xi(n),n}$$
What's that? Is not it $\det(A)$ ? If not just write $\phi$ instead of $\xi$ and see.
Supriyo
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