- #1

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- TL;DR Summary
- Let ##A## be a matrix of size ##(n,n)## where ##n\in\mathbb{N}##. Then the determinant of ##A## is equal to the determinant of ##A## transposed, to be denoted ##A^T##.

By definition, ##\det A=\sum_{p_j\in P}\textrm{sgn}(p_j)\cdot a_{1j_1}\cdot\ldots\cdot a_{nj_n}##, where ##P## denotes the set of all permutations of the ordered sequence ##(1,\ldots,n)##. Denote the number of permutations needed to map the natural ordering to ##p_j## as ##N_j##.

Now consider ##\det A^T## which is equal to:

\begin{align}\sum_{p_i\in P}\textrm{sgn}(p_i)\cdot a_{i_11}\cdot\ldots\cdot a_{i_nn}\end{align}

Note: ##i_k## denotes the element of ##p_i## at the k-th index.

To show equality, we must show that each summand in ##\det A## is also in ##\det A^T##. In other words, we must show that there is a permutation ##p_l## s.t.:

\begin{align}

\textrm{sgn}(p_l)\cdot a_{l_11}\cdot\ldots\cdot a_{l_nn}=\textrm{sgn}(p_j)\cdot a_{j_11}\cdot\ldots\cdot a_{j_nn}

\end{align}

Consider the ordered list:

\begin{align}(i_1,1),\ldots,(i_n,n)\end{align}

For each element in the list, there is an integer ##m## s.t. ##j_m=k##. It will take ##N_j## permutations in order to map this ordering to an ordering of the form:

\begin{align}(i_1',j_1),\ldots,(i_n',j_n)\end{align}

where ##(i_1',\ldots,i_n')## is the ordering obtained from permutating ##p_i## wrt the ordering ##p_j##. Bearing in mind that ##\prod_{k=1}^n a_{i_kk}\equiv \prod_{k=1}^n a_{i_k'j_k}##, we have:

\begin{align}

\textrm{sgn}(p_i)\cdot\prod_{k=1}^n a_{i_kk}=\textrm{sgn}(p_j)\cdot\textrm{sgn}(p_{i'})\cdot\prod_{k=1}^n a_{i_k'j_k}

\end{align}

It will take ##N_{i'}## permutations to map ##p_{i'}## to the natural ordering. This corresponds to ##N_{i'}## sign changes:

\begin{align}

\textrm{sgn}(p_j)\cdot\prod_{k=1}^n a_{kj_k}=\textrm{sgn}(p_{i'})\cdot\left[\textrm{sgn}(p_j)\cdot\textrm{sgn}(p_{i'})\cdot\prod_{k=1}^n a_{i_k'j_k}\right]

\end{align}

% I am asking for critique on this proof. Is it accurate? Is it understandable? Is there any unnecessary notation I used?

Now consider ##\det A^T## which is equal to:

\begin{align}\sum_{p_i\in P}\textrm{sgn}(p_i)\cdot a_{i_11}\cdot\ldots\cdot a_{i_nn}\end{align}

Note: ##i_k## denotes the element of ##p_i## at the k-th index.

To show equality, we must show that each summand in ##\det A## is also in ##\det A^T##. In other words, we must show that there is a permutation ##p_l## s.t.:

\begin{align}

\textrm{sgn}(p_l)\cdot a_{l_11}\cdot\ldots\cdot a_{l_nn}=\textrm{sgn}(p_j)\cdot a_{j_11}\cdot\ldots\cdot a_{j_nn}

\end{align}

Consider the ordered list:

\begin{align}(i_1,1),\ldots,(i_n,n)\end{align}

For each element in the list, there is an integer ##m## s.t. ##j_m=k##. It will take ##N_j## permutations in order to map this ordering to an ordering of the form:

\begin{align}(i_1',j_1),\ldots,(i_n',j_n)\end{align}

where ##(i_1',\ldots,i_n')## is the ordering obtained from permutating ##p_i## wrt the ordering ##p_j##. Bearing in mind that ##\prod_{k=1}^n a_{i_kk}\equiv \prod_{k=1}^n a_{i_k'j_k}##, we have:

\begin{align}

\textrm{sgn}(p_i)\cdot\prod_{k=1}^n a_{i_kk}=\textrm{sgn}(p_j)\cdot\textrm{sgn}(p_{i'})\cdot\prod_{k=1}^n a_{i_k'j_k}

\end{align}

It will take ##N_{i'}## permutations to map ##p_{i'}## to the natural ordering. This corresponds to ##N_{i'}## sign changes:

\begin{align}

\textrm{sgn}(p_j)\cdot\prod_{k=1}^n a_{kj_k}=\textrm{sgn}(p_{i'})\cdot\left[\textrm{sgn}(p_j)\cdot\textrm{sgn}(p_{i'})\cdot\prod_{k=1}^n a_{i_k'j_k}\right]

\end{align}

% I am asking for critique on this proof. Is it accurate? Is it understandable? Is there any unnecessary notation I used?

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