Determinant of a transposed matrix

[Eclair_de_XII]

TL;DR

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?

 

[pasmith]

Generally one uses S_N for the group of permutations of \{1, \dots, N\} and \sigma and \rho for arbitrary permutations. You can simplify your notation and argument significantly by using the fact that S_N is a group acting on \{1, \dots, N\}.

By definition <br /> \begin{split}<br /> \det A &amp;= \sum_{\sigma \in S_N} \operatorname{sgn}(\sigma) a_{1\sigma(1)} \cdots a_{N\sigma(N)}, \\<br /> \det A^T &amp;= \sum_{\rho \in S_N} \operatorname{sgn}(\rho) a_{\rho(1)1} \cdots a_{\rho(N)N}.<br /> \end{split} Your claim is then that there exists a bijection from S_N to itself such that if \sigma \mapsto \rho_\sigma then for each \sigma \in S_N we have <br /> \operatorname{sgn}(\sigma) a_{1\sigma(1)} \cdots a_{N\sigma(N)} = \operatorname{sgn}(\rho_\sigma) a_{\rho_\sigma(1)1} \cdots a_{\rho_\sigma(N)N}. (We need a bijection because we want each summand of \det A to appear exactly once in the sum for \det A^T and vice versa). Ordering the factors on the left hand side by the second index rather than the first is the way to proceed, but because \sigma is a permutation this can be done in one step: The factor with second index i has first index \sigma^{-1}(i). Two basic facts about permutation groups complete the proof.

 

[Eclair_de_XII]

Unfortunately, I've not taken abstract algebra during my undergraduate career, so I am unfamiliar with groups and would be uncomfortable attempting to implement them in my explanation/proof.

I meant to say in the last two lines that given any row-ordering $p_i$ in a summand of the determinant of the transpose, there is a row-ordering $p_{i'}$ that can be applied to $p_i$. The row-ordering $p_{i'}$ is how I described it earlier. This resulting ordering corresponds to an arbitrary term in the determinant of the original matrix.

 

[julian]

What I would write is: For every permutation $\sigma$ in the set of permutations (denoted $S_N$) there exists a unique inverse permutation $\sigma^{-1}$ in the set of permutations and where no two distinct permutations have the same inverse.

Say $\sigma (i) = j$ then

$$
a_{i \sigma(i)} = a_{ij} = a_{\sigma^{-1} (j) j}
$$

so that

$$
\prod_{i=1}^N a_{i \sigma (i)} = \prod_{i=1}^N a_{\sigma^{-1} (i) i} .
$$
If $\sigma$ is an even permutation then obviously $\sigma^{-1}$ is also an even permutation. If $\sigma$ is an odd permutation then obviously $\sigma^{-1}$ is also an odd permutation. Therefore,

$$
\text{sgn} (\sigma) = \text{sgn} (\sigma^{-1})
$$

and

$$
\text{sgn} (\sigma ) \prod_{i=1}^N a_{i \sigma (i)} = \text{sgn} ( \sigma^{-1} ) \prod_{i=1}^N a_{\sigma^{-1} (i) i} .
$$

Then

\begin{align*}
\det (A) & = \sum_{\sigma \in S_N} \text{sgn} (\sigma ) \prod_{i=1}^N a_{i \sigma (i)}
\nonumber \\
& = \sum_{\sigma \in S_N} \text{sgn} ( \sigma^{-1} ) \prod_{i=1}^N a_{\sigma^{-1} (i) i}
\nonumber \\
& = \sum_{\sigma \in S_N} \text{sgn} ( \sigma ) \prod_{i=1}^N a_{\sigma (i) i} = \det (A^T)
\end{align*}

where we have arrived at the last line by noting that for every permutation $\sigma$ in $S_N$ there exists a unique inverse permutation $\sigma^{-1}$ in $S_N$ and where no two distinct permutations have the same inverse, and so we are in effect summing over all permutations in $S_N$ in the second line.