# Determinant of a transposed matrix

• B
• Eclair_de_XII

#### Eclair_de_XII

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?

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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 $$\begin{split} \det A &= \sum_{\sigma \in S_N} \operatorname{sgn}(\sigma) a_{1\sigma(1)} \cdots a_{N\sigma(N)}, \\ \det A^T &= \sum_{\rho \in S_N} \operatorname{sgn}(\rho) a_{\rho(1)1} \cdots a_{\rho(N)N}. \end{split}$$ 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 $$\operatorname{sgn}(\sigma) a_{1\sigma(1)} \cdots a_{N\sigma(N)} = \operatorname{sgn}(\rho_\sigma) a_{\rho_\sigma(1)1} \cdots a_{\rho_\sigma(N)N}.$$ (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.

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.

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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.

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