### Proof of Properties

##### 1. Transpose

We prove that the determinant is not changed by the transpose.

If we denote the `ij`-entry of `A` by `a`_{ij}, then the `ij`-entry of `A`^{T} is `a`_{ji}. Consequently, if we change `a`_{ij} to `a`_{ji} in the explicit formula

`det`**A** = ∑_{all permutations} `sign`(`i`_{1}, `i`_{2}, ..., `i`_{n}) `a`_{i11}`a`_{i22}...`a`_{in}_{n},

then we get

`det`**A**^{T} = ∑_{all permutations} `sign`(`i`_{1}, `i`_{2}, ..., `i`_{n}) `a`_{1i1}`a`_{2i2}...`a`_{nin}.

To see why `det`**A**^{T} can be equal to `det`**A**, let us consider the 4 by 4 case. Consider a term `sign`(3, 1, 4, 2)`a`_{13}`a`_{21}`a`_{34}`a`_{42} in `det`**A**^{T}. By looking at the product of entries, we have

`a`_{13}`a`_{21}`a`_{34}`a`_{42} = `a`_{21}`a`_{42}`a`_{13}`a`_{34}.

Correspondingly, we have the term `sign`(2, 4, 1, 3)`a`_{21}`a`_{42}`a`_{13}`a`_{34} in `det`**A**. It turns out that `sign`(3, 1, 4, 2) = `sign`(2, 4, 1, 3), so that the corresponding terms in `det`**A**^{T} and `det`**A** are equal (together with the sign).

In general, a similar argument can be carried out as follows. Any term `sign`(`i`_{1}, `i`_{2}, ..., `i`_{n}) `a`_{1i1}`a`_{2i2}...`a`_{nin} in `det`**A**^{T} can be rewritten as `sign`(`i`_{1}, `i`_{2}, ..., `i`_{n}) `a`_{j11}`a`_{j22}...`a`_{jnn}, which is compared with the corresponding term `sign`(`j`_{1}, `j`_{2}, ..., `j`_{n}) `a`_{j11}`a`_{j22}...`a`_{jnn} in `det`**A**. The equality `det`**A**^{T} = `det`**A** will then be the consequence of the equality `sign`(`j`_{1}, `j`_{2}, ..., `j`_{n}) = `sign`(`i`_{1}, `i`_{2}, ..., `i`_{n}).

To prove `sign`(`j`_{1}, `j`_{2}, ..., `j`_{n}) = `sign`(`i`_{1}, `i`_{2}, ..., `i`_{n}), we first need to understand the rewriting process. In more detail, the process is the following. We start with a 2 by `n` matrix

first indices in `det`**A**^{T} ⇒ |
[ |
1 |
2 |
... |
`n` |
]. |

second indices in `det`**A**^{T} ⇒ |
`i`_{1} |
`i`_{2} |
... |
`i`_{n} |

A sequence of transpositions that changes (`i`_{1}, `i`_{2}, ..., `i`_{n}) to (1, 2, ..., `n`) can be applied to the columns of the matrix, giving us a sequence of transpositions on 2 by `n` matrices

[ |
1 |
2 |
... |
`n` |
] → ... → [ |
`j`_{1} |
`j`_{2} |
... |
`j`_{n} |
]. |
⇐ first indices in `det`**A** |

`i`_{1} |
`i`_{2} |
... |
`i`_{n} |
1 |
2 |
... |
`n` |
⇐ second indices in `det`**A** |

For the 4 by 4 example `a`_{13}`a`_{21}`a`_{34}`a`_{42}, this means the following

[ |
1 |
2 |
3 |
4 |
] → [ |
2 |
1 |
3 |
4 |
] → [ |
2 |
1 |
4 |
3 |
] → [ |
2 |
4 |
1 |
3 |
], |

3 |
1 |
4 |
2 |
1 |
3 |
4 |
2 |
1 |
3 |
2 |
4 |
1 |
2 |
3 |
4 |

where we note that the result corresponds to `a`_{21}`a`_{42}`a`_{13}`a`_{34}. The upshot is that, the sequence of transpositions that changed (3, 1, 4, 2) to (1, 2, 3, 4) also changes (1, 2, 3, 4) to (2, 4, 1, 3). Equivalently, the transpositions in the sequence

(2, 4, 1, 3) → (2, 1, 4, 3) → (2, 1, 3, 4) → (1, 2, 3, 4)

is the reverse of the transpositions in the sequence

(3, 1, 4, 2) → (1, 3, 4, 2) → (1, 3, 2, 4) → (1, 2, 3, 4).

Since the reversing does not change the number of transpositions, we conclude that `sign`(2, 4, 1, 3) = `sign`(3, 1, 4, 2).

In general, given a sequence of transpositions

(`i`_{1}, `i`_{2}, ..., `i`_{n}) → ... → (1, 2,..., `n`),

the reverse sequence of transpositions gives us

(`j`_{1}, `j`_{2}, ..., `j`_{n}) → ... → (1, 2,..., `n`).

Since the two sequences consist of the same number of transpositions, we have `sign`(`j`_{1}, `j`_{2}, ..., `j`_{n}) = `sign`(`i`_{1}, `i`_{2}, ..., `i`_{n}).