Prove that the determinant of a swap matrix is $-1$

Question

How can I prove that a swap matrix (a matrix which multiplied with another, swaps a pair of its rows/columns) has determinant $-1$?

Answer 1

There are a few ways to define the determinant, and the proof changes accordingly.

Eigenvalues

The determinant is defined to be the product of the (complex) eigenvalues to the power of their multiplicities (the dimension of the corresponding generalised eigenspace).

Let $e_i$ be the $i$th standard basis vector. In this case, if the $n \times n$ swap matrix swaps the $i$th and $j$th column ($i < j$), then the vectors $$e_1, \ldots, e_{i-1}, e_{i+1}, \ldots, e_{j-1}, e_{j+1}, \ldots, e_n$$ are all linearly independent eigenvectors corresponding to eigenvalue $1$, as is the vector $e_i + e_j$. Meanwhile, the vector $e_i - e_j$ is an eigenvector corresponding to eigenvalue $-1$.

Hence, we have eigenvalue $1$ with multiplicity at least $n - 1$, and eigenvalue $-1$ with multiplicity at least $1$. Since we have found at least $n$ linearly independent eigenvalues, we have a complete list of eigenvalues and their multiplicities. Hence, the determinant is $1^{n - 1} \cdot (-1)^1 = -1$.

Cofactor Expansion

We can use the fact that cofactor expansions can be made along any row or column, with an appropriate change of sign. Note that the diagonal entries are always counted positively. Note also that, expanding along a row that isn't being swapped, the diagonal $1$ is the only non-zero entry, and the cofactor is a swap matrix of a smaller dimension. For example, expanding along the 3rd row: $$\begin{vmatrix}1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0\end{vmatrix} = \begin{vmatrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{vmatrix}.$$ By inductively reducing, you get to the $2 \times 2$ case, which can be computed easily.

Leibniz Formula

The determinant of an $n \times n$ matrix $(a_{i,j})_{i,j = 1}^n$ can be defined as follows: $$\sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{i=1}^n a_{i, \sigma(i)},$$ where $\operatorname{sgn}(\sigma)$ returns $1$ when $\sigma$ is even, and $-1$ when $\sigma$ is odd. Note that the swap matrix can be expressed as a permutation matrix: $$a_{i,j} = \delta_{i, \tau(i)},$$ where $\delta_{i, j}$ returns $1$ when $i = j$ or $0$ otherwise (i.e. the entries of the identity matrix), and $\tau$ is a transposition. Therefore, multiplying a permutation by $\tau$ causes it to change from odd to even. Hence, \begin{align*}\sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{i=1}^n a_{i, \sigma(i)} &= \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{i=1}^n \delta_{i, \sigma \circ \tau (i)} \\ &= - \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma \circ \tau) \prod_{i=1}^n \delta_{i, \sigma \circ \tau (i)} \\ &= - \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{i=1}^n \delta_{i, \sigma(i)} \\ &= - \operatorname{det}(I) = -1. \end{align*}