The Determinant Formula and Transposition
Transposing a matrix exchanges its rows and columns. It is natural to ask whether this rearrangement changes the determinant. The answer is no, but the reason is not that every term in the determinant formula stays the same under transposition. Instead, the terms are matched by pairing a permutation with its inverse. This pairing is the key to a proof that works for every square matrix, not just for particular examples.
Let \(n\geq 1\), and let \(\mathbf{A}=(a_{ij})\in\mathbb{R}^{n\times n}\). A permutation of \(\{1,\dots,n\}\) is a bijection from that set to itself; the set of all such permutations is denoted \(S_n\). The sign \(\operatorname{sgn}(\sigma)\) of a permutation \(\sigma\) is \(1\) if it is even and \(-1\) if it is odd. In the Leibniz formula, each permutation selects one matrix entry from every row and every column, and its sign determines whether its product is added or subtracted.
For a given \(\sigma\), the corresponding product in \(\det(\mathbf{A})\) is \(a_{1,\sigma(1)}a_{2,\sigma(2)}\cdots a_{n,\sigma(n)}\). The transpose formula uses entries \(a_{\tau(i),i}\) for a permutation \(\tau\). Those are not generally the same entries for the same permutation index. The correct match will be \(\tau=\sigma^{-1}\).
Why a Permutation and Its Inverse Have the Same Sign
We first justify the sign fact needed to pair the terms. Write a permutation in one-line form as \(\sigma(1),\dots,\sigma(n)\). An inversion is a pair \(i<j\) for which \(\sigma(i)>\sigma(j)\). The sign of \(\sigma\) is \((-1)^k\), where \(k\) is its number of inversions.
Proof. We match the inversions of \(\sigma\) with those of \(\sigma^{-1}\). Suppose \(i<j\) and \(\sigma(i)>\sigma(j)\), so \((i,j)\) is an inversion of \(\sigma\). Put \(r=\sigma(j)\) and \(s=\sigma(i)\). Then \(r<s\), while \(\sigma^{-1}(r)=j>i=\sigma^{-1}(s)\). Thus \((r,s)\) is an inversion of \(\sigma^{-1}\). This construction is reversible: an inversion \((r,s)\) of \(\sigma^{-1}\), with \(r<s\) and \(\sigma^{-1}(r)>\sigma^{-1}(s)\), gives the inversion \((\sigma^{-1}(s),\sigma^{-1}(r))\) of \(\sigma\). Therefore the two permutations have the same number of inversions, and their signs, each \((-1)\) raised to that number, are equal. \(\square
Pairing the Terms in the Formula
Proof. Apply the Leibniz formula to \(\mathbf{A}^{\top}\). For each \(\tau\in S_n\), its product uses \((\mathbf{A}^{\top})_{i,\tau(i)}=a_{\tau(i),i}\). Reindex the sum by setting \(\sigma=\tau^{-1}\). In a product for this \(\tau\), set \(j=\tau(i)\). Then \(i=\tau^{-1}(j)=\sigma(j)\), so the factors can be rewritten as follows:
The lemma gives \(\operatorname{sgn}(\tau)=\operatorname{sgn}(\tau^{-1})=\operatorname{sgn}(\sigma)\). As \(\tau\) ranges over \(S_n\), its inverse \(\sigma\) also ranges over all of \(S_n\), once each. Consequently, the sum of the signed products for \(\mathbf{A}^{\top}\) is precisely the sum of the signed products in the Leibniz formula for \(\mathbf{A}\):
This proves the identity for every positive integer \(n\), including \(n=1\), where there is just the identity permutation and the determinant is the sole entry. \(\square
Worked Checks from the Formula
Worked Example: A Two-by-Two Matrix
Let \(\mathbf{A}=\begin{bmatrix}2&-3\\5&4\end{bmatrix}\). The two permutations in \(S_2\) are the identity, with sign \(1\), and the swap \((12)\), with sign \(-1\). The formula gives
Now \(\mathbf{A}^{\top}=\begin{bmatrix}2&5\\-3&4\end{bmatrix}\). Applying the same formula,
Check: both evaluations use the identity-permutation contribution \(8\) and the swap contribution \(15\); their sum is \(23\). Thus the two determinants agree.
Worked Example: Pairing All Six Terms for a Three-by-Three Matrix
Take \(\mathbf{A}=\begin{bmatrix}1&2&0\\3&-1&4\\2&1&5\end{bmatrix}\). For this example, the identity and the three swaps have signs \(1,-1,-1,-1\), respectively, and the two three-cycles have sign \(1\). The following table gives each contribution in \(\det(\mathbf{A})\) and its matching contribution in \(\det(\mathbf{A}^{\top})\). The transpose term is indexed by the inverse permutation.
| Permutation \(\sigma\) | Inverse \(\sigma^{-1}\) | Contribution in \(\det(\mathbf{A})\) | Paired contribution in \(\det(\mathbf{A}^{\top})\) |
|---|---|---|---|
| Identity | Identity | \(1\cdot1\cdot(-1)\cdot5=-5\) | \(-5\) |
| \((12)\) | \((12)\) | \(-1\cdot2\cdot3\cdot5=-30\) | \(-30\) |
| \((13)\) | \((13)\) | \(-1\cdot0\cdot(-1)\cdot2=0\) | \(0\) |
| \((23)\) | \((23)\) | \(-1\cdot1\cdot4\cdot1=-4\) | \(-4\) |
| \((123)\) | \((132)\) | \(1\cdot2\cdot4\cdot2=16\) | \(16\) |
| \((132)\) | \((123)\) | \(1\cdot0\cdot3\cdot1=0\) | \(0\) |
For instance, the contribution for \(\sigma=(123)\), which sends \(1\) to \(2\), \(2\) to \(3\), and \(3\) to \(1\), is \(a_{12}a_{23}a_{31}=2\cdot4\cdot2=16\). Its matching transpose term uses \(\sigma^{-1}=(132)\). Adding all six contributions gives
Check: the total on each side is \(-5-30-4+16=-39+16=-23\), with the zero contributions leaving the sum unchanged. In particular, the cycle contributions match because the transpose pairs \((123)\) with its inverse \((132)\), rather than requiring those two cycle terms to agree individually.
Worked Example: A Sparse Matrix with One Nonzero Term
Consider \(\mathbf{B}=\begin{bmatrix}0&2&0\\0&0&3\\4&0&0\end{bmatrix}\). A nonzero product must select the entry in column 2 of row 1, column 3 of row 2, and column 1 of row 3. This is the permutation \((123)\), whose sign is \(1\). Every other permutation selects at least one zero entry. Therefore
The transpose is \(\mathbf{B}^{\top}=\begin{bmatrix}0&0&4\\2&0&0\\0&3&0\end{bmatrix}\). Its only nonzero product selects columns \(3,1,2\) in rows \(1,2,3\), the inverse cycle \((132)\), also with sign \(1\). Hence
Check: both nonzero products multiply the same three entries, \(2\), \(3\), and \(4\), so both determinants equal \(24\).
Common Pitfall: Matching the Same Permutation
A tempting but incorrect proof is to compare the term indexed by \(\sigma\) in \(\det(\mathbf{A})\) with the term indexed by that same \(\sigma\) in \(\det(\mathbf{A}^{\top})\), and claim they are always equal. The cycle terms in the three-by-three example show why this fails. For \(\sigma=(123)\), the term in \(\det(\mathbf{A})\) is \(1\cdot2\cdot4\cdot2=16\). The term indexed by \((123)\) in \(\det(\mathbf{A}^{\top})\), however, is zero: its corresponding product in entries of \(\mathbf{A}\) is \(a_{21}a_{32}a_{13}=3\cdot1\cdot0=0\). These two individual terms do not agree.
The correction is to pair \((123)\) with its inverse \((132)\). The transpose term indexed by \((132)\) is \(a_{31}a_{12}a_{23}=2\cdot2\cdot4=16\), which matches the original contribution. The full sums agree because this inverse pairing works for every permutation and preserves its sign, not because each term keeps the same index.
What the Identity Lets Us Conclude
The determinant can be computed from the entries arranged as rows or from the entries arranged as columns: the two arrangements are transposes, and their determinants are equal. The proof gives a precise reason for this symmetry, term by term after reindexing. It does not assert that each individual permutation term is unchanged under transposition, and it does not by itself provide rules for how other matrix operations affect determinants. Keeping these distinctions in view prevents a correct identity from being extended into an unsupported claim.
Check Your Understanding
Use the permutation formula and the inverse-pairing argument to answer these questions.
- For \(\mathbf{C}=\begin{bmatrix}3&1\\-2&5\end{bmatrix}\), compute \(\det(\mathbf{C})\) and \(\det(\mathbf{C}^{\top})\) separately.
- Explain why the sign of a permutation and its inverse is the same, using inversions.
- In the proof of the theorem, if \(\tau=\sigma^{-1}\), rewrite \(\prod_{i=1}^{n}a_{\tau(i),i}\) using \(\sigma\).
- For the matrix in the six-term example, why does the \((123)\) term in \(\det(\mathbf{A}^{\top})\) not match the \((123)\) term in \(\det(\mathbf{A})\)? Which transpose term does match it?
- State the determinant identity for a square matrix and identify the permutation pairing that proves it.
