# Leibniz formula for determinants

In algebra, the Leibniz formula, named in honor of Gottfried Leibniz, expresses the determinant of a square matrix in terms of permutations of the matrix elements. If $A$ is an $n\times n$ matrix, where $a_{ij}$ is the entry in the $i$-th row and $j$-th column of $A$, the formula is
    $\det(A)=\sum _{\tau \in S_{n}}\operatorname {sgn}(\tau )\prod _{i=1}^{n}a_{i\tau (i)}=\sum _{\sigma \in S_{n}}\operatorname {sgn}(\sigma )\prod _{i=1}^{n}a_{\sigma (i)i}$
where $\operatorname {sgn}$ is the sign function of permutations in the permutation group $S_{n}$, which returns $+1$ and $-1$ for even and odd permutations, respectively.
Another common notation used for the formula is in terms of the Levi-Civita symbol and makes use of the Einstein summation notation, where it becomes
    $\det(A)=\epsilon _{i_{1}\cdots i_{n}}{a}_{1i_{1}}\cdots {a}_{ni_{n}},$
which may be more familiar to physicists.
Directly evaluating the Leibniz formula from the definition requires $\Omega (n!\cdot n)$ operations in general—that is, a number of operations asymptotically proportional to $n$ factorial—because $n!$ is the number of order-$n$ permutations. This is impractically difficult for even relatively small $n$. Instead, the determinant can be evaluated in $O(n^{3})$ operations by forming the LU decomposition $A=LU$ (typically via Gaussian elimination or similar methods), in which case $\det A=\det L\cdot \det U$ and the determinants of the triangular matrices $L$ and $U$ are simply the products of their diagonal entries. (In practical applications of numerical linear algebra, however, explicit computation of the determinant is rarely required.) See, for example, Trefethen & Bau (1997). The determinant can also be evaluated in fewer than $O(n^{3})$ operations by reducing the problem to matrix multiplication, but most such algorithms are not practical.

## Formal statement and proof

Theorem. There exists exactly one function $F:M_{n}(\mathbb {K} )\rightarrow \mathbb {K}$ which is alternating multilinear w.r.t. columns and such that $F(I)=1$.
Proof.
Uniqueness: Let $F$ be such a function, and let $A=(a_{i}^{j})_{i=1,\dots ,n}^{j=1,\dots ,n}$ be an $n\times n$ matrix. Call $A^{j}$ the $j$-th column of $A$, i.e. $A^{j}=(a_{i}^{j})_{i=1,\dots ,n}$, so that $A=\left(A^{1},\dots ,A^{n}\right).$
Also, let $E^{k}$ denote the $k$-th column vector of the identity matrix.
Now one writes each of the $A^{j}$'s in terms of the $E^{k}$, i.e.
    $A^{j}=\sum _{k=1}^{n}a_{k}^{j}E^{k}$.
As $F$ is multilinear, one has
    ${\begin{aligned}F(A)&=F\left(\sum _{k_{1}=1}^{n}a_{k_{1}}^{1}E^{k_{1}},\dots ,\sum _{k_{n}=1}^{n}a_{k_{n}}^{n}E^{k_{n}}\right)=\sum _{k_{1},\dots ,k_{n}=1}^{n}\left(\prod _{i=1}^{n}a_{k_{i}}^{i}\right)F\left(E^{k_{1}},\dots ,E^{k_{n}}\right).\end{aligned}}$
From alternation it follows that any term with repeated indices is zero. The sum can therefore be restricted to tuples with non-repeating indices, i.e. permutations:
    $F(A)=\sum _{\sigma \in S_{n}}\left(\prod _{i=1}^{n}a_{\sigma (i)}^{i}\right)F(E^{\sigma (1)},\dots ,E^{\sigma (n)}).$
Because F is alternating, the columns $E$ can be swapped until it becomes the identity. The sign function $\operatorname {sgn}(\sigma )$ is defined to count the number of swaps necessary and account for the resulting sign change. One finally gets:
    ${\begin{aligned}F(A)&=\sum _{\sigma \in S_{n}}\operatorname {sgn}(\sigma )\left(\prod _{i=1}^{n}a_{\sigma (i)}^{i}\right)F(I)\\&=\sum _{\sigma \in S_{n}}\operatorname {sgn}(\sigma )\prod _{i=1}^{n}a_{\sigma (i)}^{i}\end{aligned}}$
as $F(I)$ is required to be equal to $1$.
Therefore no function besides the function defined by the Leibniz Formula can be a multilinear alternating function with $F\left(I\right)=1$.
Existence: We now show that F, where F is the function defined by the Leibniz formula, has these three properties.
Multilinear:
    ${\begin{aligned}F(A^{1},\dots ,cA^{j},\dots )&=\sum _{\sigma \in S_{n}}\operatorname {sgn}(\sigma )ca_{\sigma (j)}^{j}\prod _{i=1,i\neq j}^{n}a_{\sigma (i)}^{i}\\&=c\sum _{\sigma \in S_{n}}\operatorname {sgn}(\sigma )a_{\sigma (j)}^{j}\prod _{i=1,i\neq j}^{n}a_{\sigma (i)}^{i}\\&=cF(A^{1},\dots ,A^{j},\dots )\\\\F(A^{1},\dots ,b+A^{j},\dots )&=\sum _{\sigma \in S_{n}}\operatorname {sgn}(\sigma )\left(b_{\sigma (j)}+a_{\sigma (j)}^{j}\right)\prod _{i=1,i\neq j}^{n}a_{\sigma (i)}^{i}\\&=\sum _{\sigma \in S_{n}}\operatorname {sgn}(\sigma )\left(\left(b_{\sigma (j)}\prod _{i=1,i\neq j}^{n}a_{\sigma (i)}^{i}\right)+\left(a_{\sigma (j)}^{j}\prod _{i=1,i\neq j}^{n}a_{\sigma (i)}^{i}\right)\right)\\&=\left(\sum _{\sigma \in S_{n}}\operatorname {sgn}(\sigma )b_{\sigma (j)}\prod _{i=1,i\neq j}^{n}a_{\sigma (i)}^{i}\right)+\left(\sum _{\sigma \in S_{n}}\operatorname {sgn}(\sigma )\prod _{i=1}^{n}a_{\sigma (i)}^{i}\right)\\&=F(A^{1},\dots ,b,\dots )+F(A^{1},\dots ,A^{j},\dots )\\\\\end{aligned}}$
Alternating:
    ${\begin{aligned}F(\dots ,A^{j_{1}},\dots ,A^{j_{2}},\dots )&=\sum _{\sigma \in S_{n}}\operatorname {sgn}(\sigma )\left(\prod _{i=1,i\neq j_{1},i\neq j_{2}}^{n}a_{\sigma (i)}^{i}\right)a_{\sigma (j_{1})}^{j_{1}}a_{\sigma (j_{2})}^{j_{2}}\\\end{aligned}}$
For any $\sigma \in S_{n}$ let $\sigma '$ be the tuple equal to $\sigma$ with the $j_{1}$ and $j_{2}$ indices switched.
    ${\begin{aligned}F(A)&=\sum _{\sigma \in S_{n},\sigma (j_{1})<\sigma (j_{2})}\left[\operatorname {sgn}(\sigma )\left(\prod _{i=1,i\neq j_{1},i\neq j_{2}}^{n}a_{\sigma (i)}^{i}\right)a_{\sigma (j_{1})}^{j_{1}}a_{\sigma (j_{2})}^{j_{2}}+\operatorname {sgn}(\sigma ')\left(\prod _{i=1,i\neq j_{1},i\neq j_{2}}^{n}a_{\sigma '(i)}^{i}\right)a_{\sigma '(j_{1})}^{j_{1}}a_{\sigma '(j_{2})}^{j_{2}}\right]\\&=\sum _{\sigma \in S_{n},\sigma (j_{1})<\sigma (j_{2})}\left[\operatorname {sgn}(\sigma )\left(\prod _{i=1,i\neq j_{1},i\neq j_{2}}^{n}a_{\sigma (i)}^{i}\right)a_{\sigma (j_{1})}^{j_{1}}a_{\sigma (j_{2})}^{j_{2}}-\operatorname {sgn}(\sigma )\left(\prod _{i=1,i\neq j_{1},i\neq j_{2}}^{n}a_{\sigma (i)}^{i}\right)a_{\sigma (j_{2})}^{j_{1}}a_{\sigma (j_{1})}^{j_{2}}\right]\\&=\sum _{\sigma \in S_{n},\sigma (j_{1})<\sigma (j_{2})}\operatorname {sgn}(\sigma )\left(\prod _{i=1,i\neq j_{1},i\neq j_{2}}^{n}a_{\sigma (i)}^{i}\right)\underbrace {\left(a_{\sigma (j_{1})}^{j_{1}}a_{\sigma (j_{2})}^{j_{2}}-a_{\sigma (j_{1})}^{j_{2}}a_{\sigma (j_{2})}^{j_{_{1}}}\right)} _{=0{\text{, if }}A^{j_{1}}=A^{j_{2}}}\\\\\end{aligned}}$
Thus if $A^{j_{1}}=A^{j_{2}}$ then $F(\dots ,A^{j_{1}},\dots ,A^{j_{2}},\dots )=0$.
Finally, $F(I)=1$:
    ${\begin{aligned}F(I)&=\sum _{\sigma \in S_{n}}\operatorname {sgn}(\sigma )\prod _{i=1}^{n}I_{\sigma (i)}^{i}=\sum _{\sigma \in S_{n}}\operatorname {sgn}(\sigma )\prod _{i=1}^{n}\operatorname {\delta } _{i,\sigma (i)}\\&=\sum _{\sigma \in S_{n}}\operatorname {sgn}(\sigma )\operatorname {\delta } _{\sigma ,\operatorname {id} _{\{1\ldots n\}}}=\operatorname {sgn}(\operatorname {id} _{\{1\ldots n\}})=1\end{aligned}}$
Thus the only alternating multilinear functions with $F(I)=1$ are restricted to the function defined by the Leibniz formula, and it in fact also has these three properties. Hence the determinant can be defined as the only function $\det :M_{n}(\mathbb {K} )\rightarrow \mathbb {K}$ with these three properties.