در جبر ، فرمولِ لایبنیتس ، دترمینانِ یک ماتریس مربعی
A
=
(
a
i
j
)
i
,
j
=
1
,
…
,
n
{\displaystyle A=(a_{ij})_{i,j=1,\dots ,n}}
را نشان میدهد. فرمول به افتخارِ گوتفرید لایبنیتس ریاضیدانِ آلمانی نامگذاری شدهاست و عبارت است از:
det
(
A
)
=
∑
σ
∈
S
n
sgn
(
σ
)
∏
i
=
1
n
a
i
,
σ
(
i
)
{\displaystyle \det(A)=\sum _{\sigma \in S_{n}}\operatorname {sgn}(\sigma )\prod _{i=1}^{n}a_{i,\sigma (i)}}
برایِ یک ماتریسِ n ×n که sgn تابع علامتِ مخصوصِ جایگشتها در گروه جایگشتهایِ S n است که مقدارِ +۱ یا -۱ را به ترتیب برایِ جایگشتهای زوج یا جایگشتهایِ فرد برمیگرداند.
یک شیوهی رایج دیگر از نشانگذاری آن، استفاده از نماد لوی-چیویتا است که با قرارداد جمعزنی اینشتین به رابطهی زیر تبدیل میشود:
det
(
A
)
=
ϵ
i
1
⋯
i
n
A
1
i
1
⋯
A
n
i
n
,
{\displaystyle \det(A)=\epsilon ^{i_{1}\cdots i_{n}}{A}_{1i_{1}}\cdots {A}_{ni_{n}},}
که در بین فیزیکدانان کاربرد بیشتری دارد.
قضیه
تنها یک تابع چون
F
:
M
n
(
K
)
⟶
K
{\displaystyle F:{\mathfrak {M}}_{n}(\mathbb {K} )\longrightarrow \mathbb {K} }
وجود دارد که پادمتقارن است، نسبت به ستونها خطی است و مقدار آن به ازایِ ماتریس همانی برابر است با ۱:
F
(
I
)
=
1
{\displaystyle F(I)=1}
.
یکتایی: فرض میکنیم که
F
{\displaystyle F}
چنین تابعی باشد و قرار میدهیم
A
=
(
a
i
j
)
i
=
1
,
…
,
n
j
=
1
,
…
,
n
{\displaystyle A=(a_{i}^{j})_{i=1,\dots ,n}^{j=1,\dots ,n}}
که یک ماتریسِ
n
×
n
{\displaystyle n\times n}
است.
A
j
{\displaystyle A^{j}}
را به عنوانِ
j
{\displaystyle j}
ام از ماتریسِ
A
{\displaystyle A}
میخوانیم به عنوان مثال
A
j
=
(
a
i
j
)
i
=
1
,
…
,
n
{\displaystyle A^{j}=(a_{i}^{j})_{i=1,\dots ,n}}
. بنابراین داریم:
A
=
(
A
1
,
…
,
A
n
)
.
{\displaystyle A=\left(A^{1},\dots ,A^{n}\right).}
همچنین قرارداد میکنیم که
E
k
{\displaystyle E^{k}}
ستونِ
k
{\displaystyle k}
ام از ماتریسِ یکه را نشان دهد.
حال میتوانیم هر یک از ستونهایِ
A
j
{\displaystyle A^{j}}
را بر حسب
E
k
{\displaystyle E^{k}}
نمایش دهیم:
A
j
=
∑
k
=
1
n
a
k
j
E
k
{\displaystyle A^{j}=\sum _{k=1}^{n}a_{k}^{j}E^{k}}
.
از آنجا که
F
{\displaystyle F}
خطی است داریم:
F
(
A
)
=
F
(
∑
k
1
=
1
n
a
k
1
1
E
k
1
,
A
2
,
…
,
A
n
)
=
∑
k
1
=
1
n
a
k
1
1
F
(
E
k
1
,
A
2
,
…
,
A
n
)
=
∑
k
1
=
1
n
a
k
1
1
∑
k
2
=
1
n
a
k
2
2
F
(
E
k
1
,
E
k
2
,
A
3
,
…
,
A
n
)
=
∑
k
1
,
k
2
=
1
n
(
∏
i
=
1
2
a
k
i
i
)
F
(
E
k
1
,
E
k
2
,
A
3
,
…
,
A
n
)
=
⋯
=
∑
k
1
,
…
,
k
n
=
1
n
(
∏
i
=
1
n
a
k
i
i
)
F
(
E
k
1
,
…
,
E
k
n
)
.
{\displaystyle {\begin{aligned}F(A)&=F\left(\sum _{k_{1}=1}^{n}a_{k_{1}}^{1}E^{k_{1}},A^{2},\dots ,A^{n}\right)\\&=\sum _{k_{1}=1}^{n}a_{k_{1}}^{1}F\left(E^{k_{1}},A^{2},\dots ,A^{n}\right)\\&=\sum _{k_{1}=1}^{n}a_{k_{1}}^{1}\sum _{k_{2}=1}^{n}a_{k_{2}}^{2}F\left(E^{k_{1}},E^{k_{2}},A^{3},\dots ,A^{n}\right)\\&=\sum _{k_{1},k_{2}=1}^{n}\left(\prod _{i=1}^{2}a_{k_{i}}^{i}\right)F\left(E^{k_{1}},E^{k_{2}},A^{3},\dots ,A^{n}\right)\\&=\cdots \\&=\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}}}
از پادمتقارن بودن ماتریس نتیجه میگیریم که اگر
k
1
=
k
2
{\displaystyle k_{1}=k_{2}}
آنگاه:
F
(
…
,
E
k
1
,
…
,
E
k
2
,
…
)
=
−
F
(
…
,
E
k
2
,
…
,
E
k
1
,
…
)
F
(
…
,
E
k
1
,
…
,
E
k
2
,
…
)
=
−
F
(
…
,
E
k
1
,
…
,
E
k
2
,
…
)
F
(
…
,
E
k
1
,
…
,
E
k
2
,
…
)
=
0
{\displaystyle {\begin{aligned}\\F\left(\dots ,E^{k_{1}},\dots ,E^{k_{2}},\dots \right)&=-F\left(\dots ,E^{k_{2}},\dots ,E^{k_{1}},\dots \right)\\F\left(\dots ,E^{k_{1}},\dots ,E^{k_{2}},\dots \right)&=-F\left(\dots ,E^{k_{1}},\dots ,E^{k_{2}},\dots \right)\\F\left(\dots ,E^{k_{1}},\dots ,E^{k_{2}},\dots \right)&=0\end{aligned}}}
As the above sum takes into account all the possible choices of ordered
n
{\displaystyle n}
-tuples
(
k
1
,
…
,
k
n
)
{\displaystyle \left(k_{1},\dots ,k_{n}\right)}
، and because
k
i
1
=
k
i
2
{\displaystyle k_{i_{1}}=k_{i_{2}}}
implies that F is zero, the sum can be reduced from all tuples to جایگشت as
∑
σ
∈
S
n
(
∏
i
=
1
n
a
σ
(
i
)
i
)
F
(
E
σ
(
1
)
,
…
,
E
σ
(
n
)
)
.
{\displaystyle \sum _{\sigma \in {\mathfrak {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
{\displaystyle E}
can be swapped until it becomes the identity. The sign function
sgn
(
σ
)
{\displaystyle \operatorname {sgn}(\sigma )}
is defined to count the number of swaps necessary and account for the resulting sign change. One finally gets:
F
(
A
)
=
∑
σ
∈
S
n
sgn
(
σ
)
(
∏
i
=
1
n
a
σ
(
i
)
i
)
F
(
I
)
=
∑
σ
∈
S
n
sgn
(
σ
)
∏
i
=
1
n
a
σ
(
i
)
i
{\displaystyle {\begin{aligned}F(A)&=\sum _{\sigma \in {\mathfrak {S}}_{n}}\operatorname {sgn}(\sigma )\left(\prod _{i=1}^{n}a_{\sigma (i)}^{i}\right)F(I)\\&=\sum _{\sigma \in {\mathfrak {S}}_{n}}\operatorname {sgn}(\sigma )\prod _{i=1}^{n}a_{\sigma (i)}^{i}\end{aligned}}}
as
F
(
I
)
{\displaystyle F(I)}
is required to be equal to
1
{\displaystyle 1}
.
Therefore no function besides the function defined by the Leibniz Formula is a multilinear alternating function with
F
(
I
)
=
1
{\displaystyle 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 :
F
(
A
0
,
…
,
c
A
j
,
…
)
=
∑
σ
∈
S
n
sgn
(
σ
)
c
a
σ
(
j
)
j
∏
i
=
1
,
i
≠
j
n
a
σ
(
i
)
i
=
c
∑
σ
∈
S
n
sgn
(
σ
)
a
σ
(
j
)
j
∏
i
=
1
,
i
≠
j
n
a
σ
(
i
)
i
=
c
F
(
A
0
,
…
,
A
j
,
…
)
F
(
A
0
,
…
,
b
+
A
j
,
…
)
=
∑
σ
∈
S
n
sgn
(
σ
)
(
b
j
+
a
σ
(
j
)
j
)
∏
i
=
1
,
i
≠
j
n
a
σ
(
i
)
i
=
∑
σ
∈
S
n
sgn
(
σ
)
(
(
b
j
∏
i
=
1
,
i
≠
j
n
a
σ
(
i
)
i
)
+
(
a
σ
(
j
)
j
∏
i
=
1
,
i
≠
j
n
a
σ
(
i
)
i
)
)
=
(
∑
σ
∈
S
n
sgn
(
σ
)
b
j
∏
i
=
1
,
i
≠
j
n
a
σ
(
i
)
i
)
+
(
∑
σ
∈
S
n
sgn
(
σ
)
∏
i
=
1
n
a
σ
(
i
)
i
)
=
F
(
A
0
,
…
,
b
,
…
)
+
F
(
A
0
,
…
,
A
j
,
…
)
{\displaystyle {\begin{aligned}F(A^{0},\dots ,cA^{j},\dots )&=\sum _{\sigma \in {\mathfrak {S}}_{n}}\operatorname {sgn}(\sigma )ca_{\sigma (j)}^{j}\prod _{i=1,i\neq j}^{n}a_{\sigma (i)}^{i}\\&=c\sum _{\sigma \in {\mathfrak {S}}_{n}}\operatorname {sgn}(\sigma )a_{\sigma (j)}^{j}\prod _{i=1,i\neq j}^{n}a_{\sigma (i)}^{i}\\&=cF(A^{0},\dots ,A^{j},\dots )\\\\F(A^{0},\dots ,b+A^{j},\dots )&=\sum _{\sigma \in {\mathfrak {S}}_{n}}\operatorname {sgn}(\sigma )\left(b_{j}+a_{\sigma (j)}^{j}\right)\prod _{i=1,i\neq j}^{n}a_{\sigma (i)}^{i}\\&=\sum _{\sigma \in {\mathfrak {S}}_{n}}\operatorname {sgn}(\sigma )\left(\left(b_{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 {\mathfrak {S}}_{n}}\operatorname {sgn}(\sigma )b_{j}\prod _{i=1,i\neq j}^{n}a_{\sigma (i)}^{i}\right)+\left(\sum _{\sigma \in {\mathfrak {S}}_{n}}\operatorname {sgn}(\sigma )\prod _{i=1}^{n}a_{\sigma (i)}^{i}\right)\\&=F(A^{0},\dots ,b,\dots )+F(A^{0},\dots ,A^{j},\dots )\\\\\end{aligned}}}
Alternating :
F
(
…
,
A
j
1
,
…
,
A
j
2
,
…
)
=
∑
σ
∈
S
n
sgn
(
σ
)
(
∏
i
=
1
,
i
≠
j
1
,
i
≠
j
2
n
a
σ
(
i
)
i
)
a
σ
(
j
1
)
j
1
a
σ
(
j
2
)
j
2
{\displaystyle {\begin{aligned}F(\dots ,A^{j_{1}},\dots ,A^{j_{2}},\dots )&=\sum _{\sigma \in {\mathfrak {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}}}
Now let
ω
{\displaystyle \omega }
be the tuple equal to
σ
{\displaystyle \sigma }
with the
j
1
{\displaystyle j_{1}}
and
j
2
{\displaystyle j_{2}}
indices switched. It follows from the definition of
sgn
{\displaystyle \operatorname {sgn} }
that
sgn
(
σ
)
=
−
s
g
n
(
ω
)
{\displaystyle \operatorname {sgn}(\sigma )=-sgn(\omega )}
.
F
(
…
,
A
j
1
,
…
,
A
j
2
,
…
)
=
∑
ω
∈
S
n
−
sgn
(
ω
)
(
∏
i
=
1
,
i
≠
j
1
,
i
≠
j
2
n
a
ω
(
i
)
i
)
a
ω
(
j
1
)
j
1
a
ω
(
j
2
)
j
2
=
−
F
(
…
,
A
j
2
,
…
,
A
j
1
,
…
)
{\displaystyle {\begin{aligned}\\F(\dots ,A^{j_{1}},\dots ,A^{j_{2}},\dots )&=\sum _{\omega \in {\mathfrak {S}}_{n}}-\operatorname {sgn}(\omega )\left(\prod _{i=1,i\neq j_{1},i\neq j_{2}}^{n}a_{\omega (i)}^{i}\right)a_{\omega (j_{1})}^{j_{1}}a_{\omega (j_{2})}^{j_{2}}\\&=-F(\dots ,A^{j_{2}},\dots ,A^{j_{1}},\dots )\\\\\end{aligned}}}
Finally،
F
(
I
)
=
1
{\displaystyle F(I)=1}
:
F
(
I
)
=
∑
σ
∈
S
n
sgn
(
σ
)
∏
i
=
1
n
I
σ
(
i
)
i
=
∑
σ
=
(
1
,
2
,
…
,
n
)
∏
i
=
1
n
I
i
i
=
1
{\displaystyle {\begin{aligned}\\F(I)&=\sum _{\sigma \in {\mathfrak {S}}_{n}}\operatorname {sgn}(\sigma )\prod _{i=1}^{n}I_{\sigma (i)}^{i}\\&=\sum _{\sigma =(1,2,\dots ,n)}\prod _{i=1}^{n}I_{i}^{i}\\&=1\end{aligned}}}
Thus the only functions which are multilinear alternating with
F
(
I
)
=
1
{\displaystyle 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
(
K
)
⟶
K
{\displaystyle \det :{\mathfrak {M}}_{n}(\mathbb {K} )\longrightarrow \mathbb {K} }
with these three properties.