فرمول لایبنیتس برای دترمینان

در جبر، فرمولِ لایبنیتس، دترمینانِ یک ماتریس مربعی ${\displaystyle A=(a_{ij})_{i,j=1,\dots ,n}}$ را نشان می‌دهد. فرمول به افتخارِ گوتفرید لایبنیتس ریاضی‌دانِ آلمانی نام‌گذاری شده‌است و عبارت است از:

${\displaystyle \det(A)=\sum _{\sigma \in S_{n}}\operatorname {sgn}(\sigma )\prod _{i=1}^{n}a_{i,\sigma (i)}}$

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:

${\displaystyle \det(A)=\epsilon ^{i_{1}\cdots i_{n}}{A}_{1i_{1}}\cdots {A}_{ni_{n}},}$

که در بین فیزیک‌دانان کاربرد بیشتری دارد.

قضیه تنها یک تابع چون

${\displaystyle F:{\mathfrak {M}}_{n}(\mathbb {K} )\longrightarrow \mathbb {K} }$

وجود دارد که پادمتقارن است، نسبت به ستون‌ها خطی است و مقدار آن به ازایِ ماتریس همانی برابر است با ۱: ${\displaystyle F(I)=1}$.

یکتایی: فرض می‌کنیم که ${\displaystyle F}$ چنین تابعی باشد و قرار می‌دهیم ${\displaystyle A=(a_{i}^{j})_{i=1,\dots ,n}^{j=1,\dots ,n}}$ که یک ماتریسِ ${\displaystyle n\times n}$ است. ${\displaystyle A^{j}}$ را به عنوانِ ${\displaystyle j}$ام از ماتریسِ ${\displaystyle A}$ می‌خوانیم به عنوان مثال ${\displaystyle A^{j}=(a_{i}^{j})_{i=1,\dots ,n}}$. بنابراین داریم: ${\displaystyle A=\left(A^{1},\dots ,A^{n}\right).}$

هم‌چنین قرارداد می‌کنیم که ${\displaystyle E^{k}}$ ستونِ ${\displaystyle k}$ام از ماتریسِ یکه را نشان دهد.

حال می‌توانیم هر یک از ستون‌هایِ ${\displaystyle A^{j}}$را بر حسب ${\displaystyle E^{k}}$ نمایش دهیم:

${\displaystyle A^{j}=\sum _{k=1}^{n}a_{k}^{j}E^{k}}$.

از آن‌جا که ${\displaystyle F}$ خطی است داریم:

${\displaystyle {\begin{aligned}F(A)&=F\left(\sum _{k_{1}=1}^{n}a_{k_{1}}^{1}E^{k_{1}},A^{2},\dots ,A^{n}\right)\\&=\sum _{k_{1}=1}^{n}a_{k_{1}}^{1}F\left(E^{k_{1}},A^{2},\dots ,A^{n}\right)\\&=\sum _{k_{1}=1}^{n}a_{k_{1}}^{1}\sum _{k_{2}=1}^{n}a_{k_{2}}^{2}F\left(E^{k_{1}},E^{k_{2}},A^{3},\dots ,A^{n}\right)\\&=\sum _{k_{1},k_{2}=1}^{n}\left(\prod _{i=1}^{2}a_{k_{i}}^{i}\right)F\left(E^{k_{1}},E^{k_{2}},A^{3},\dots ,A^{n}\right)\\&=\cdots \\&=\sum _{k_{1},\dots ,k_{n}=1}^{n}\left(\prod _{i=1}^{n}a_{k_{i}}^{i}\right)F\left(E^{k_{1}},\dots ,E^{k_{n}}\right).\end{aligned}}}$

از پادمتقارن بودن ماتریس نتیجه می‌گیریم که اگر ${\displaystyle k_{1}=k_{2}}$ آن‌گاه:

${\displaystyle {\begin{aligned}\\F\left(\dots ,E^{k_{1}},\dots ,E^{k_{2}},\dots \right)&=-F\left(\dots ,E^{k_{2}},\dots ,E^{k_{1}},\dots \right)\\F\left(\dots ,E^{k_{1}},\dots ,E^{k_{2}},\dots \right)&=-F\left(\dots ,E^{k_{1}},\dots ,E^{k_{2}},\dots \right)\\F\left(\dots ,E^{k_{1}},\dots ,E^{k_{2}},\dots \right)&=0\end{aligned}}}$

As the above sum takes into account all the possible choices of ordered ${\displaystyle n}$-tuples ${\displaystyle \left(k_{1},\dots ,k_{n}\right)}$، and because ${\displaystyle k_{i_{1}}=k_{i_{2}}}$ implies that F is zero, the sum can be reduced from all tuples to جایگشت as

${\displaystyle \sum _{\sigma \in {\mathfrak {S}}_{n}}\left(\prod _{i=1}^{n}a_{\sigma (i)}^{i}\right)F(E^{\sigma (1)},\dots ,E^{\sigma (n)}).}$

Because F is alternating, the columns ${\displaystyle E}$ can be swapped until it becomes the identity. The sign function ${\displaystyle \operatorname {sgn}(\sigma )}$ is defined to count the number of swaps necessary and account for the resulting sign change. One finally gets:

${\displaystyle {\begin{aligned}F(A)&=\sum _{\sigma \in {\mathfrak {S}}_{n}}\operatorname {sgn}(\sigma )\left(\prod _{i=1}^{n}a_{\sigma (i)}^{i}\right)F(I)\\&=\sum _{\sigma \in {\mathfrak {S}}_{n}}\operatorname {sgn}(\sigma )\prod _{i=1}^{n}a_{\sigma (i)}^{i}\end{aligned}}}$

as ${\displaystyle F(I)}$ is required to be equal to ${\displaystyle 1}$.

Therefore no function besides the function defined by the Leibniz Formula is a multilinear alternating function with ${\displaystyle F\left(I\right)=1}$.

Existence: We now show that F, where F is the function defined by the Leibniz formula, has these three properties.

Multilinear:

${\displaystyle {\begin{aligned}F(A^{0},\dots ,cA^{j},\dots )&=\sum _{\sigma \in {\mathfrak {S}}_{n}}\operatorname {sgn}(\sigma )ca_{\sigma (j)}^{j}\prod _{i=1,i\neq j}^{n}a_{\sigma (i)}^{i}\\&=c\sum _{\sigma \in {\mathfrak {S}}_{n}}\operatorname {sgn}(\sigma )a_{\sigma (j)}^{j}\prod _{i=1,i\neq j}^{n}a_{\sigma (i)}^{i}\\&=cF(A^{0},\dots ,A^{j},\dots )\\\\F(A^{0},\dots ,b+A^{j},\dots )&=\sum _{\sigma \in {\mathfrak {S}}_{n}}\operatorname {sgn}(\sigma )\left(b_{j}+a_{\sigma (j)}^{j}\right)\prod _{i=1,i\neq j}^{n}a_{\sigma (i)}^{i}\\&=\sum _{\sigma \in {\mathfrak {S}}_{n}}\operatorname {sgn}(\sigma )\left(\left(b_{j}\prod _{i=1,i\neq j}^{n}a_{\sigma (i)}^{i}\right)+\left(a_{\sigma (j)}^{j}\prod _{i=1,i\neq j}^{n}a_{\sigma (i)}^{i}\right)\right)\\&=\left(\sum _{\sigma \in {\mathfrak {S}}_{n}}\operatorname {sgn}(\sigma )b_{j}\prod _{i=1,i\neq j}^{n}a_{\sigma (i)}^{i}\right)+\left(\sum _{\sigma \in {\mathfrak {S}}_{n}}\operatorname {sgn}(\sigma )\prod _{i=1}^{n}a_{\sigma (i)}^{i}\right)\\&=F(A^{0},\dots ,b,\dots )+F(A^{0},\dots ,A^{j},\dots )\\\\\end{aligned}}}$

Alternating:

${\displaystyle {\begin{aligned}F(\dots ,A^{j_{1}},\dots ,A^{j_{2}},\dots )&=\sum _{\sigma \in {\mathfrak {S}}_{n}}\operatorname {sgn}(\sigma )\left(\prod _{i=1,i\neq j_{1},i\neq j_{2}}^{n}a_{\sigma (i)}^{i}\right)a_{\sigma (j_{1})}^{j_{1}}a_{\sigma (j_{2})}^{j_{2}}\\\end{aligned}}}$

Now let ${\displaystyle \omega }$ be the tuple equal to ${\displaystyle \sigma }$ with the ${\displaystyle j_{1}}$ and ${\displaystyle j_{2}}$ indices switched. It follows from the definition of ${\displaystyle \operatorname {sgn} }$ that ${\displaystyle \operatorname {sgn}(\sigma )=-sgn(\omega )}$.

${\displaystyle {\begin{aligned}\\F(\dots ,A^{j_{1}},\dots ,A^{j_{2}},\dots )&=\sum _{\omega \in {\mathfrak {S}}_{n}}-\operatorname {sgn}(\omega )\left(\prod _{i=1,i\neq j_{1},i\neq j_{2}}^{n}a_{\omega (i)}^{i}\right)a_{\omega (j_{1})}^{j_{1}}a_{\omega (j_{2})}^{j_{2}}\\&=-F(\dots ,A^{j_{2}},\dots ,A^{j_{1}},\dots )\\\\\end{aligned}}}$

Finally، ${\displaystyle F(I)=1}$:

${\displaystyle {\begin{aligned}\\F(I)&=\sum _{\sigma \in {\mathfrak {S}}_{n}}\operatorname {sgn}(\sigma )\prod _{i=1}^{n}I_{\sigma (i)}^{i}\\&=\sum _{\sigma =(1,2,\dots ,n)}\prod _{i=1}^{n}I_{i}^{i}\\&=1\end{aligned}}}$

Thus the only functions which are multilinear alternating with ${\displaystyle F(I)=1}$ 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

${\displaystyle \det$ :{\mathfrak {M}}_{n}(\mathbb {K} )\longrightarrow \mathbb {K} }

with these three properties.

• "Determinant", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
• Lloyd N. Trefethen and David Bau، Numerical Linear Algebra (SIAM، 1997).