Expansion Theorem for Determinants

== Theorem ==
Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$.
Let $D = \map \det {\mathbf A}$ be the determinant of $\mathbf A$:
:$\ds \map \det {\mathbf A} := \sum_{\lambda} \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{k \map \lambda k} } = \sum_\lambda \map \sgn \lambda a_{1 \map \lambda 1} a_{2 \map \lambda 2} \cdots a_{n \map \lambda n}$
where:
:the summation $\ds \sum_\lambda$ goes over all the $n!$ permutations of $\set {1, 2, \ldots, n}$
:$\map \sgn \lambda$ is the sign of the permutation $\lambda$.
Let $a_{p q}$ be an element of $\mathbf A$.
Let $A_{p q}$ be the cofactor of $a_{p q}$ in $D$.
Then:
:$(1): \quad \ds \forall r \in \closedint 1 n: D = \sum_{k \mathop = 1}^n a_{r k} A_{r k}$
:$(2): \quad \ds \forall c \in \closedint 1 n: D = \sum_{k \mathop = 1}^n a_{k c} A_{k c}$
Thus the value of a determinant can be found either by:
:multiplying all the elements in a row by their cofactors and adding up the products
or:
:multiplying all the elements in a column by their cofactors and adding up the products.
The identity:
:$\ds D = \sum_{k \mathop = 1}^n a_{r k} A_{r k}$
is known as the expansion of $D$ in terms of row $r$, while:
:$\ds D = \sum_{k \mathop = 1}^n a_{k c} A_{k c}$
is known as the expansion of $D$ in terms of column $c$.
=== Corollary ===
== Proof ==
Because of Determinant of Transpose, it is necessary to prove only one of these identities.
Let:
:$D = \begin {vmatrix}
a_{1 1} & \cdots & a_{1 k} & \cdots & a_{1 n} \\
 \vdots & \ddots & \vdots & \ddots & \vdots \\
a_{r 1} & \cdots & a_{r k} & \cdots & a_{r n} \\
 \vdots & \ddots & \vdots & \ddots & \vdots \\
a_{n 1} & \cdots & a_{n k} & \cdots & a_{n n}
\end {vmatrix}$
First, note that from Determinant with Row Multiplied by Constant, we have:
:$\begin{vmatrix}
a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
 \vdots & \vdots & \ddots & \vdots \\
a_{r 1} & 0 & \cdots & 0 \\
 \vdots & \vdots & \ddots & \vdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n n}
\end {vmatrix} = a_{r 1} \begin {vmatrix}
a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
 \vdots & \vdots & \ddots & \vdots \\
 1 & 0 & \cdots & 0 \\
 \vdots & \vdots & \ddots & \vdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n n}
\end {vmatrix}$
and similarly:
:$\begin {vmatrix}
a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
 \vdots & \vdots & \ddots & \vdots \\
 0 & a_{r 2} & \cdots & 0 \\
 \vdots & \vdots & \ddots & \vdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n n}
\end {vmatrix} = a_{r 2} \begin {vmatrix}
a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
 \vdots & \vdots & \ddots & \vdots \\
 0 & 1 & \cdots & 0 \\
 \vdots & \vdots & \ddots & \vdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n n}
\end{vmatrix}$
and so on for the whole of row $r$.
From Determinant as Sum of Determinants:
:$\begin {vmatrix}
a_{1 1} & \cdots & a_{1 k} & \cdots & a_{1 n} \\
 \vdots & \ddots & \vdots & \ddots & \vdots \\
a_{r 1} & \cdots & a_{r k} & \cdots & a_{r n} \\
 \vdots & \ddots & \vdots & \ddots & \vdots \\
a_{n 1} & \cdots & a_{n k} & \cdots & a_{n n}
\end {vmatrix} = \ds \sum_{k \mathop = 1}^n \paren {a_{r k} \begin {vmatrix}
a_{1 1} & \cdots & a_{1 k} & \cdots & a_{1 n} \\
 \vdots & \ddots & \vdots & \ddots & \vdots \\
 0 & \cdots & 1 & \cdots & 0 \\
 \vdots & \ddots & \vdots & \ddots & \vdots \\
a_{n 1} & \cdots & a_{n k} & \cdots & a_{n n}
\end {vmatrix} }$
Consider the determinant:
:$\begin{vmatrix}
 a_{1 1} & \cdots & a_{1 \paren {k - 1} } & a_{1 k} & a_{1 \paren {k + 1} } & \cdots & a_{1 n} \\
 \vdots & \ddots & \vdots & \ddots & \vdots & \ddots & \vdots \\
a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} \paren {k - 1} } & a_{\paren {r - 1} k} & a_{\paren {r - 1} \paren {k + 1} } & \cdots & a_{\paren {r - 1} n} \\
 0 & \cdots & 0 & 1 & 0 & \cdots & 0 \\
a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} \paren {k - 1} } & a_{\paren {r + 1} k} & a_{\paren {r + 1} \paren {k + 1} } & \cdots & a_{\paren {r + 1} n} \\
 \vdots & \ddots & \vdots & \ddots & \vdots & \ddots & \vdots \\
 a_{n 1} & \cdots & a_{n \paren {k - 1} } & a_{n k} & a_{n \paren {k + 1} } & \cdots & a_{n n}
\end {vmatrix}$
Exchange rows $r$ and $r - 1$, then (the new) row $r - 1$ with row $r - 2$, until finally row $r$ is at the top.
Row $1$ will be in row $2$, row $2$ in row $3$, and so on.
This is permuting the rows by a $k$-cycle of length $r$.
Call that $k$-cycle $\rho$.
Then from Parity of K-Cycle:
:$\map \sgn \rho = \paren {-1}^{r - 1}$
Thus:
:$\begin {vmatrix}
 0 & \cdots & 1 & \cdots & 0 \\
 a_{1 1} & \cdots & a_{1 k} & \cdots & a_{1 n} \\
 \vdots & \ddots & \vdots & \ddots & \vdots \\
a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} k} & \cdots & a_{\paren {r - 1} n} \\
a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} k} & \cdots & a_{\paren {r + 1} n} \\
 \vdots & \ddots & \vdots & \ddots & \vdots \\
 a_{n 1} & \cdots & a_{n k} & \cdots & a_{n n}
\end {vmatrix} = \paren {-1}^{r - 1} \begin {vmatrix}
 a_{1 1} & \cdots & a_{1 k} & \cdots & a_{1 n} \\
 \vdots & \ddots & \vdots & \ddots & \vdots \\
a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} k} & \cdots & a_{\paren {r - 1} n} \\
 0 & \cdots & 1 & \cdots & 0 \\
a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} k} & \cdots & a_{\paren {r + 1} n} \\
 \vdots & \ddots & \vdots & \ddots & \vdots \\
 a_{n 1} & \cdots & a_{n k} & \cdots & a_{n n}
\end {vmatrix}$
The same argument can be applied to columns.
Thus:
:$\begin {vmatrix}
 1 & 0 & \cdots & 0 & 0 & \cdots & 0 \\
 a_{1 k} & a_{1 1} & \cdots & a_{1 \paren {k - 1} } & a_{1 \paren {k + 1} } & \cdots & a_{1 n} \\
 \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\
a_{\paren {r - 1} k} & a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} \paren {k - 1} } & a_{\paren {r - 1} \paren {k + 1} } & \cdots & a_{\paren {r - 1} n} \\
a_{\paren {r + 1} k} & a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} \paren {k - 1} } & a_{\paren {r + 1} \paren {k + 1} } & \cdots & a_{\paren {r + 1} n} \\
 \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\
 a_{n k} & a_{n 1} & \cdots & a_{n \paren {k - 1} } & a_{n \paren {k + 1} } & \cdots & a_{n n}
\end {vmatrix} = \paren {-1}^{k-1}\begin {vmatrix}
 0 & \cdots & 1 & \cdots & 0 \\
 a_{1 1} & \cdots & a_{1 k} & \cdots & a_{1 n} \\
 \vdots & \ddots & \vdots & \ddots & \vdots \\
a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} k} & \cdots & a_{\paren {r - 1} n} \\
a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} k} & \cdots & a_{\paren {r + 1} n} \\
 \vdots & \ddots & \vdots & \ddots & \vdots \\
 a_{n 1} & \cdots & a_{n k} & \cdots & a_{n n}
\end {vmatrix}$
and so:
:$\begin{vmatrix}
 1 & 0 & \cdots & 0 & 0 & \cdots & 0 \\
 a_{1 k} & a_{1 1} & \cdots & a_{1 \paren {k - 1}} & a_{1 \paren {k + 1} } & \cdots & a_{1 n} \\
 \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\
a_{\paren {r - 1} k} & a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} \paren {k - 1} } & a_{\paren {r - 1} \paren {k + 1} } & \cdots & a_{\paren {r - 1} n} \\
a_{\paren {r + 1} k} & a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} \paren {k - 1} } & a_{\paren {r + 1} \paren {k + 1} } & \cdots & a_{\paren {r + 1} n} \\
 \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\
 a_{n k} & a_{n 1} & \cdots & a_{n \paren {k - 1} } & a_{n \paren {k + 1} } & \cdots & a_{n n}
\end {vmatrix} = \paren {-1}^{r + k} \begin {vmatrix}
 a_{1 1} & \cdots & a_{1 k} & \cdots & a_{1 n} \\
 \vdots & \ddots & \vdots & \ddots & \vdots \\
a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} k} & \cdots & a_{\paren {r - 1} n} \\
 0 & \cdots & 1 & \cdots & 0 \\
a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} k} & \cdots & a_{\paren {r + 1} n} \\
 \vdots & \ddots & \vdots & \ddots & \vdots \\
 a_{n 1} & \cdots & a_{n k} & \cdots & a_{n n}
\end {vmatrix}$
Then:
:$\paren {-1}^{r + k} \begin {vmatrix}
 a_{1 1} & \cdots & a_{1 \paren {k - 1} } & a_{1 \paren {k + 1} } & \cdots & a_{1 n} \\
 \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\
a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} \paren {k - 1} } & a_{\paren {r - 1} \paren {k + 1} } & \cdots & a_{\paren {r - 1} n} \\
a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} \paren {k - 1} } & a_{\paren {r + 1} \paren {k + 1} } & \cdots & a_{\paren {r + 1} n} \\
 \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\
 a_{n 1} & \cdots & a_{n \paren {k - 1} } & a_{n \paren {k + 1} } & \cdots & a_{n n}
\end{vmatrix}$
is $A_{r k}$, the cofactor of $a_{r k}$ in $D$.
But from Determinant with Unit Element in Otherwise Zero Row, we have:
:$\begin {vmatrix}
 1 & 0 & \cdots & 0 \\
b_{2 1} & b_{2 2} & \cdots & b_{2 n} \\
 \vdots & \vdots & \ddots & \vdots \\
b_{n 1} & b_{n 2} & \cdots & b_{n n}
\end {vmatrix} = \begin {vmatrix}
b_{2 2} & \cdots & b_{2 n} \\
 \vdots & \ddots & \vdots \\
b_{n 2} & \cdots & b_{n n}
\end {vmatrix}$
Assembling all the pieces derived above, the result follows.
== Examples ==
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercises -- Second Set
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace: Proposition $1.12$
* 
* 
* 
Category:Expansion Theorem for Determinants
Category:Determinants
Category:Named Theorems