Laplace's Expansion Theorem

Theorem

Let $D$ be the determinant of order $n$.

Let $r_1, r_2, \ldots, r_k$ be integers such that:

• $1 \le k < n$;
• $1 \le r_1 < r_2 < \cdots < r_k \le n$.

Let $D \left({r_1, r_2, \ldots, r_k | u_1, u_2, \ldots, u_k}\right)$ be an order-$k$ minor of $D$.

Let $\tilde D \left({r_1, r_2, \ldots, r_k | u_1, u_2, \ldots, u_k}\right)$ be the cofactor of $D \left({r_1, r_2, \ldots, r_k | u_1, u_2, \ldots, u_k}\right)$.

Then $\displaystyle D = \sum_{1 \le u_1 < \cdots < u_k \le n} D \left({r_1, r_2, \ldots, r_k | u_1, u_2, \ldots, u_k}\right) \tilde D \left({r_1, r_2, \ldots, r_k | u_1, u_2, \ldots, u_k}\right)$.

A similar result applies for columns.

Proof

Let us define $r_{k+1}, r_{k+2}, \ldots, r_n$ such that

• $1 \le r_{k+1} < r_{k+2} < \cdots < r_n \le n$
• $\rho = \left({r_1, r_2, \ldots, r_n}\right)$ is a permutation on $\N^*_n$.

Let $\sigma = \left({s_1, s_2, \ldots, s_n}\right)$ be a permutation on $\N^*_n$.

Then by Permutation of Determinant Indices we have:

	\(\displaystyle \)	\(\displaystyle D\)	\(=\)	\(\displaystyle \sum_\sigma \operatorname{sgn} \left({\rho}\right) \operatorname{sgn} \left({\sigma}\right) \prod_{j=1}^n a_{\rho \left({j}\right) \sigma \left({j}\right)}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_\sigma \left({-1}\right)^{\sum_{i=1}^k \left({r_i + s_i}\right)} \operatorname{sgn} \left({\rho \left({r_1,\ldots, r_k}\right)}\right) \operatorname{sgn} \left({\sigma \left({s_1,\ldots, s_k}\right)}\right) \operatorname{sgn} \left({\rho \left({r_{k+1},\ldots, r_n}\right)}\right) \operatorname{sgn} \left({\sigma \left({s_{k+1},\ldots, s_n}\right)}\right) \prod_{j=1}^n a_{\rho \left({j}\right) \sigma \left({j}\right)}\)	\(\displaystyle \)

We can get all the permutations $\sigma$ exactly once by separating the numbers $1, \ldots, n$ in all possible ways into a set of $k$ and $n-k$ numbers.

We let $\left({s_1,\ldots, s_k}\right)$ vary over the first set and $\left({s_{k+1},\ldots, s_n}\right)$ over the second set.

So the summation over all $\sigma$ can be replaced by:

• $\left({u_1, \ldots, u_n}\right) = \sigma \left({1, \ldots, n}\right)$
• $u_1 < u_2 < \cdots < u_k, u_{k+1} < u_{k+2} < \cdots < u_n$
• $\left({s_1,\ldots, s_k}\right) = \sigma \left({u_1, \ldots, u_k}\right)$
• $\left({s_{k+1},\ldots, s_n}\right) = \sigma \left({u_{k+1}, \ldots, u_n}\right)$.

Thus we get:

	\(\displaystyle \)	\(\displaystyle D\)	\(=\)	\(\displaystyle \sum_{\sigma \left({u_1, \ldots, u_n}\right)} \left({-1}\right)^{\sum_{i=1}^k \left({r_i + u_i}\right)} \sum_{\sigma \left({u_1, \ldots, u_k}\right)} \operatorname{sgn} \left({\rho \left({r_1,\ldots, r_k}\right)}\right) \operatorname{sgn} \left({\sigma \left({s_1,\ldots, s_k}\right)}\right)\prod_{j=1}^k a_{\rho \left({j}\right) \sigma \left({j}\right)}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\times\)	\(\displaystyle \sum_{\sigma \left({u_{k+1}, \ldots, u_n}\right)} \operatorname{sgn} \left({\rho \left({r_{k+1},\ldots, r_n}\right)}\right) \operatorname{sgn} \left({\sigma \left({s_{k+1},\ldots, s_n}\right)}\right)\prod_{j=k+1}^n a_{\rho \left({j}\right) \sigma \left({j}\right)}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{\sigma \left({u_1, \ldots, u_n}\right)} \left({-1}\right)^{\sum_{i=1}^k \left({r_i + u_i}\right)}\begin{vmatrix}a_{r_1 u_1} & \cdots & a_{r_1 u_k} \\ \vdots & \ddots & \vdots \\ a_{r_k u_1} & \cdots & a_{r_k u_k}\end{vmatrix}\times \begin{vmatrix}a_{r_{k+1} u_{k+1} } & \cdots & a_{r_{k+1} u_n} \\ \vdots & \ddots & \vdots \\ a_{r_n u_{k+1} } & \cdots & a_{r_n u_n}\end{vmatrix}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{\sigma \left({u_1, \ldots, u_n}\right)} \left({-1}\right)^{\sum_{i=1}^k \left({r_i + u_i}\right)} D \left({r_1, \ldots, r_k \vert u_1, \ldots, u_k}\right) \times D \left({r_{k+1}, \ldots, r_n \vert u_{k+1}, \ldots, u_n}\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{\sigma \left({u_1, \ldots, u_n}\right)} D \left({r_1, \ldots, r_k \vert u_1, \ldots, u_k}\right) \times \tilde D \left({r_1, \ldots, r_k \vert u_1, \ldots, u_k}\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{1 \le u_1 < \cdots < u_k \le n} D \left({r_1, \ldots, r_k \vert u_1, \ldots, u_k}\right) \tilde D \left({r_1, \ldots, r_k \vert u_1, \ldots, u_k}\right) \sum_{u_{k+1}, \ldots, u_n} 1\)	\(\displaystyle \)

That last inner sum extends over all integers which satisfy:

• $\left({u_1, \ldots, u_n}\right) = \sigma \left({1, \ldots, n}\right)$
• $u_1 < u_2 < \cdots < u_k, u_{k+1} < u_{k+2} < \cdots < u_n$.

But for each set of $u_1, \ldots, u_k$, then the integers $u_{k+1}, \ldots, u_n$ are clearly uniquely determined.

So that last inner sum equals 1 and the theorem is proved.

The result for columns follows from Determinant of Transpose.

Comment

This gives us an expansion of the determinant $D$ in terms of $k$ specified rows.

We form all possible order-$k$ minors of $D$ which involve all of these rows, and multiply each of them by their cofactors.

The sum of these products is equal to $D$.

We note that when $k=1$ this becomes the Expansion Theorem for Determinants.

Source of Name

This entry was named for Pierre-Simon de Laplace.