proof of cofactor expansion

Let $M\in mat_{N}(K)$ be a $n\times n$-matrix with entries from a commutative field $K$. Let $e_{1},\ldots,e_{n}$ denote the vectors of the canonical basis of $K^{n}$. For the proof we need the following

Lemma: Let $M_{ij}^{*}$ be the matrix generated by replacing the $i$-th row of $M$ by $e_{j}$. Then

 $\det{M_{ij}^{*}}=(-1)^{i+j}\det{M_{ij}}$

where $M_{ij}$ is the $(n-1)\times(n-1)$-matrix obtained from $M$ by removing its $i$-th row and $j$-th column.

Proof.

By adding appropriate of the $i$-th row of $M_{ij}^{*}$ to its remaining rows we obtain a matrix with 1 at position $(i,j)$ and 0 at positions $(k,j)$ ($k\neq i$). Now we apply the permutation

 $(12)\circ(23)\circ\dots\circ((i-1)i)$

to rows and

 $(12)\circ(23)\circ\dots\circ((j-1)j)$

to columns of the matrix. The matrix now looks like this:

• Row/column 1 is the vector $e_{1}$;

• under row 1 and right of column 1 is the matrix $M_{ij}$.

Since the determinant has changed its sign $i+j-2$ times, we have

 $\det{M_{ij}^{*}}=(-1)^{i+j}\det{M_{ij}}.$

Note also that only those permutations $\pi\in S_{n}$ are for the computation of the determinant of $M_{ij}^{*}$ where $\pi(i)=j$. ∎

Now we start out with

 $\displaystyle\det{M}$ $\displaystyle=\sum\limits_{\pi\in S_{n}}\mathrm{sgn}\pi\left(\prod_{j=1}^{n}m_% {j\pi(j)}\right)$ $\displaystyle=\sum_{k=1}^{n}m_{ik}\left(\sum\limits_{\pi\in S_{n}\mid\pi(i)=k}% \mathrm{sgn}\pi\left(\prod\limits_{1\leq j\leq i}m_{j\pi(j)}\right)\cdot 1% \cdot\left(\prod\limits_{i\leq j\leq n}m_{j-\pi(j)}\right)\right).$

From the previous lemma, it follows that the associated with $M_{ik}$ is the determinant of $M_{ij}^{*}$. So we have

 $\det{M}=\sum_{k=1}^{n}M_{ik}\left((-1)^{i+k}\det{M_{ik}}\right).$
Title proof of cofactor expansion ProofOfCofactorExpansion 2013-03-22 13:22:08 2013-03-22 13:22:08 Thomas Heye (1234) Thomas Heye (1234) 13 Thomas Heye (1234) Proof msc 15A15 Laplace expansion