For every matrix $A \in M_n(K)$: $\det(A) = \sum_{\sigma \in S} \operatorname{sgn}(\sigma)a_{\sigma(1),1} \cdots a_{\sigma(n),n}$.

Proof: Consider $B = (b_{ij}) \in M_n(K)$. Then $ C := AB \in M_n(K) $ with $C = (C_1,\dots,C_n) $ and $C_k = b_{1k}A_1 + \cdots + b_{nk}A_n$. Using the linearity of the determinant we get: $\underline{\det(C_1,\dots,C_n) = \sum_{i_1,i_2,\dots,i_n} b_{i_1,1}b_{i_2,2}\cdots b_{i_n,n} \det(A_{i_1},\dots,A_{i_n}) }$, where all $i_j$ are independently range from $1$ to $n$.

If $i_k = i_{\ell}$, matrix $(A_{i_1},\dots,A_{i_n})$ will have two identical columns and therefore $\det(A_{i_1},\dots,A_{i_n}) = 0$. Only the terms with $\sigma = (i_1,\dots,i_n)$ a permutation of $\{1,\dots,n\}$ appear in the sum. Using $\det(A_{i_1},\dots,A_{i_n}) = \operatorname{sgn}(\sigma)\det(A)$, we get:

$\underline{\det(C_1,\dots,C_n) = \det(A_{1},\dots,A_{n}) \sum_{\sigma\in S} \operatorname{sgn}(\sigma)b_{\sigma(1),1}b_{\sigma(2),2}\cdots b_{\sigma(n),n} }$. With $S$ the set of all permutations of $\{1,\dots,n\}$.

The given statement can be proven by setting $A = I_n$ in the above.

I'm having troubles understanding where the underlined parts come from, especially the first expression for $\det(C)$. Hopefully, someone could help me.

  • 1
    $\begingroup$ First expression is just multilinearity of the determinant. $\endgroup$ – Dietrich Burde Jan 12 at 9:18

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