# Determinant: question on proof

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.

• First expression is just multilinearity of the determinant. – Dietrich Burde Jan 12 at 9:18