Exercise:
Show that the determinant is unique.

Solution:
Let D:M_ntimes nKlongrightarrow K be an n-linear alternating function. Let A pmatrix hdots & alpha_ & hdots hdots & vdots & hdots hdots & alpha_n & hdots pmatrix in M_ntimes nK. Denote I pmatrix & hdots & hdots & vdots & hdots & hdots & pmatrix pmatrix hdots & epsilon_ & hdots hdots & vdots & hdots hdots & epsilon_n & hdots pmatrix so epsilon_...epsilon_n is the standard basis K_textrow^n. Clearly alpha_i_j^n Aijepsilon_j leq ileq n. Longrightarrow DADalpha_...alpha_nDleft_j^n Ajepsilon_j alpha_...alpha_n right left_j^n Aj Depsilon_jalpha_...alpha_n right left_j^n Aj Depsilon_j_k^n Ak epsilon_k...alpha_n right left_jk Aj Ak epsilon_jepsilon_k...alpha_n right ... _k_...k_nAk_ Ak_... Ank_n Depsilon_k_epsilon_k_...epsilon_k_n where each of the indices k_...k_n runs between and n. So far we haven't used the fact that D is alternating. Now since D is alternating Depsilon_k_...epsilon_k_n whenever two of the indices k_...k_n are equal. So we are erested only in ordered ! sequences k_...k_n in which k_iin ...n forall i and forall k_ineq k_j forall ineq j. Such a sequence is called a permutation of ...n or a permutation of degree n. Alternatively we can think of a permutation of degree n as a bijective function sigma :...nlongrightarrow ...n. Then the sequence k_...k_nsigma sigma ...sigma n is a permutation as defined earlier. Notation sigma i : sigmai. In other words: a permutation of deg n is a rordering of a list of n elements ...nlongrightarrow sigma ...sigma n. We can rewrite as: DA _sigmaAsigma Asigma ... Ansigma nDepsilon_sigma epsilon_sigma ...epsilon_sigma n where sigma runs over all permutations of degree n.