# $\varphi_{i_1}\wedge\dots\wedge\varphi_{i_k}(v_1,\dots,v_k)$ as a determinant

Let $e_1\dots,e_n$ be the standard basis of $\mathbf R^n$ and let $\varphi_1\dots,\varphi_n$ be the dual basis. I need to show that $\varphi_{i_1}\wedge\dots\wedge\varphi_{i_k}(v_1,\dots,v_k)$ is the determinant of the $k\times k$ minor of $(v_1,\dots, v_k)^T$ obtained by selecting columns $i_1,\dots, i_k$.

By definition, $$\varphi_{i_1}\wedge\dots\wedge\varphi_{i_k}(v_1,\dots,v_k)=\sum_{\sigma \in S_k} \operatorname{sgn}\sigma\cdot\varphi_{i_1}(v_{\sigma(1)})\dots\varphi_{i_k}(v_{\sigma(k)}),$$ but I don't know what simplifications I should do further.

• That formula with the sign and the permutations is the determinant. – Trevor Gunn Apr 2 '18 at 23:27
• What definition of determinant are you using? – mucciolo Apr 2 '18 at 23:31
• @TrevorGunn It certainly looks like the the determinant, but it's not obvious to me that this is the determinant of the matrix described. – user531232 Apr 2 '18 at 23:35
• @mucciolo I'm allowed to use any definition; I guess the one involving permutations would be the most appropriate. – user531232 Apr 2 '18 at 23:36
• If $v_i = (v_{i,1},\dots,v_{i,n})$ then $\varphi_j(v_i) = v_{i,j}$. – Trevor Gunn Apr 3 '18 at 0:57