Talk:Completion of Nondegenerate Bases

Comment to User:Usagiop
There is already a condition on $\tuple {v_1, \ldots, v_k}$, as they are stated to be a nondegenerate $k$-tuple. So it is impossible that $v_1 = v_2$, as $\set {v_1, v_2}$ is required to span a 2-dimensional subspace (or at least, that's how I interpret the definition of nondegenerate $k$-tuple ). So I think that the statement of this theorem is correct. --Anghel (talk) 14:50, 27 January 2023 (UTC)


 * Yes, my mistake. --Usagiop (talk) 15:25, 27 January 2023 (UTC)