If $\mathcal V:=\{\vec v_i\}$ and $\mathcal U:=\{\vec u_i\}$ are bases that is $\mathcal V_i:=\{\vec v_1,...,\vec u_i,...\vec v_n\}$ a base?

Question

So let be $V$ a vector space of finite dimension and let be $\mathcal V:=\{\vec v_i:i=1,...,n\}$ and $\mathcal U:=\{\vec u_i:i=1,...,n\}$ two different bases so that there exist $i=1,...,n$ such that $u_i\neq v_j$ for each $j=1,...,n$. Therefore I asked if the collection $\mathcal V_i:=\{\vec v_1,...,\vec v_{i-1},\vec u_i,\vec v_{i+1},...\vec v_n\}$ is a base too and I answered to me that it could be not because it seems to me that it is possible that the vector $\vec u_i$ lies in the subspace spanned by the vector $\vec v_1,...,\vec v_{i-1},\vec v_{i+1},...,\vec v_n$ but unfortunately I did not able to find a counterexample. So could someone clarify if $\mathcal U$ is effectively a base, please?

Answer 1

Suppose that $V=\Bbb R^3$, that$$\mathcal U=\{(0,1,1),(1,0,1),(1,1,0)\}\quad\text{and that}\quad\mathcal V=\{(1,0,0),(0,1,0),(0,0,1)\}.$$Then $\mathcal U\cap\mathcal V=\emptyset$, but if the first element of $\mathcal V$ gets replaced by the first element of $\mathcal U$, then you don't have a basis anymore.