Linear algebra, basis

Question

Let $ B = \langle v_1, ..., v_n\rangle, \space C = \langle u_1, ..., u_n\rangle $ be some different bases of $ \mathbb{F^{n}} $ .

Suppose that there exists a vector

$$ w \in \mathbb{F^{n}} $$

which satisfies

$$ [w]_B = [w]_C $$

What can we say then about B, C or even the vector itself? I figured that this situation is possible when $w$ is a part of the the basis itself, namely if there exists some $1 \leq i \leq n$ , such that $ u_i = v_i = w $ then it has to imply that

$$ [w]_B = [w]_C = e_i $$

Does anyone see any other situations when this is possible?

Answer 1

This condition is always satisfied by $0$.

A nontrivial case of this condition exists if and only if $\{u_i-v_i\}_{i=1}^n$ is not a basis (in particular linearly dependent).

If $\{u_i-v_i\}_{i=1}^n$ is not a basis, then choose $\{a_i\}_{i=1}^n \subseteq \mathbb{F}$ not all zero such that $\sum_{i=1}^n a_i(u_i-v_i) =0$, which gives $\sum_{i=1}^na_iu_i=\sum_{i=1}^na_iv_i$.

If $\sum_{i=1}^na_iu_i=\sum_{i=1}^na_iv_i$ for $\{a_i\}_{i=1}^n \subseteq \mathbb{F}$ not all zero, then $\sum_{i=1}^n a_i(u_i-v_i) =0$, which proves $\{u_i-v_i\}_{i=1}^n$ is not a basis.

Answer 2

Fix $B, C$. The vectors $v \in \mathbb{F}^n$ with the property that there exists $w$ such that $[w]_B = [w]_C = v$ are precisely $0$ and the eigenvectors of the change of basis matrix $Q$ between bases $B$ and $C$ associated to the eigenvalue $1$ (a.k.a the fixed space of $Q$, a.k.a. $\ker(Q - I)$).