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Let $V$ be an arbitrary $n$ dimensional vector space over $\mathbb{F}$, and let $\beta_1$ and $\beta_2$ be bases for $V$. Take elements $s_1, s_2, \cdots, s_m$ of $V$ with $m < n$, and let $S = \{[s_1]_{\beta_1}, [s_2]_{\beta_1}, \cdots, [s_m]_{\beta_1}\}$ be the set of coordinate vectors of $s_i$ relative to $\beta_1$.

Now take $[\sigma]_{\beta_1} \in span(S)$. There exists some $v \in V$ such that the n-tuple of scalars representing $[\sigma]_{\beta_1}$ is the same as the n-tuple of scalars representing $[v]_{\beta_2}$. Does this imply that $[v]_{\beta_1} \in span(S)$? Note that $span(S)$ is the set of all linear combinations of the coordinate vectors $[s_i]_{\beta_1}$.

Is there any general reading I can find studying the properties of vectors of the same space relative to different bases and their interactions?

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Try as a counterexample: $n=2,m=1$, $\beta_1=\{e_1,e_2\}$, $\beta_2=\{e_2,e_1\}$, $s_1=\sigma=e_1$. – Simon Markett Nov 26 '12 at 18:23
@SimonMarkett, while the example is clear, it's not good as $\beta_1$, $\beta_2$ are sets and are thus equal. – Karolis Juodelė Nov 26 '12 at 18:29
@KarolisJuodelė Hmm, yes. I suppose I should have added that the bases are ordered, however, so the error is on my part. – providence Nov 26 '12 at 18:30
@KarolisJuodelė, actually I thought about it for a second. But then I figured instead of introducing a new notation for ordered sets I was safe assuming that it is clear that bases are always ordered. – Simon Markett Nov 26 '12 at 18:33
up vote 1 down vote accepted

Let $\beta_1 = \{e_1, e_2, \dots e_n\}$ and $\beta_2 = \{e_1 + e_2, e_2, \dots e_n\}$. Let $s_1 = e_1$. Now $(1, 0, \dots 0)_{\beta_1} = e_1 \in span(S) = span(\{e_1\})$. Also $[e_1]_{\beta_2} = (1, -1, 0, \dots 0)_{\beta_2}$. But clearly, $(1, -1, 0, \dots 0)_{\beta_1} = e_1 - e_2 \notin span(\{e_1\}) = \{ce_1 : c \in \mathbb{R} \}$.

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No. Try with $V:=\Bbb R^2$, $\beta_1$ is the standard basis and $\beta_2:=\left\langle\pmatrix{1\\1}, \pmatrix{0\\1}\right\rangle$, and consider $m=1$, $s_1=\pmatrix{1\\1}$.

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