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Take the underlying field as $\mathbb{R}$. Suppose $\beta_1, \cdots, \beta_n$ are in the span of $\alpha_1, \cdots, \alpha_m$. Suppose that $n > m$, then prove that $\beta_1, \cdots, \beta_n$ are linearly dependent.

I think this is somewhat trivial since the dimension of $\{\beta_j\}$ is less than $\{\alpha_i\}$. However, this problem occurs before the introduction of basis and dimension. So I think there should be a more elementary way to show it.

Edit: Sorry for the typo, it should be $n > m$, instead of $m > n$. I have fixed it in the statement.

I think about it for a while, I think we could use some "Replacement".

Suppose the contrary, that is, $\beta_j$'s are linearly independent. WLOG, we could assume $\{\alpha_1, \cdots, \alpha_m\}$ to be linearly independent, otherwise we could further extract a linearly independent set from it without changing its span. Then since $\beta_1 \in \text{span} \{\alpha_i\}$, we could write a combination $k_1\alpha_1 + \cdots + k_m \alpha_m = \beta_1$, with at least one $k_j \neq 0$. WLOG assume $k_1 \neq 0$. Then $\{\beta_1, \alpha_2, \cdots, \alpha_m\}$ is linearly independent, and its span is the same as $\{\alpha_1, \alpha_2, \cdots, \alpha_m\}$. Using the linear independence of $\beta_j$'s, we could inductively replace $\alpha_2$ by $\beta_2$, and continue this process. Then a subset $\{\beta_1, \cdots, \beta_m\}$ could spen all $\{\beta_1, \cdots, \beta_n\}$, a contradiction to the linear independence.

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  • $\begingroup$ You mean $m < n$? Otherwise, the claim is not true in general. $\endgroup$ – Dirk May 3 at 15:01
  • $\begingroup$ Don't you mean $m<n$? If e.g. $\alpha_1,\alpha_2$ are independent then so is $\alpha_1$ (case where $n=1<2=m$ and $\beta_1=\alpha_1$). $\endgroup$ – drhab May 3 at 15:01
  • $\begingroup$ Let $S$ be the subspace spanned by the $\alpha $. What can you say about dim $S$? $\endgroup$ – Lozenges May 3 at 15:21
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Here is a counterexample.

Take $\beta_1 = \alpha_1$ and $\beta_2 = \alpha_2$ assuming you have $\mathcal{B} = \left\{ \alpha_1, \alpha_2, \alpha_3 \right\}$. Surely the $\{\beta_k\}$ are linearly independent.

You likely mean $m < n$...

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  • $\begingroup$ Thanks, you are right. It is $m < n$! $\endgroup$ – mathdoge May 3 at 16:19
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The fact that $\beta_1,...,\beta_n$ lie in the span of $\alpha_1,...,\alpha_m$ does not imply that they are linearly independent. For instance, if $\beta_k=k\beta_1$ and $\beta_1$ belongs to the span of $\alpha_i$, then the hypothesis hold but the thesis is false.

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    $\begingroup$ Thanks, you are right. I made a typo, it should be $m < n$! $\endgroup$ – mathdoge May 3 at 16:19

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