# A question in linear algebra: prove that $\{\beta_j\}$ is linearly dependent.

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.

• You mean $m < n$? Otherwise, the claim is not true in general. – Dirk May 3 at 15:01
• 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 $\beta_1=\alpha_1$). – drhab May 3 at 15:01
• Let $S$ be the subspace spanned by the $\alpha$. What can you say about dim $S$? – Lozenges May 3 at 15:21

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