If $\text{span } \{v_1, v_2\} = \text{span } \{w\}$ then the set $\{v_1, v_2\}$ is linearly dependent. [vectors]

It must follow that $w = v_1$ or $w = v_2$. Thus in the first case we have $c_3v_1 = c_1v_1 + c_2v_2$, contradiction.

Is this enough?

  • $\begingroup$ The first sentence of your argument is invalid. Given non-unit scalars $\alpha_1,\alpha_2$, we can have $v_1=\alpha_1w$, $v_2=\alpha_2w$, and the hypothesis of the statement are obeyed. $\endgroup$ – Aweygan Feb 5 '17 at 21:29
  • $\begingroup$ by contradiction, if $\{v_1,v_2\}$ is lin. ind. we have that $\{v_1,v_2\}$ ist basis for $span(\{w\})$, therefore $dim_K(span(\{w\}))=2$... $\endgroup$ – mle Feb 5 '17 at 22:10

$ v_1=a_1 w , v_2 = a_2 w$. $ a_2 v_1 + (-a_1)v_2 = 0 $

  • $\begingroup$ One should add the case when $a_1=a_2=0$, but it's easy. $\endgroup$ – egreg Feb 5 '17 at 21:41

The span of $\{v_1,v_2\}$ certainly contains $v_1$ and $v_2$, so if $\operatorname{span}\{v_1,v_2\} = \operatorname{span}\{w\}$, then there are scalars $a_1$ and $a_2$ such that $$a_1w=v_1 \;\text{and}\; a_2w=v_2\,.$$ Or to say this another way $$w=\frac{1}{a_1}v_1 \;\text{and}\; w=\frac{1}{a_2}v_2\,.$$

But then we have $$\frac{1}{a_1}v_1 - \frac{1}{a_2}v_2 = 0\,,$$ which is a problem.


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