In our textbook I've seen this claim:

$v_1, v_2, \cdots , v_r$ span $R^n$ if and only if $v_1+\alpha *v_2, v_2, \cdots, v_r$ span $R^n$ for some $\alpha \in \mathbb{R}$

Now I've tried to prove it to myself and I'm having some difficulty with the left-to-right direction.

For the first direction I assumed: $v_1+\alpha *v_2, v_2, \cdots, v_r$ span $R^n$ so every vector in $R^n$ can be represented as a linear combination of those vectors, meaning a general vector $w$ in $R^n$ can be written as:

$\alpha_1 *(v_1 + \alpha*v_2 ) + \alpha_2 *v_2 + \cdots + \alpha_r*v_r = w$

$\Rightarrow \alpha_1 * v_1 + (\alpha_1*\alpha + \alpha_2)*v_2 + \cdots + \alpha_r*v_r = w$

meaning that $v_1, v_2, \cdots , v_r$ also span $R^n$

For the other direction I assumed that $v_1, v_2, \cdots , v_r$ span $R^n$ so for some $w \in R^n$ :

$\alpha_1 * v_1 + \alpha_2 * v_2 +\cdots+\alpha_r*v_r = w$

$\Rightarrow \alpha_1 *v_1 + (a+b)*v_2 +\cdots+\alpha_r*v_r = w$

$\Rightarrow \alpha_1 * v_1 +a*v_2 + b*v_2 +\cdots+\alpha_r*v_r = w$

$\Rightarrow \alpha_1 * (v_1 + \frac{a}{\alpha_1}*v_2) + b*v_2 + \cdots+\alpha_r*v_r = w$

which would prove that $v_1+\alpha *v_2, v_2, \cdots, v_r$ span $R^n$ but I have no guarantee that $\alpha_1\ne0$ or that such a combination where $\alpha_1\ne0$ even exists (as far as I know, maybe that's what I'm missing here)

any guidance would be appreciated!

  • $\begingroup$ The other implication is wrong unless $v_1+\alpha v_2\neq 0$. So you just have to chose one $\alpha$ such that it is true and call $v'_1=v_1+\alpha v_2$. $\endgroup$ – Tom-Tom Jun 25 '14 at 12:22
  • $\begingroup$ could you elaborate? what do you mean by the other implication (I'm not familiar with the word implication in this context)? If you're referring to the last lines as I wrote they are what I tried to assume but as I said couldn't since I have no guarantee that $\alpha_1\ne0$ $\endgroup$ – byentropy Jun 25 '14 at 12:27

Clearly $$ \langle v_{1} + \alpha v_{2}, v_{2}, \dots , v_{r} \rangle \subseteq \langle v_{1}, v_{2}, \dots , v_{r} \rangle, $$ where $\langle \cdots \rangle$ denotes the linear span.

Conversely $$ v_{1} = (v_{1} + \alpha v_{2}) - \alpha v_{2} \in \langle v_{1} + \alpha v_{2}, v_{2}, \dots , v_{r} \rangle, $$ so the other inclusion also holds.

  • $\begingroup$ Thanks, this certainly helped me see I was going at it the other way around, but I'm still wondering could I have also proved it as I tried? by reaching to $\alpha_1 * (v_1+\frac{\alpha}{\alpha_1}v_2)$ could I have known or assured that $\alpha_1\ne0$ as well? $\endgroup$ – byentropy Jun 25 '14 at 13:02
  • $\begingroup$ @byentropy, following your approach, if you can write an arbitrary $v$ as $c_1 v_1 + c_2 v_2 + \dots$, then you can also write it as $v = c_1 (v_1 + \alpha v_2) + (c_2 - c_1 \alpha) v_2 + \dots$. $\endgroup$ – Andreas Caranti Jun 25 '14 at 17:58

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