# Let $v_1 v_2$ be vectors, prove span{$v_1, v_2$} = span{$v_1, sv_1+tv_2$} where $s,t$ are real numbers

I know in order to prove that they are equal, I need to show that they are subsets of each other, I have managed to show that span{$v_1,sv_1+tv_2$} is a subset of span{$v_1,v_2$} but I'm having trouble showing that span{$v_1,v_2$} is a subset of span{$v_1,sv_1+tv_2$}

Edit: I've tried this but im not sure if it is correct span{$v_1,v_2$} can be written as a vector equation $x=av_1+bv_2$, $a,b \in \Bbb{R}$ then is we add and subtract $sv_1$ it yields $x=av_1+bv_2+sv_1-sv_1$ $a,b,s \in \Bbb{R}$ and then we can manipulate this to get $x=(a-s)v_1+\frac{b}{t}(tv_2)+b(sv_1)$ $a,b,s,t \in \Bbb{R}$ and therefore span{$v_1,v_2$} is a subset of span{$v_1, v_1+v_2$}

• I'm guessing $t \neq 0$? – Theo Bendit Jan 6 '18 at 16:52
• yes, would it be possible to add $sv_1$ and subtract $sv_1$ and work algebraiclly from there – Skrrrrrtttt Jan 6 '18 at 16:54
• your edit is almost correct, see the answers, you need to have $\text{someting}(tv_2+sv_1)$, they have to be together – Holo Jan 6 '18 at 17:05
• @Skrrrrrtttt no, if you take the $b$ out you get $b(\frac1t(tv_2)+sv_1)$. You should add and subtract $\frac stv_1$ so you can factor out $\frac bt$ to get to the right form(this is what I did in my answer) – Holo Jan 6 '18 at 17:10

## 3 Answers

Actually you need to have $t \ne 0$ for it to hold.

It suffices to prove that $$v_1, v_2 \in \operatorname{span}\{v_1, sv_1 + t v_2\}$$

Clearly $v_1 \in \operatorname{span}\{v_1, sv_1 + t v_2\}$.

For $v_2$ we have:

$$v_2 = \frac{1}t(sv_1 + tv_2) - \frac{s}{t} v_1 \in \operatorname{span}\{v_1, sv_1 + t v_2\}$$

If $t = 0$, then for linearly independent $v_1, v_2$ we have $$\operatorname{span} \{v_1, v_2\} \ne \operatorname{span}\{v_1\} = \operatorname{span}\{v_1, sv_1\}$$

so it does not hold.

Note that $$v_2=t^{-1}[(sv_1+tv_2)-sv_1]\in\text{span}\{v_1,sv_1+tv_2\}$$ ($t\neq 0$) and $$v_1\in\text{span}\{v_1,sv_1+tv_2\}$$ so $$\text{span}\{v_1, v_2\}\subseteq \text{span}\{v_1,sv_1+tv_2\}$$ by minimality.

if $x\in\text{span}\{v_1,v_2\},(t\ne0)$ then $x=av_1+bv_2=av_1+b(\frac 1ttv_2)=av_1+b(\frac stv_1-\frac stv_1+\frac 1ttv_2)=av_1+b(\frac{1}t(sv_1 + tv_2) - \frac{s}{t} v_1)=av_1+\frac{b}t(sv_1 + tv_2) - b\frac{s}{t} v_1=(a-b\frac{s}{t})v_1+\frac{b}t(sv_1 + tv_2)\in\text{span}\{v_1,sv_1+tv_2\}$