0
$\begingroup$

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{$v_1, v_1+v_2$}

$\endgroup$
  • 2
    $\begingroup$ I'm guessing $t \neq 0$? $\endgroup$ – Theo Bendit Jan 6 '18 at 16:52
  • $\begingroup$ yes, would it be possible to add $sv_1$ and subtract $sv_1$ and work algebraiclly from there $\endgroup$ – Skrrrrrtttt Jan 6 '18 at 16:54
  • $\begingroup$ your edit is almost correct, see the answers, you need to have $\text{someting}(tv_2+sv_1)$, they have to be together $\endgroup$ – Holo Jan 6 '18 at 17:05
  • 1
    $\begingroup$ @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) $\endgroup$ – Holo Jan 6 '18 at 17:10
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.