$V$ is vector space. Let $v_1,v_2,v_3 in V$. Assume that $v_3$ is not linear combination of $v_1$ and $v_2$ .Then set $\{v_1,v_2\}$ linearly independent if and only if $\{v_1+v_3, v_2+v_3\}$ is linearly independent

My idea

$\Rightarrow $ suppose $\{v_1,v_2\}$ are L.I

let suppose $a(v_1+v_2)+b(v_2+v_3)=0 \Rightarrow av_1+(a+b)v_2+bv_3=0$

then $a,b=0$ if not then $v_3$ is linear combination of $v_1$ and $v_2$

Am I right if i am can you help me other part

thank you.........


Forward direction:

$$a(v_1+\color{red}{v_3})+b(v_2+v_3)=0$$ $$av_1+bv_2+(a+b)v_3=0$$

Hence we have $a+b=0$, and hence


and we conclude that $a=b=0$ due to linear independence of $\{ v_1, v_2\}$.

Give the other direction a try.


The other direction is not true as illustrated by the other post.

  • $\begingroup$ ..thank you,, can you tell me first direction is i am right $\endgroup$ – Inverse Problem Feb 17 '18 at 6:30
  • $\begingroup$ you made a mistake, that is why I colored the mistake. $\endgroup$ – Siong Thye Goh Feb 17 '18 at 6:32
  • $\begingroup$ ..got it..thank you once again $\endgroup$ – Inverse Problem Feb 17 '18 at 6:34
  • $\begingroup$ ....but i facong problem with other direction $\endgroup$ – Inverse Problem Feb 17 '18 at 6:35
  • $\begingroup$ edited my solution, try to complete it? $\endgroup$ – Siong Thye Goh Feb 17 '18 at 6:44

The forward direction is indeed true: If $v_1$, $v_2$ are linearly independent, then for any $v_3$ not a linear combination of $v_1, v_2$, the vectors $v_1 +v_3$ and $v_2+v_3$ are linearly independent.

Indeed, for any $a$ and $b$, $a(v_1+v_3) + b(v_2+v_3) = av_1 + bv_2 + (a+b)v_3$. However, $av_1 + bv_2$ is nonzero and not equal to $-(a+b)v_3$, as $v_3$ is not a linear combination of $v_1$ and $v_2$.

The backwards direction is not true: Take $v_1 = 2v_2 = (1,0)$ and $v_3 = (0,1)$. Then $v_1$ and $v_2$ are not linearly independent but the vectors $v_1+v_3$ and $v_2+v_3$ are.


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