Prove the linear independence of $\{v_1,v_2+v_2 \}\,$? 
Prove that if $\{v_1,v_2\}$ are linearly independent (LI), then also $\{v_1,v_1+v_2\}$ are linearly independent.
Is the converse true?

I have proved the first part, but now i am stuck on whether the converse is true.
that is, if the $ \{ v_1,v_1+v_2 \} $ are LI, so are $\{v_1,v_2\}$.
How can i prove this?
Thanks in advance.
 A: The reverse can be proved with the usual definition of linear independence.
Let $c_1,c_2\in\mathbb{R}$ be such that $c_1 v_1+c_2v_2=0$. Then
$$0=c_1v_1+c_2v_2=c_1v_1+c_2v_2+c_2v_1-c_2v_1=(c_1-c_2)v_1+c_2(v_1+v_2)$$
Because $v_1,v_1+v_2$ are linearly independent, we must conclude $c_1-c_2=c_2=0$, which leads to $c_1=c_2=0$ and proves $v_1,v_2$ are linearly independent.
A: $\{v_1 ,v_1 +v_2 \}$ L.I $\implies \{v_1,v_2 \}$ L.I
Proof : Assume the contrary that $\{v_1,v_2 \}$  L.D
Then, $v_2 =\lambda v_1$ for some scalar $\lambda$
Then $v_1+ v_2 = (1+\lambda ) v_1$
And this implies $\{v_1 ,v_1 +v_2 \}$ is linearly dependent.
A: Both answers by @Suzane And @S.G are  great.
This is a bit different and a more general way to prove this for $n$ vectors.
Take the linearly independent set $\mathcal{B}=\{v_{1},v_{1}+v_{2},v_{1}+v_{2}+v_{3},...,v_{1}+v_{2}+..+v_{n}\}$
But denote it by $\mathcal{B}=\{w_{1},w_{2},...,w_{n}\}$.
Then $v_{1}=w_{1}$,
$v_{2}=w_{2}-w_{1},$
$\vdots$
$v_{k}=w_{k}-w_{k-1}\,,2\leq k\leq n$
So if we were to write the matrix of this transformation wrt to basis $w_{1},w_{2},...,w_{n}$ it would be:-
$$\begin{bmatrix}
1&-1&0&\cdots &0\\
0& 1 &-1&\cdots &0\\
0&0&1&-1\cdots &0\\
\vdots\\
0&0&\cdots&&-1\\
0&0&\cdots&&1
\end{bmatrix}$$
It is a simple induction to show that the determinant of the matrix is $1$. Hence $\{v_{1},v_{2},...,v_{n}\}$ are linearly independent as the invertible map takes basis to basis.
A: $\because\{v_1,v_1+v_2\}$ are LI;
$\therefore \alpha.v_1+\beta. (v_1+v_2)=0$ holds if $\alpha=\beta=0$
$ \implies (\alpha+\beta)v_1+\beta.v_2=0$ holds if $\alpha=\beta=0$
or, $c_1.v_1+c_2.v_2=0$ holds if $\alpha=\beta=0$ where $c_1=\alpha+\beta, c_2=\beta$
Equivalently, $c_1.v_1+c_2.v_2=0$ holds if $c_1=c_2=0$
