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(Prove) If $\{v_1, v_2, v_3 \}$ if linearly independent then $\{v_1, v_1 + v_2, v_1 + v_2 + v_3 \}$ is linearly independent as well.

By definition we have solution $a = b = c = 0$ if $av_1 + bv_2 + cv_3 = 0$.

Goal is to show $d = e = f = 0$ for $d(v_1) + e(v_1 + v_2) + f(v_1 + v_2 + v_3) = 0$

So this means $dv_1 + ev_2 + fv_3 = -ev_1 - fv_1 - fv_2 - fv_3$

Matching coefficients, $d = -e - f, e = -f, f = -f$ so this means $f=0, e = 0, d = 0 - 0 = 0$ so we have $d = e = f = 0$ as required.

But I never used the hypothesis? So there has to be something wrong?

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    When you wrote "matching coefficient" you used that $\;v_1,v_2,v_3\;$ are linearly independent, otherwise it is **not true** that corresponding coefficients are equal. – 2017-01-08
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    maybe it's more understandable if you group it on one side... $(d+e+f)v_1+(e+f)v_2+(f)v_3=0$ so all coefficients must be $0$ (where you use the assumption) and you get equivalent system of equations – 2017-01-08
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    @DonAntonio, so If I have two vectors $av_1 = bv_1$, is it not true in general that $a = b$? – 2017-01-08
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    @Gaandmit Of course not. For example, $\;3\cdot0=345\cdot0\;$, but $\;3\neq345\;$ . But what you **actually should want to ask** is: "if $\;a_1v_1+\ldots +a_nv_n=b_1v_1+\ldots+b_nv_n\;$ , then it is not true in general $\;a_1=b_1,\,a_2=b_2,\,\ldots,a_n=b_n\;$ ?" And the answer is: yes, it is **not true** in general, *only* if $\;v_1,...,v_n\;$ are linearly independent. – 2017-01-08
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    @DonAntonio, oh wow! Thanks for teaching me this. So we cant do algebra as we would with polynomials for example as with vectors? – 2017-01-08
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    @Gaandmit I'm not sure what you mean, but if you mean that we cannot equal them coefficient-coefficient then that is true... **only** when we have a linearly independent set, e.g. a basis, we can do that, as determining a basis is like determining a coordinate set and then we can work with that as if had coordinate vectors, say...or also with polynomials. – 2017-01-08

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Another proof, without computing whatever:

Consider the subspace $V$ with basis $(v_1,v_2,v_3)$. The system $(v_1, v_1+v_2,v_1+v_2+v_3)$ is another basis of $V$ since its matrix in the former basis is $\;\begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix}$.

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Since $v_i$, $1\leq i\leq 3$ are linearly dependent hence for a given vector $u$, $av_1+bv_2+cv_3=u$ implies that the scalars $a$,$b$ and $c$ are unique. This fact have already been used by you in the step where you have compared coefficients.

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Suppose that $\{v_1, v_2, v_3 \}$ is linearly independent in some vector space , say V .

To show that the list $\{v_1, v_1 + v_2, v_1 + v_2 + v_3 \}$ is linearly independent , we need to find coefficients : $ a_1, a_2 $ and $a_3 $ elements from the set of scalars such that

$a_1v_1 + a_2(v_1 + v_2) + a_3(v_1 + v_2 + v_3) = 0$ , which can further written as

$(a_1+a_2+a_3)v_1+(a_2+a_3)v_2+a_3v_3=0$

Now , since $\{v_1, v_2, v_3 \}$ is linearly independent , there exists saclars : $ b_1 , b_2 $ and $ b_3$ such that

$b_1v_1+b_2v_2+b_3v_3=0$ if $b_1=b_2=b_3=0$

Upon comparison , we have

$a_1+a_2+a_3=b_1=0$
$a_2+a_3=b_2=0$
$a_3=b_3=0$

Consequently , we have a system of linear equations

$a_1+a_2+a_3=0$
$a_2+a_3=0$
$a_3=0$
, whose solution is $ a_1=a_2=a_3=0$

Thus , since $ a_1=a_2=a_3=0$ , $\{v_1, v_1 + v_2, v_1 + v_2 + v_3 \}$ is linearly independent .

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    If, there is any sort of misunderstanding, please, do let me know. – 2017-01-08