Show the set of vectors is Linearly independent

Question

(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?

Answer 1

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 $\;\begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix}$.

Answer 2

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.

Answer 3

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 .