Need help proving this, I struggle with proofs overall so feel free to nitpick:
Let $\{v_1, … , v_n\} \in V$. Show that if $\{v_1, … , v_n\}$ is linearly independent, then $\{v_1, v_1 + v_2, v_1 + v_2 + v_3, … , v_1 + … + v_n\}$ is linearly indepedent.
Proof:
Assume $\{v_1, … , v_n\} \in V$ is linearly indepedent.
Linear independence implies for $c_1v_1+…+c_nv_n=0$ for all $v\in V$ and $c_i \in R$ $c_1 = c_2 = … = c_n = 0$
Conceptually I understand why $\{v_1, v_1 + v_2, v_1 + v_2 + v_3, … , v_1 + … + v_n\}$ is linearly indepedent. I am just unsure how to continue the proof other than to say each sum is in the initial sets linear combination which we know is linearly independent so adding the sums onto each other would just create another linear combination where the only solution to the homogenous system is 0 for all scalar $c_i$.
Best Answer
Suppose that $\{v_1,v_2,\dots,v_n\}$ is linearly independent. To show linear independence of $\{v_1,v_1+v_2,v_1+v_2+v_3,\dots,v_1+v_2+\dots+v_n\}$ take a linear combination and set it equal to zero, then prove that the coefficients are necessarily zero. Observe that if, \begin{align} a_1(v_1) + a_2(v_1+v_2) + \dots + a_n(v_1+v_2+\dots+v_n) &= 0\\ \Rightarrow \left(\sum_{i=1}^n a_i\right) v_1 + \left(\sum_{i=2}^n a_i\right) v_2 + \dots + a_nv_n &= 0 \end{align} Which by linear independence of $\{v_1,v_2,\dots,v_n\}$ implies that $\left(\sum_{i=1}^n a_i\right), \left(\sum_{i=2}^n a_i\right),\dots, a_n$ are all equal zero, from which it's easy to see that $a_1,a_2,\dots,a_n$ are all zero as well so we are done.