$C=\{x_1e_1,x_1e_1+x_2e_2,\dots,x_1e_1+\dots+x_ne_n\}$ linearly indipendent?

$V$ is a vector space over a field $F$ with a basis $B=\{e_1,\dots,e_n\}$, $x_1,x_2,\dots,x_n\in F$. Let $$C=\{x_1e_1,x_1e_1+x_2e_2,\dots,x_1e_1+\dots+x_ne_n\}$$

Then I need to say which of the followings are correct.

1. $C$ is linearly indipendent $\Rightarrow x_i\ne0\forall i$

2. $x_i\ne 0\forall i\Rightarrow C$ is a linearly indipendent set.

3. The linear span of $C$ is $V$ implies that $x_i\ne 0\forall i$

4. $x_i\ne 0\forall i\Rightarrow$ the linear span of $C$ is $V$

I calculated and I have understood that $1$ and $2$ are $\Leftrightarrow$ condition and hence both are true.

for $3$ if one of $x_k=0$ then we will miss $e_k$ in the span by this set and hence $V$ will not be the span of $C$ so again I am sure $3,$ are $\Leftrightarrow$ condition. Thanks for correcting me.

• 3 is an obvious consequence of 1 and the same for 4 and 2. – MBYagbasan Nov 21 '14 at 13:48
• If you use the symbol $\forall$, then don't place it as an afterthought. In a formula it should always come before the expression it applies to. So for instance in 1.: ...$\Rightarrow(\forall i:x_i\neq0)$. Or in this special case you could avoid the symbol altogether by multiplying: ...$\Rightarrow x_1\ldots x_n\neq0$. – Marc van Leeuwen Nov 21 '14 at 14:08

Assume that there exist $k_1,k_2,\ldots,k_n$ that make $$k_1(x_1e_1)+k_2(x_1e_1+x_2e_2)+\ldots+k_n(x_1e_1+\ldots+x_ne_n)=0$$
We can get $$\sum_{i=1}^{n}k_ix_1e_1+\ldots+k_nx_ne_n=0$$
Beacuse of the linear independence of $e_j$, we have $$\underbrace{\begin{bmatrix} x_1 &x_1 &\cdots &x_1 \\ &x_2 & \ldots &x_2 \\ & &\ddots &\vdots \\ & & &x_n \end{bmatrix}}_{=\mathbf{A}} \begin{bmatrix} k_1 \\ k_2\\ \vdots \\ k_n \end{bmatrix}=0$$
When $\mathbf{A}$ is nonsingular, the vectors are linear independent beacause the linear equation has only solution $0$.