linear algebra - $C$ is linearly independent implies $x_ineq 0$ for all $i$.

Let $V$ denote a vector space over a field $F$ with a basis $B=\{e_1,e_2,\ldots e_n\}$. Let $x_1,x_2\ldots x_n\in F$.

Let $C=\{x_1e_1,x_1e_1+x_2e_2,\ldots ,x_1e_1+\ldots +x_ne_n\}$. Then

1. $C$ is linearly independent implies $x_i\neq 0$ for all $i$.


2. $x_i\neq 0$ for every $i$ implies that $C$ is a linearly independent set.


3. The linear span of $C$ is $V$ implies that $x_i\neq 0$ for all $i$.


4. $x_i\neq 0$ for every $i$ implies that the linear span of $C$ is $V$.

My try:

1.Unable to do this problem.

2.Let us consider

$c_1x_1e_1+c_2(x_1e_1+x_2e_2)+\ldots +c_n(x_1e_1+\ldots +x_ne_n)=0\implies e_1(c_1x_1+c_2x_1+\ldots c_nx_1)+e_2(c_2x_2+\ldots c_nx_2)+\ldots c_nx_ne_n=0$

Since $\{e_i\}$ forms a base and $x_i\neq 0$ so $c_n=c_{n-1}=c_1=0$.
So $C$ is linearly independent.

3.We know that a basis is minimal generating set.If $x_i=0$ for some $i$ and $\text{span} C=V$ and then $\{x_1,x_2,\ldots x_n\}\setminus \{x_i\}$ generates $V$ which is false.

4.Since $x_i\neq 0 $ and $\{e_i\}$ spans $V$ so does $C$.

Please check my explanations and suggest some help for $1$.

Looking forward to your help.

Answer

For both $(1)$ and $(2)$, consider the equation

$$ c_1 (x_1 e_1) + c_2 (x_1 e_1 + x_2 e_2) + \dots + c_{n-1} (x_1 e_1 + \dots + x_{n-1} e_{n-1}) + c_n (x_1 e_1 + \dots + x_n e_n) = \\

(c_1 x_1 + \dots + c_n x_n) e_1 + (c_2 x_2 + \dots + c_{n} x_{n}) e_2 + \dots + (c_{n-1} x_{n-1} + c_n x_n) e_{n-1} + (c_n x_n) e_n = 0. $$

Since the $\{e_i\}$ are linearly independent, this system has a non-trivial solution $(c_1, \dots, c_n) \neq (0,\dots,0)$ if and only if the system of equations

$$ c_1 x_1 + \dots + c_n x_n = 0,\\
c_2 x_2 + \dots + c_n x_n = 0, \\
\vdots \\
c_{n-1} x_{n-1} + c_n x_n = 0, \\
c_n x_n = 0, $$

have a non-trivial solution. If some $x_i = 0$ then the system of equations doesn't involve the variable $c_i$ and so $(c_1, \dots, c_i, \dots, c_n) = (0,\dots,1,\dots,0)$ is a non-trivial solution to the equation and $C$ is linearly dependent. If all the $x_i \neq 0$ then the last equation $x_n c_n = 0$ together with $x_n \neq 0$ implies that $c_n = 0$ and then the one-before-last equation reads $c_{n-1} x_{n-1} = 0$ which again together with $x_{n-1} \neq 0$ implies that $c_{n-1} = 0$ and so on. Note that this is pretty much your argument for $(2)$, but I felt that is didn't provide enough details as to why $c_n = \dots = c_1 = 0$.

For $(3)$, if some $x_i = 0$ then $e_i$ doesn't appear in $C$ and so $\operatorname{span} C$ is a subspace of $\operatorname{span} \{ e_1, \dots, e_{i-1}, e_{i+1}, \dots, e_{n} \}$ which is not the whole of $V$.

For $(4)$, if all the $x_i \neq 0$ then by $(2)$ we know that $C$ is linearly independent and consists of $n$ elements in an $n$ dimensional space and so must be a spanning set and a basis.