Exercise about linear independence A) Let $V$ a vector space over $\mathbb{R}$ and $T: V \rightarrow V$ a linear mapping, such that $T^n=0$ for a natural $n \geq 2$. If $x \in V$, then the set $\{x, Tx, T^2x, \dots, T^{n-1}x \}$ is linearly independent if and only if $T^{n-1}x \neq 0$.
B) Let $V$ a vector space over $\mathbb{R}$ and $n \in \mathbb{N}$ odd. If the vectors $x_1, \dots,x_n$are linearly independent, then the same stands also for the vectors $x_1+x_2, x_2+x_3, \dots, x_{n-1}+x_n, x_n+x_1$.
$$$$
I have done the following:
For $(A)$:
for the direction $\Rightarrow$
The set $\{x, Tx, T^2x, \dots, T^{n-1}x \}$ is linearly independent:
$c_0x +c_1 Tx +c_2 T^2x+ \dots+c_{n-1} T^{n-1}x=0 \Rightarrow c_0=c_1=c_2= \dots=c_{n-1}=0$
$c_0x +c_1 Tx +c_2 T^2x+ \dots+c_{n-1} T^{n-1}x=0 \overset{\cdot T}{\Rightarrow} \\ c_0Tx +c_1 T^2x +c_2 T^3x+ \dots+c_{n-1} T^{n}x=0 \Rightarrow \\ c_0Tx +c_1 T^2x +c_2 T^3x+ \dots+c_{n-2}T^{n-1} =0 \overset{\cdot T}{\Rightarrow} \\ c_0T^2x +c_1 T^3x +c_2 T^4x+ \dots+c_{n-2}T^{n} =0 \Rightarrow  \\ c_0T^2x +c_1 T^3x +c_2 T^4x+ \dots+c_{n-3}T_{n-1} =0 \overset{\cdot T}{\Rightarrow} \\ c_0T^3x +c_1 T^4x +c_2 T^5x+ \dots+c_{n-3}T_{n} =0 \Rightarrow \\ c_0T^3x +c_1 T^4x +c_2 T^5x+ \dots+c_{n-4}T^{n-1}=0 \overset{\cdot T}{\Rightarrow} \dots \Rightarrow c_0T^{n-2}x+c_1T^{n-1}=0 \overset{\cdot T}{\Rightarrow} \\ c_0T^{n-1}x+c_1T^{n}=0 \Rightarrow \\ c_0T^{n-1}x=0 \overset{c_0=0}{\Rightarrow} T^{n-1}x \neq 0$
Is this correct??
for the direction $\Leftarrow$
Knowing that $T^{n-1}x \neq 0$ we have to show that 
$c_0x +c_1 Tx +c_2 T^2x+ \dots+c_{n-1} T^{n-1}x=0 \Rightarrow c_1=c_2=\dots c_{n-1}=0$, right? But how could I do that??
For $(B)$:
$x_1, \dots,x_n$are linearly independent:
$c_1 x_1+c_2x_2+ \dots c_nx_n=0 \Rightarrow c_1=c_2= \dots =c_n=0 \ \ \ (*)$
$a_1(x_1+x_2)+a_2(x_2+x_3)+\dots+a_{n-1}(x_{n-1}+x_n)+a_n(x_n+x_1)=0 \Rightarrow \\ (a_1+a_n)x_1+(a_1+a_2)x_2+\dots(a_{n-1}+a_n)x_n=0 \overset{(*)}{\Rightarrow} \\ a_1+a_n=a_1+a_2= \dots =a_{n-1}+a_n=0$
Can we conclude from here that $a_1=a_2= \dots =a_{n-1}=a_n=0$ ??

To prove (A), it is clear that if $T^{n-1} x = 0$, then the set $\{x, Tx, T^2x, \dots, T^{n-1}x \}$ is linearly dependent.
Conversely, suppose the set is linearly dependent, and let
$$
c_{0} x + c_{1} Tx + \dots + c_{n-1} T^{n-1} x = 0\tag{eq}
$$
be a relation where not all coefficients are zero.
If $c_{0} = c_{1} = \dots = c_{k-1} = 0$, and $c_{k} \ne 0$, apply $T^{n-1-k}$ to (eq) to get
$$
c_{k} T^{n-1} x = 0,
$$
so that $T^{n-1} x = 0$ as $c_{k} \ne 0$.

As to (B), note that if
$$
a_{1} (x_{1} + x_{2}) + a_{2} (x_{2} + x_{3}) + \dots + a_{n} (x_{n} + x_{1}) = 0,
$$
then
$$
(a_{n} + a_{1}) x_{1} + (a_{1} + a_{2} ) x_{2} + \dots + (a_{n-1} + a_{n}) x_{n} = 0,
$$
so that
$$
a_{n} + a_{1} = a_{1} + a_{2} = \dots = a_{n-1} + a_{n} = 0.\tag{eq2}
$$
Now note that since $n$ is odd
$$
0 = (a_{1} + a_{2}) - (a_{2} + a_{3}) + \dots  - (a_{n-1} + a_{n}) + (a_{n} + a_{1})  = 2 a_{1},
$$
so $a_{1} = 0$ and thus, by (eq2), all $a_{i}$ are zero.
