Linear Independence in Quotient Space Recently, I have been reading ''Introductory Real Analysis'' by A.N. Kolmogorov (Translated by R.A. Silverman). I am puzzled by a statement made in the proof of Theorem 2, Section 13.4:
Theorem 2: Suppose $W$ is a subspace of a linear space $V$. It follows $\text{dim}(V/W) = n$ iff $\{ v_{1}, \dots, v_{n} \}$ is a linearly independent set in $V$ and each $v \in V$ has a unique representation of the form $v = \alpha_{1} v_{1} + \dots + \alpha_{n} v_{n} + w$ such that $\alpha_{1}, \dots, \alpha_{n} \in \mathbb{C}$ and $w \in W$.
In the proof of this theorem, the author claims that the linear independence of $\{ v_{1}+W, \dots, v_{n}+W \}$ follows from the linear independence of $\{ v_{1}, \dots, v_{n} \}$. It is later claimed that the linear independence of $\{ v_{1}, \dots, v_{n} \}$ follows from that of $\{ v_{1}+W, \dots, v_{n}+W \}$ during the proof of the reverse implication. I am struggling to show these claims.
Attempt: Suppose $\{ v_{1}, \dots, v_{n} \}$ is a linearly independent set in $V$. It follows $\alpha_{1} v_{1} + \dots + \alpha_{n} v_{n} = 0 \Leftrightarrow \alpha_{1} = \dots = \alpha_{n} = 0$. Further: 
$$\alpha_{1} v_{1} + \dots + \alpha_{n} v_{n} = 0$$
$$\Rightarrow (\alpha_{1} v_{1} + \dots + \alpha_{n} v_{n}) + W = 0+W$$
$$\Rightarrow \alpha_{1} (v_{1}+W) + \dots + \alpha_{n} (v_{n}+W) = 0+W$$
Since $\alpha_{1} = \dots = \alpha_{n} = 0$ from the linear independence of $\{ v_{1}, \dots, v_{n} \}$, does this necessarily imply $\{ v_{1}+W, \dots, v_{n}+W \}$ is a linearly independent set? I certainly see that $\alpha_{1} = \dots = \alpha_{n} = 0$ is a solution, but I do not believe this implies it is the only solution.
If $\{ v_{1}+W, \dots, v_{n}+W \}$ is a linearly independent set, then $\alpha_{1} (v_{1}+W) + \dots + \alpha_{n} (v_{n}+W) = 0+W \Leftrightarrow \alpha_{1} = \dots = \alpha_{n} = 0$. Further, 
$$\alpha_{1} (v_{1}+W) + \dots + \alpha_{n} (v_{n}+W) = 0+W$$
$$\Rightarrow (\alpha_{1} v_{1} + \dots + \alpha_{n} v_{n}) + W = 0+W$$
$$\Rightarrow \alpha_{1} v_{1} + \dots + \alpha_{n} v_{n} \in W$$
I don't see how I can proceed any further in this direction. I would greatly appreciate any guidance.
 A: It is not right to say that the linear independence of the images follows simply from the linear independence of $v_1,\ldots,v_n$. It does follow, however, once you combine that with the uniqueness clause of the expressions. 
One of the things being asserted is true (that if $v_1+W,\ldots,v_n+W$ are linearly independent in $V/W$, then $v_1,\ldots,v_n$ are linearly independent in $W$). The other is not.
We have (in what follows, a list with a repeated vector is automatically linearly dependent).

Theorem. Let $V$ and $Z$ be vector spaces, and let $T\colon V\to Z$ be a linear transformation. 
  
  
*
  
*If $v_1,\ldots,v_n$ are vectors in $V$ such that $T(v_1),\ldots,T(v_n)$ are linearly independent in $Z$, then $v_1,\ldots,v_n$ are linearly independent in $V$. 
  
*$T$ is one-to-one if and only if whenever $v_1,\ldots,v_n$ are linearly independent in $V$, then $T(v_1),\ldots,T(v_n)$ are linearly independent in $Z$.

Proof. 1. Suppose $T(v_1),\ldots,T(v_n)$ are linearly indepedent, and let $\alpha_1,\ldots,\alpha_n$ be scalars such that
$$\alpha_1v_1+\cdots\alpha_nv_n=0.$$
We want to show that $\alpha_1=\cdots=\alpha_n=0$. 
Applying $T$ we have $$0 = T(0) = T(\alpha_1v_1+\cdots+\alpha_nv_n) = \alpha_1T(v_1)+\cdots + \alpha_nT(v_n).$$
Since $T(v_1),\ldots,T(v_n)$ are linearly independent, it follows that $\alpha_1=\cdots=\alpha_n=0$, as desired.


*Suppose $T$ is one-to-one, and let $v_1,\ldots,v_n$ be linearly independent in $V$. To show that $T(v_1),\ldots,T(v_n)$ are linearly independent, assume that $\alpha_1T(v_1)+\cdots+ \alpha_nT(v_n)=0$ in $Z$. We want to show that $\alpha_1=\cdots=\alpha_n = 0$.


Now we have $$0 = \alpha_1T(v_1)+\cdots+\alpha_n(Tv_n) = T(\alpha_1v_1+\cdots+\alpha_nv_n).$$
Since $T$ is one-to-one, this implies that $\alpha_1v_1+\cdots+\alpha_nv_n = 0$. And since $v_1,\ldots,v_n$ are linearly independent, then $\alpha_1=\cdots=\alpha_n=0$, as desired.
Conversely, assume that whenever $v_1,\ldots,v_n$ are linearly independent in $V$, then $T(v_1),\ldots,T(v_n)$ are linearly independent in $Z$. To prove that $T$ is one-to-one, let $v$ be in the kernel of $T$. We want to show that $v=0$. If $v\neq 0$, then $v$ is linearly independent, so $T(v)$ is linearly independent, so $T(v)\neq 0$. Thus, if $T(v)=0$, then $v=0$. This proves $T$ is one-to-one. $\Box$
So their assertion that if $v_1+W,\ldots,v_n+W$ is linearly independent then $v_1,\ldots,v_n$ is linearly independent holds, by using the above theorem with $T\colon V\to W/V$ being the canonical projection, $T(v) = v+W$. 
Their assertion that if $v_1,\ldots,v_n$ is linearly independent in $V$ then $v_1+W,\ldots,v_n+W$ is linearly independent in $W$ is false unless $W=0$, because the canonical projection is not one-to-one. So just take an element $w\in W$ with $w\neq 0$. Then $w$ is linearly independent, but $w+W=0+W$ is not.
You are trying to prove that if $v_1,\ldots,v_n$ are linearly independent and satisfy the uniqueness clause, then $v_1+W,\ldots,v_n+W$ are linearly independent in $V/W$. But you are starting with the wrong linear combination.
To prove that $v_1+W,\ldots,v_n+W$ is linearly independent, you need to start with them, not with $v_1,\ldots,v_n$. So, to that end, let $\alpha_1,\ldots,\alpha_n$ be scalars such that 
$$\alpha_1(v_1+W) + \cdots + \alpha_n(v_n+W) = \mathbf{0}.$$
This means that
$$(\alpha_1v_1+\cdots+\alpha_nv_n) + W = 0+W$$
and hence that there exists $w\in W$ such that $\alpha_1v_1+\cdots + \alpha_nv_n = w$. Therefore,
$$\alpha_1v_1+\cdots+\alpha_nv_n - w = \mathbf{0}\text{ in }V.$$
Now, we can invoke the uniqueness clause. Since we can express $\mathbf{0}$ as 
$$0 = 0v_1+\cdots+0v_n + 0$$
it follows by the uniqueness clause that $w=0$ and that $\alpha_1=\cdots=\alpha_n=0$, proving that $v_1+W,\ldots,v_n+W$ is linearly independent. 

Aside: it is weird that they specifically say $v_1,\ldots,v_n$ are linearly independent, because the uniqueness clause implies linear independence, as noted above. 
