Trouble with problem #13 3.A Linear Algebra Done Right I am having an incredibly difficult time understanding this proof. I will break down my confusion on every step, and hopefully some of this confusion will clear up.
Problem: Suppose $v_1,...,v_m$ is a linearly dependent list of vectors in $V$. Suppose also that $W \neq \{0\}$. Prove there exist $w_1,...,w_m \in W$ such that no $T \in \mathcal{L}(V,W)$ satisfies $Tv_k=w_k$ for each $k=1,...,m$
Proof (not my attempt):

*

*There exist scalars $a_i$ not all $0$ such that $ \sum a_iv_i=0$.

*Suppose $a_k \neq 0$.

*Pick any $w_k \neq 0$ in $W$ and let $w_i=0$ for  $ i \neq k$.

*If there exists a linear map $T:V \to W$ such that $Tv_i=w_i$ for all $i$ then $0=T( \sum a_iv_i)= \sum a_iT(v_i)=a_kw_k$ which is a contradiction. Hence no such $T$ exists.

I have numbered all the sentences and will display my confusion.
My understanding of the problem: If $v_1,...,v_m$ are linearly dependent and $W \neq \{0\}$ we are supposed to show there exists some $w's$ such that no transformation exists:
we need to show there exist $w's \in W$ such that there is no transformation $T:V \rightarrow W$ such that $T(a_1v_1+\dots +a_mv_m)=a_1w_1+\dots +a_mw_m$
Now to the proof above.
Sentence 1.
This is the only part of the proof I understand. Since $v_1,...,v_m$ are linearly dependent there exist scalars $a_1,...,a_m$ not all zero such that $a_1v_1+\dots +a_mv_m=0$
Sentence 2.
"Suppose $a_k \neq 0$" To my understanding this means $a_k$ is a coefficient in this linearly combination that is nonzero. To my understanding $k \in \{1,...,m\}$
Sentence 3.
This is the part I get lost at.  "Pick any $w_k \neq 0$ in $W$ and let $w_i=0$ for  $ i \neq k$"
What does Pick any $w_k \neq 0$ in $W$ mean? To my understanding this means this is a nonzero vector that the transformation is supposed to map to.
In other words $T(a_1v_1+\dots + a_mv_m)=a_kw_k$
The next sentence states
Let $w_i=0$ for $i\neq k$ To my understanding this means all the vectors $w_i$ where $i \neq k$ are getting mapped to the zero vector.
Sentence 4.
This is where I become lost I have no idea what is going on here
"If there exists a linear map $T:V \to W$ such that $Tv_i=w_i$ for all $i$ then $0=T( \sum a_iv_i)= \sum a_iT(v_i)=a_kw_k$ which is a contradiction. Hence no such $T$ exists."
I have a very hard time understanding this. I understand that the proof is using the linearity of the transformation and applying this transformation to both sides of the linear dependence relation $0=a_1v_m+\dots +a_mv_m$ I do not understand why they are doing this. Also is the result that $T(a_1v_1+\dots +a_mv_m)=a_kw_k$ due to the fact that this($w_k$) was the only $w$ that was chosen, to get mapped to a nonzero vector? I understand that the contradiction comes from assuming $a_k,w_k \neq 0$ and getting them equal to $0$ in the final transformation applied to the linear dependence relation. Any help on this like breaking the proof down into something understandable would be much appreciated.I also think I am misunderstanding exactly what this question is asking.
 A: Maybe it will help to think about it this way: our goal is to find some vectors $w_1,w_2,\ldots,w_m$ that satisfy the desired conditions. Nobody claims that there's a unique way to find them. But as long as we can find something that works, we've achieved our goal.

What does "Pick any $w_k\neq0$ in $W$" mean?

Literally what it says: pick any nonzero vector from $W$ (which is possible since we're given that $W\neq\{0\}$), and give it any name we want — and we want to call it $w_k$. But the actual intention behind this statement is that, as you said yourself, we want this vector to be the image of $v_k$. The author of this proof didn't say that explicitly here, but in the end we're going to assume that $Tv_k=w_k$.

Let $w_i=0$ for $i\neq k$.

Kinda the same thing. We want to find some vectors $w_1,w_2,\ldots,w_m$. We've already decided what our $w_k$ is. Now we're making an executive decision that the rest of them are going to be zero.

To my understanding this means all the vectors $w_i$ where $i\neq k$ are getting mapped to the zero vector.

Yes, you're absolutely right! Once again, that's the assumed intention here.

"If there exists a linear map $T:V\to W$ such that $Tv_i=w_i$ for all $i$ then $0=T\left(\sum a_iv_i\right)=\sum a_iT(v_i)=a_kw_k$ which is a contradiction. Hence no such $T$ exists." … I do not understand why they are doing this.

To prove the desired claim by contradiction. We picked certain vectors $w_1,w_2\ldots,w_m$ that we believe satisfy what we want, so let's demonstrate that they indeed do. The claim is that there's no $T$ with certain properties. Assume to the contrary that such $T$ exists. Then such $T$ would satisfy $Tv_k=w_k$ for all $k$. However, this would imlpy that
$$0=T(0)=T\left(\sum a_iv_i\right)=\sum a_iT(v_i)=a_kw_k\neq0,$$
which is a contradiction. So such $T$ cannot exist.
A: Sentence 2: yes, $k$ is any index in $\{1,\dots,m\}$ for which the coefficient in the dependency $\sum a_i v_i$ is non-zero. There can be more than one such $k$, but you pick and fix only one. 
Sentence 3 should be read as: pick $m$ vectors $w_1$, $\dots$, $w_m$ in $W$ such that $w_k\neq 0$ and all others are 
$0$. This is allowed under the assumptions of the problem since $W\neq\{0\}$ and the $w_i$'s are not assumed to be distinct.
It is written like that to make it clear how $T$ acts on the vectors $v_1,\dots,v_m$. This can be rewritten without indices as: let $w\neq 0$ (which exists because $W\neq\{0\}$) and suppose that there exists a linear operator $T$ such that $T(v_k)=w$ and $T(v_j)=0$ for $j\neq k$.
For Sentence 4 there is no reason other than to force a contradiction. In problems of the type "Prove, under some assumptions, that an object doesn't exist" it is a standart proof technique to assume that such an object does exist under those assumptions, and derive something absurd, like in this case that $0$ is equal to something non-zero.
A: In the first part of the proof you are just saying that some $v_k$ is a linear combination of the remaining $v_i$, $i\neq k$. This is because of linear dependence.
Therefore (linearity) $T(v_k)$ is a linear combination of the remaining $T(v_i)$.
So, in particular, if the remaining $T(v_i)$ are zero, then $T(v_k)$ must also be zero. This last sentence holds for any linear map $T$. Therefore there is no linear map that would map all the remaining $v_i$ to zero, but map $v_k$ to any nonzero vector $w_k$. That is your choice of $w$s.
A: We need to show that we can pick vectors $w_1,w_2,\dots,w_m$ from $W$ so that for any transformation $T: V \to W$, at least one of the $w$ vectors (which we'll call $w_k$, with corresponding vector $v_k$) has $Tv_k \neq w_k$
Now, since the $v$ vectors are dependent, we can write $$a_1 v_1 + a_2v_2 + \dots + a_mv_m = 0$$
where at least one of the $a$'s is nonzero. (This is Sentence 1)
Suppose that we say that a non-zero $a$ occurs at the $k$th index, i.e. $a_k \neq 0$. (This is Sentence 2) 
Remember, this $a_k$ corresponds to a $v_k$, and we need to find a $w_k$ corresponding to $v_k$ as well.
Choose $w_k$ to be any non-zero vector in $W$ we want, and set all of the other $w$'s to the zero vector ($w_i = 0$ for $i \neq k$). (This is Sentence 3)
Now, we know that $a_1 v_1 + a_2v_2 + \dots + a_m v_m = 0$ so any linear transformation $T$ must satisfy
$$0 = T(0) = T(a_1 v_1 + a_2v_2 + \dots + a_m v_m)=a_1T(v_1) + a_2T(v_2) + \dots + a_mT(v_m)$$
$$= a_1 w_1 + a_2 w_2 + \dots + a_m w_m$$
(This is the first part of Sentence 4)
Since we've chosen $w_k \neq 0$ and $w_i = 0$ for $i \neq k$ (Sentence 3) that means that nearly every term on the right disappears and we get:
$$0 = a_k w_k$$
Which is a contradiction, since by Sentence 2, $a_k \neq 0$, and by Sentence 3, $w_k \neq 0$, so $a_k w_k \neq 0$. (This is the end of Sentence 4, and the proof)
A: I really had to think about what this proposition said.  Here is a simple example.
Lets make  $V = W = \mathbb R$
Now we need a non-independent sent of real numbers... any two non-zero numbers will do.
$v_1 = 1, v_2 = 2$
The set of linear transformations on this vector space take on the form $T(v) = a v$
There exist real numbers $w_1,w_2$ such that there is no linear map such that $a = w_1, 2a = w_2$
Well that looks pretty obvious, we just have to choose $w_2 \ne 2w_1$
Can we generalize this to higher dimensional vector spaces?
We have $v_1,\cdots, v_m$ as a linearly dependent set of vectors.  There exists a set of scalars such that.
$a_1v_1 + \cdots + a_mv_m= 0$
We can rewrite this to say that there is a set of scalars such that
$a_1v_1 + \cdots + a_{m-1}v_{m-1}= v_m$
Suppose we have a transformation such that $T(v_k) = w_k$ for $k<m.$ 
Then $T(v_m) = a_1T(v_1) +\cdots + a_{m-1}T(v_{m-1}) = a_1w_1 + \cdots +a_{m-1}w_{m-1}$
And we have the freedom to choose $w_m\ne a_1w_1 + \cdots +a_{m-1}w_{m-1} $ 
