Is $Y$ necessarily a subspace of $V^n$? The question is given below:


My guess for the answer is yes.

My  trial for justification is:
I know that for $Y$ to be a subspace of $V^n$ it must satisfy that $\forall r \in F$ and $\forall u,v \in Y$, we must have that $ru + v \in Y$. And we are sure that this is true because all vectors in $Y$ are linearly dependent.
Am I correct?
 A: $\newcommand{\o}{\mathbf 0}$(Note : $\dim V = \infty$ if $V$ is infinite dimensional)
Hint 1 : If $2 \leq n \leq \dim V$, then let $\{e_1,...e_n\}$ be a linearly independent subset of $V$. Consider $(\o,\o , ... , e_i, ...,\o) = f_i \in V^n$, where $\o$ denotes the zero vector.


*

*Are each of the $f_i \in Y$? Why/Why not?

*Is $\sum_{i=1}^n f_i \in Y$? Why/Why not?

*Can you conclude if $Y$ is a subspace or not from the above?
Hint 2 : If $n > \dim V$, then why is $Y=V^n$?

I think you have not understood the question properly.
So $V$ is a vector space. You want me to assume it is finite dimensional, so I will assume that. We call the elements of $V$ as vectors.
Now, what is $V^n$? Note that $V^n$, is the set of all $n$-tuples of vectors in $V$. That is, if say $v_1,v_2,...,v_n$ are vectors in $V$, then the element $(v_1,v_2,...,v_n)$ is an element of $V^n$. 
For example, if $\o$ denotes the zero vector in 
$V$, then $(\o,\o,...,\o)$ is an element of $V^n$, where all the $n$ entries are $\o$.
Note that $V^n$ is a vector space, under component wise addition and scalar multiplication. Also, there is no relation between $n$ and the dimension of $V$ for now.
What is $Y$? $Y$ is a subset of $V^n$ which is defined as follows : take an element of $V^n$. As I mentioned before, it has $n$ entries, which are all vectors, say $v_1,...,v_n$ (they need not be distinct, remember that).
Now, take the set ${v_1,...,v_n}$. This is either linearly independent or linearly dependent in $V$. If it is linearly dependent in $V$, then $(v_1,...,v_n) \in Y$. Otherwise it does not belong in $Y$.
Let us take an example. Is $(\o,\o,...,\o) \in Y$? Yes, because the elements are linearly independent : see, we need a non-zero linear combination of $\o,\o,...,\o$ to be zero. Non-zero means at least one of the coefficients must be non-zero. But this can be done easily , because :
$$
1\o + 0\o + ... + 0\o = \o
$$
which is a non-zero linear combination of the elements of the tuple and equals zero. Therefore, $(\o,\o,...,\o) \in Y$.
Consider the element $(\o,v_2,...,v_n) \in V^n$. I claim it is in $Y$. Why? Because :
$$
1 \o + 0v_2+...+0v_n = \o
$$
is a non-zero linear combination of the entries which equals zero. Therefore , $(\o,v_2,...,v_n) \in Y$ for all choices of $v_2,...,v_n$. 
From here, you should be able to see the following lemma :

Let $(v_1,v_2,...,v_n) \in V^n$. If at least one of the $v_i$ is the zero vector ,then $(v_1,v_2,...,v_n) \in Y$.

Why? Well, think what I did when the zero was in the first position. What non-zero linear combination would I take if the zero was in the $i$th position, for example?
With the lemma behind us, we now need to recall a common result :

Let $W$ be a subspace of dimension $m$. Then, any set of $m+1$ or more elements in $W$ is linearly dependent.

Now, let $n > \dim V$. Then, any element of $V^n$ contains at least $\dim V + 1$ entries, so must be linearly dependent. That is, $Y = V^n$, because I just showed that every element of $V^n$ must be in $Y$. So of course, in this case, $Y$ is a subspace.
Finally, what happens when $2 \leq n \leq \dim V$? Well, there we have the following result :

If $W$ is a vector space and $p \leq \dim W$ then there exists a subset $\{v_1,...,v_p\}$ of $W$ which is linearly independent.

With this I claim $Y$ is not a subspace. Why? We will show that $Y$ is not closed under addition : recall that closed under addition means (or can shown to be equivalent to) that for every $w_1,...,w_n \in Y$, we have that $w_1+...+w_n \in Y$. We will find vectors $w_1,...,w_n$ such that each one is in $Y$ but their sum is not in $Y$.
What we do for this is use the common result : since $n \leq \dim V$, we get a set $\{e_1,...,e_n\} \subset V$ which is linearly independent. This set has nothing to do with the standard basis vectors in $\mathbb R^n$, which are also often labelled $e_i$.
Now, we create the elements $f_i\in V^n$ ,for $1 \leq i \leq n$, as follows : $f_i$ needs to have $n$ vector entries. Let us take all entries except the $i$th entry as $\o$, and the $i$th entry as $e_i$. In other words, $f_i = (\o,\o,...,\o,\underbrace{e_i}_{\text{$i$th entry}},\o,...,\o)$. Convince yourself that $f_i$ is actually an element of $V^n$, and then from the lemma, an element of $Y$.
Now, what is $\sum_{i=1}^n f_i$? The addition is component wise, so we add up the first components of the $f_i$, then the second components of the $f_i$, and the corresponding answers become the components of $\sum_{i=1}^n f_i$, which we call $f \in Y^n$.
For example, the first entries of the $f_i$ are all $\o$, except that of $f_1$ whose first entry is $e_1$. The sum of all these is $e_1+\o+\o+...+\o = e_1$, so the first entry of $f$ is $e_1$.
Similarly, the second entries of the $f_i$ are all $\o$, expect for $f_2$ whose first entry is $e_2$. The sum of all these is $\o+e_2+...+\o = e_2$. Thus the second entry of $f$ is $e_2$.
Now, do the rest and convince yourself that $f = (e_1,e_2,...,e_n)$.
But then , when you take the entries of $f$ as a set , you get $\{e_1,...,e_n\}$ which is linearly independent by assumption. Therefore, $f \notin Y$.
Which shows that $Y$ is not a subspace.
(Note : Have you seen where I used $n > 1$ above? It was crucial).
