A subset of n vectors is linearly independent iff it spans V I am trying to solve #8 from this PDF: http://www.math.purdue.edu/~lai37/MA353/M353P1.pdf
The solutions are given in http://www.math.purdue.edu/~lai37/MA353/MP1soln.pdf
I am not really seeing the rational behind some of the arguments. For example, why must $n+1$ vectors in $V$ be linearly dependent if $\text{dim}(V) = n$? Does it just follow naturally from the definition of the dimension of a vector space and needs no further explanation?
Also, when it is trying to represent $0 \in V$, why is "this impossible"? What is impossible here? Why can't $a=0$?
Thank you
 A: Firstly, if a step in a proof is not obvious to you, even if it does follow naturally from definitions, it always needs further explanation. 
In this case, the statement that $n+1$ elements of a $n$-dimensional vector space are linearly dependent is not obvious at all unless you can assume the result of exercise 8. So I agree with you that the proof does not make much sense, it is just circular reasoning. 
To prove that $n+1$ vectors of an $n$-dimensional vector space are linearly dependent, apply induction on $n$. For $n=0$ and $n=1$ the statement is clearly true. Now assume $n > 1$, let $V$ be an $n$-dimensional vector space and let $a_1, \ldots, a_{n+1} \in V$ be given. We know that there exists a basis $b_1,\ldots,b_n$ of $V$. For each $1 \leq i \leq n+1$ we can write $a_i = c_{i1} b_1 + \ldots + c_{in} b_n$, for certain scalars $c_{ij}$. 
Now consider the $n-1$ dimensional vector space $V' = Span(b_1,\ldots,b_{n-1})$ and the $n+1$ elements $a'_1, \ldots, a'_{n+1}$ defined by $a'_i = a_i - c_{in} b_n$. By the induction hypothesis $a'_1, \ldots a'_n$ are linearly dependent over $V'$, and so there exist scalars $d_1, \ldots, d_n$ such that $d_1 a'_1 + \ldots + d_n a'_n = 0$. Similarly, the $n$ vectors $a'_2, \ldots, a'_{n+1}$ are linearly dependent over $V'$, and so there exist scalars $e_2, \ldots, e_{n+1}$ such that $e_2 a'_2 + \ldots + e_{n+1} a'_{n+1} = 0$.
Looking back in $V$, we see that $d_1 a_1 + \ldots + d_n a_n = D b_n$ for some scalar $D$, and $e_2 a_2 + \ldots + e_{n+1} a_{n+1} = E b_n$ for some scalar $E$. If $D=0$, we have shown linear dependence and we are done. If not, then we find the linear dependence $\frac{-E d_1}{D} a_1 + (\frac{- E d_2 }{D}+ e_2) a_2 + (\frac{- E d_3 }{D}+ e_3) a_3 + \ldots + (\frac{- E d_n }{D}+ e_n) a_n + e_{n+1} a_{n+1} = 0$.
As for your second question, if $a = 0$ then we can write $a_1 v_1 + \ldots + a_n v_n = 0$, which contradicts the assumption that $v_1, \ldots, v_n$ are linearly independent.
By the way, another strange thing in the proof I noted was the part where they prove that $0 \in Span(S)$. This is true because $0$ is equal to the empty sum, and it does not follow from the fact that $Span(S)$ is a subspace. You cannot show that $Span(S)$ is a subspace without explicitly showing that it contains $0$.
A: The dimension of $V$ or $\text{dim}(V)$ is the unique size of a linearly independent spanning set of vectors in $V$. As it turns out, the minimum size of a spanning set of vectors in $V$ and the maximum size of a linearly independent set of vectors in $V$ is the same which we'll call $\text{dim}(V)$. Furthermore, any set of $\text{dim}(V)$ linearly independent vectors also spans $V$ and any spanning set of size $\text{dim}(V)$ is linearly independent. (These are not definitions and can be proved.)
So to answer the first part of your question, $n + 1$ vectors in $V$ must be linearly dependent if $\text{dim}(V) = n$ because $n$ is the maximum size of any linearly independent set of vectors in $V$. Whenever you have a set of vectors whose size differs from the dimension of the vector space, the set cannot be both spanning and linearly independent. If the set is larger than the dimension of $V$, it cannot be linearly independent. And if the set is smaller than the dimension of $V$, it cannot be spanning.
For the second part of your question, $a$ cannot be $0$ because if it were, we could use the supposition that $S$ is linearly independent to prove the existence of a set of $n + 1$ vectors which is linearly independent in a vector space of dimension $n$.
