understand of span, linear independence and basis by using dimension Before start explaining what makes me confused, I'm sorry about my poor English. I'm not good at English. lol

If $V$ is Finite-dimensional vector space, let $\{v_1,v_2,\dots,v_n\}$ is arbitrary basis of $V$.


(a) The set of vectors that have more than $n$ is linearly dependent set.


(b) The set of vectors that have less than $n$ can't span $V$

I wanna know that (b) means if we have more than $n$ vectors when the basis of the V is $n$, we can just span V.
I'm confused because I'm not sure about the concept of span and linear combination exactly.

If $S=\{v_1,v_2,\dots,v_r\}$, $S'=\{w_1,w_2,\dots,w_k\}$ is a vector set that is included in vector space $V$, if and only if $span\{v_1,v_2,\dots,v_r\}=span\{w_1,w_2,\dots,w_k\}$ is that each vector of S is linear combination of $w_1,w_2,\dots,w_k$ and also each vector of $S'$ is linear combination of $v_1,v_2,\dots,v_r$.

I wanna know this sentence is right.
If there are **n**-times basis* *(I mean the number of the basis vector is n)* and n≦r, n≦k then S,S' can span vector space V and also S,S' is linearly dependent set.
 A: A set of $n$ vectors in an $n$ dimensional vector space $V$ needs not be a base for $V$. Think of $\Bbb R^2$ and $\{(1,0),(2,0)\}$. We need the vectors to be linearly independent to actually be a base of $V$.
If the vector space has dimension $n$ in any set of $r>n$ vectors there will be at most $n$ that are linearly independent, but the actual number of linearly independent vectors will be between $1$ and $n$. If there exist precisely $n$ linearly independent vectors, those will form a basis of $V$ and will span $V$. Any base of the vector space spans it.
A: On the matter of your first question (i.e., "Must a set of more than $n$ vectors span a vector space of dimension $n$?"), I believe that you are misinterpreting the logic. The statement (b) says that, "If $|S| < n,$ then the vectors of $S$ cannot span $V.$" Put another way, it says that, "If the vectors of $S$ span $V,$ then we must have that $|S| \geq n.$" This equivalence is due to the fact that the implication $P \implies Q$ is logically equivalent to the contrapositive $(\neg Q) \implies (\neg P).$ One can see this by typing (p=>q)<=>(!q=>!p) into this truth table generator from Stanford.
Unfortunately, it is not true that every set $S$ of at least $n$ vectors in an $n$-dimensional vector space $V$ must span $V.$ One of the silliest examples is to look at the set $S$ that consists of $m \geq n$ copies of the same vector $v.$ In this case, we have that
\begin{align*}
\operatorname{span} \{v, v, \dots, v\} &= \{c_1 v + c_2 v + \cdots + c_m v \,|\, c_1, c_2, \dots, c_m \text{ are scalars}\} \\ \\
&= \{(c_1 + c_2 + \cdots + c_m) v \,|\, c_1, c_2, \dots, c_m \text{ are scalars}\} \\ \\
&= \{cv \,|\, c \text{ is a scalar}\} \\ \\
&= \operatorname{span} \{v\}. \end{align*}
By definition, one vector cannot span an $n$-dimensional vector space (whenever $n > 1$).
