I have a basic doubt. Can we say that a set of vectors span the entire vector space iff they are linearly independent ? Do they need to satisfy any other property ?
Let $V$ be a vector space.
There are two reasons why your proposed characterization fails:
(i) Not every linearly independent set spans the vector space; so the “if” clause fails; and
(ii) Not every set of vectors that spans $V$ is linearly independent; so the “only if” clause fails.
So, unfortunately, you get caught both coming and going…
For an example of (i), take $\{(1,0)\}$ in $\mathbb{R}^2$; for an example of (ii), take all of $V$ (or $\{(1,0), (0,1), (1,1)\}$ in $\mathbb{R}^2$).
However:
If $V$ is finite dimensional, with $\dim(V)=n$, then a set of $n$ vectors of $V$ spans $V$ if and only if they are linearly independent (this may be what you were aiming for or trying to recall).
More generally, if $\beta$ is a linearly independent set of vectors which is maximal among linearly independent subsets (that is, if we add a vector to $\beta$ that is not already in $\beta$, the resulting set is linearly dependent), then $\beta$ is a basis. So a linearly independent set of vectors span $V$ if and only if no strictly larger set of vectors is linearly independent.
Dually, if $\beta$ is a set that spans $V$ and is such that no proper subset of $\beta$ spans $V$, then $\beta$ is linearly independent. So a spanning set for $V$ is linearly independent if and only if no strictly smaller set of vectors spans $V$.
2 and 3 both hold even in the case where $V$ is infinite dimensional.
This is clearly false. If we take fewer than $\dim(V)$ vectors in $V$, they can’t span $V$ regardless of whether they are linearly independent or not. In fact the empty set $\{\,\,\,\,\,\,\}\subset V$ is technically linearly independent, but it won’t span the vector space unless the vector space is zero-dimensional.
However, if we have a finite-dimensional vector space $V$, then given a set $\{v_1,\ldots,v_k\}$ of $k=\dim(V)$ vectors, then they will span if and only if they are linearly independent.
Linear independence doesn’t mean that they span the entire vector space. For instance, $(1,0,0)$ and $(0,0,1)$ are linearly independent but they do not span $\mathbb{R}^3$. Also, $(1,0,0),(0,1,0),(0,0,1),(2,3,5)$ are not linearly independent but they span $\mathbb{R}^3$. A set of vectors span the entire vector space iff the only vector orthogonal to all of them is the zero vector. (As Gerry points out, the last statement is true only if we have an inner product on the vector space.)
Let $V$ be a vector space. Vectors $\{v_i\}$ are called generators of $V$ if they span $V$. A set of linearly independent generators is called a basis of $V$. The important feature of a basis is that every vector in $V$ can be written uniquely as a linear combination of the basis vectors.
An important basic result in the theory of vector spaces says that a set of generators of a vector space $V$ always contains a basis as a subset.
Thus, we may conclude that a set of vectors spans $V$ if and only if contains a basis.
When the vectors $v_i$ are given in terms of a previously fixed basis $\{e_j\}_{j=1,\ldots,n}$, namely
$$
v_i=\sum_{j=1}^na_{ij}e_j,
$$
we thus get the following possibly useful criterion:
$$
\hbox{${v_i}$ span $V$}\quad\iff\quad
\hbox{the matrix $(a_{ij})$ has rank $n$}.
$$