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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 ?

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In addition to that you need enough vectors in your set! For example the set $\{(1,1,0)\}$ of vectors of $\mathbf{R}^3$ is linearly independent, but only generates a 1-dimensional subspace. If your space is finite dimensional, then it suffices to check that, in addition to linear independence, the number of vectors in your set equals the dimension of the space. In an infinite dimensional space you need to show that any vector is a linear combination of vectors in your linearly independent set. – Jyrki Lahtonen Jun 21 '11 at 5:57

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 $\mathbb{R}^2$).


  1. 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).

  2. 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.

  3. 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.

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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.

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Provided, of course, that $k$ is finite. – Gerry Myerson Jun 21 '11 at 6:07
@Gerry: Good catch, thanks! – Zev Chonoles Jun 21 '11 at 6:11

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.)

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This of course presumes that the vector space in question has a concept of orthogonality attached to it. Not every vector space comes equipped with this useful gadget. – Gerry Myerson Jun 21 '11 at 6:06
@Gerry: Ah! True. I shall edit it accordingly. – user17762 Jun 21 '11 at 6:07
@All: Yeah that is my doubt. Supposing that a set has "enough linearly independent vectors " , then can we say we can "span" the entire vector space with linear combination of this set of vectors . – Amit Jun 21 '11 at 6:12
@Amit: What do you mean by "enough"? If you are talking of a finite dimensional vector space, then $n$ linearly independent vectors would span the vector space. For an infinite dimensional vector space, things get a bit trickier. – user17762 Jun 21 '11 at 6:15
Yeah I agree, i am talking about finite dimensional vector space. I thought that for a set of vectors to span the vector space, they need to be "Basis" of that vector space. – Amit Jun 21 '11 at 6:30

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: $$ \hbox{${v_i}$ span $V$}\quad\iff\quad \hbox{the matrix $(a_{ij})$ has rank $n$}. $$

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