# If $\dim(V) = n$, is every spaning set $\{v_1,v_2,\ldots,v_n\}$ a basis for $V$?

Okay, so I need help clearing things up.

Let $V$ be a vector space and $dim(V)=n$.

Does it mean that every Spanning set $\{ v_1,v_2,v_3,\ldots,v_n \}$ is necessarily a basis for V?

What if $\{ v_1,\ldots,v_n\}$ is linearly dependent? It is still a spanning set, right? And there's no way its a basis for $V$, right?

• There is no linearly dependent spanning set of $n$ vectors in a vector space of dimension $n$. – Christoph Mar 3 '15 at 13:22

A set is called a basis for $V$ if and only if the vectors in that set:

• are linearly independent,
• span $V$.

If $\dim V = n$, then any set of $n$ linearly independent vectors will span $V$, and thus be a basis for $V$.

If a set of $n$ vectors is linearly dependent, then it cannot span an $n$-dimensional space; it can't be a basis.

Addendum: you must be careful with conclusions only based on the number of vectors. If $\dim V = n$, then any set containing $m$ vectors with $m>n$ is definitely linearly dependent; it may or may not span $V$. Any set containing $k$ vectors with $k<n$ will definitely not span $V$; it may or may not be linearly (in)dependent.

• Indeed we can say that a basis is a maximal set of linearly independent vectors. The cardinality of this set is the dimension of the space. With this definition, you can prove by Zorn lemma that every vector space admits a basis. – Oscar Mar 3 '15 at 13:17

If the set is linearly dependent, then they can span at most an $n-1$-dimensional space.
At least one vector can be written in terms of the others, so a linear combination of the $n$ vectors can be written as a linear combination of the other $n-1$ vectors.
You are right, they are not a basis because they are linearly dependent.

Assume that a spanning set $S$ for a vector space of dimension $n$ is not linearly independent.

Then we can write at least one of the vectors in $S$ as a linear combination of the others:

$$v_j=\sum_i \lambda_iv_i.$$

Now, if $S$ is a spanning set, every element of $u\in V$ can be written as a linear combination of elements from $S$

$$u=\sum_i a_i v_i,$$ but rewriting $v_j$ in terms of the other $v_i$ shows that every vector $u\in V$ can be written as a linear combination of $n-1$ vectors:

$$u=\sum_{i\neq j}a_i'v_i.$$

Hence there is a basis of $V$ size less than or equal to $n-1$. This contradicts $\dim V=n$ and so this spanning set must be linearly independent.