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

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.