# Dimension of the span of a set of vectors

1. Oct 20, 2015

### MostlyHarmless

My linear algebra is a bit rusty.

Let $A=\{\bar{v}_1, \dots, \bar{v}_1\}$ be a set of vectors in $R^n$. Can dim(span$(A))=n$ without spanning $R^n$?

I guess I'm unclear on how to interpret the dimension of the span of a set of vectors.

2. Oct 20, 2015

### andrewkirk

The span of a set of vectors is a vector space. There cannot be a proper n-dimensional subspace of an n-dimensional vector space. Any n-dimensional subspace must be the whole thing.

That is one area where vector spaces differ from modules.

3. Oct 20, 2015

### Staff: Mentor

Are there n vectors in A? Your subscripts all appear to be 1. Presumably you meant $\{\bar{v}_1, \dots, \bar{v}_n\}$.

4. Oct 21, 2015

### MostlyHarmless

Yeah that second, 1 should have been an m. Not necessarily an n. (at least I don't think it should necessarily be an n).

5. Oct 21, 2015

### MostlyHarmless

Ok, this is exactly what I needed to know!

I've not gotten to modules yet. I'm taking a second graduate Algebra class next semester though. I'm told we will finally get into them then.

6. Oct 21, 2015

### Staff: Mentor

Then you should specify some condition on m. If m < n, then your set of vectors could not possibly span $\mathbb{R}^n$, since dim(span(A)) $\le m < n$. If $m \ge n$, the set of vectors might span $\mathbb{R}^n$, or might not.

7. Oct 21, 2015

### MostlyHarmless

The only condition I wanted on m is that it was not necessarily equal to n. So that it was truly an arbitrary set of vectors in $R^n$. That way the question was just: "Given an arbitrary set of vectors, A, such that dim(span(A)) = n. Does A necessarily span $R^n$?" This is a more concise way of asking the question.. I believe the two questions are equivalent though. If not, this is definitely the question I mean to ask.

Which the answer appears to be yes... as per andrewkirk.

Edit: I guess the condition that dim(span(A))=n forces the condition that $m\geq n$.

8. Oct 21, 2015

Yes