# Relation between number of components in vectors and the dimension of their span.

We have $$n$$ vectors each with $$k$$ components. Let $$V$$ be their span and $$r$$ be the dimension of $$V$$. Then $$r \leq k$$.

I encountered this in a paper I am reading. I am not able to see why this is true(probably due to my not so sound linear algebra background).

Can anyone help me in understanding why the argument is true? Thanks.

• A spanning set can be reduced to a basis, so the dimension is less than or equal to the number of vectors you started out with. May 20, 2020 at 9:17

Let $$\Bbb F$$ be the underlying field. A vector with $$k$$ components is a vector in $$\Bbb F^k$$, which is a vector space of dimension $$k$$. If $$V$$ is the span of $$n$$ vectors in $$\Bbb F^k$$, then it is a subspace of $$\Bbb F^k$$: in particular, its dimension is less or equal to the dimension of $$\Bbb F^k$$, that is, $$k$$.
Denote the $$n$$ vectors by $$v_1, \cdots, v_n \in W$$ (where $$W$$ is the vector space you're working with). To say that each vector has $$k$$ components is the same as saying that $$W$$ is $$k$$-dimensional, that is, that there exist $$k$$ linearly independent vectors $$e_1, \cdots, e_k$$ such that their span is equal to $$W$$ and that no combination of less than $$k$$ vectors can possibly span $$W$$ (and of course, also that any combination of more than $$k$$ vectors is linearly dependent). So the result you want follows by the very definition of dimension and basis.