# The subspace $S ⊆ \mathbb{R}^n$ has linearly independent vectors $u_1,…u_k$. Show that any basis for $S$ must have at least $k$ vectors.

Let $S ⊆ \mathbb{R}^n$ be a subspace. Say that $u_1,.....u_k ∈ S$ are linearly independent vectors. Show that any basis for S must have at least k vectors.

"Say that $u_1,.....u_k ∈ S$ are linearly independent vectors." From what i understand, this means that the solution set of these set of vector is a trivial solution.

"Show that any basis for S must have at least k vectors." In order for S to be a basis two conditions must hold: 1. $S$ is linearly independent. (as stated) 2. $S$ spans $V$. (is the key to proving this question lies with this statement?)

I need help understanding the question and also proving the question....

• What is the matrix is not a square matrix, then the usage of determinant would not application for this question? – teo93 Nov 5 '14 at 8:36
• What you are being asked easily implies the dimension theorem (any two finite bases of the same space have the same number of vectors). If you are not already a bit beyond the level of the definitions, this does not seem a fair thing to ask (any proof of this important theorem is necessarily somewhat complicated, involving induction at the least). Unless you are supposed to know a number of things about dimension (but which depend on mentioned theorem). – Marc van Leeuwen Nov 5 '14 at 15:27

A basis is a l.i. generating set. The set $\{u_1,\cdots,u_k\}$ is a l.i. generating set of $\text{span}\{u_1,\cdots,u_k\}$, i.e. a basis. But $\text{span}\{u_1,\cdots,u_k\}$ is a subspace of $S$, so $$k=\dim\text{span}\{u_1,\cdots,u_k\}\le\dim S.$$