# Proof that the all bases of a subspace V consist of the same number of vectors?

My textbook explains the proof, which I don't understand:

"Consider two bases $v_1...v_p$ and $w_1...w_q$ of V. Since the vectors $v$ are linearly independent and the vectors $w$ span V..."

How exactly does $w$ span V?

The book then says the same for vectors $v$, that $v$ spans V and hence $p=q$, but I don't really understand how you can assume that given two sets of vectors that are basis, one set must span V.

• What is the definition of a basis? – Rodrigo Dias Jan 27 '18 at 19:57
• @rldias the first thing that comes to mind is: a set of n linearly independent vectors in the space $R^n$? Is that right? – Goldname Jan 27 '18 at 19:58
• It works only for $n$ dimentional vector spaces.. see the definition here en.m.wikipedia.org/wiki/Basis_(linear_algebra) – Rodrigo Dias Jan 27 '18 at 20:00
• @rldias Ok I just reread the definition in the textbook. Basically, vectors form a basis of V if they span V and are linearly independent. However, I still don't understand the proof – Goldname Jan 27 '18 at 20:01
• I think we proved it by showing that that the rank of a set of vectors (the number of linear independent vectors) doesn't change when transforming them like in matrix transformations, then argued that the rank of the rows is equal to the rank of the columns and then argued that thus the dimension of the vector space is unique, since you can map basis on to each other linearly. I would be interested how you did it – Felix B. Jan 27 '18 at 20:07

A subset $B$ of a vector space $V$ is a basis for $V$ if it spans $V$ and is linearly independent.
If $\{v_i\}_1^p$ and $\{w_k\}_1^q$ are bases of $V$, then:
1. Both $\{v_i\}$ and $\{w_k\}$ span $V$,
2. Both $\{v_i\}$ and $\{w_k\}$ are linearly independent sets.
By definition, a basis for a (say real) vector space $V$ consists of a family of vectors $(v_1,\ldots,v_q)$ that are $\mathbb{R}$-linearly independant and that span the vector space V. The fact that they span $V$ means any vector $v\in V$ can be written as a linear sum of elements of the basis, which means there exists reals $a_1,\ldots,a_q$ scalars such that $v=a_1v_1+\ldots + a_qv_q$.