Define:
A basis for a vector space \(V\) is a linearly independent subset of \(V\) that spans \(V\).
Every basis for the same vector space \(V\) has the identical number of vectors, which equals to \(dim(V)\).
Proof:
Let \(V_0 = \{\boldsymbol{v_1}, ..., \boldsymbol{v_n}\}\) and \(W = \{\boldsymbol{w_1}, ..., \boldsymbol{w_m}\}\) are both basis of \(V\).
(i)
Assume \(n > m\). Since \(\{\boldsymbol{v_1}, ..., \boldsymbol{v_n}\}\) is a basis for \(V\), every element in \(V\) can be expressed as a linear combination of \(\{\boldsymbol{v_1}, ..., \boldsymbol{v_n}\}\). Particularly, let
\[\boldsymbol{w_i} = a_{1i}\boldsymbol{v_1}+ ... + a_{mi}\boldsymbol{v_m}, \forall\ 1 \leq i \leq m\], i.e.
\[\begin{bmatrix}\boldsymbol{w_1} &...& \boldsymbol{w_m}\end{bmatrix} = \begin{bmatrix}\boldsymbol{v_1} &...& \boldsymbol{v_n}\end{bmatrix}\begin{bmatrix}\boldsymbol{a_{11}} &...& \boldsymbol{a_{1n}} \\ ... \\ \boldsymbol{a_{m1}} &...& \boldsymbol{a_{mn}} \end{bmatrix}\], which means:
\[W = V_0A\]Here \(A\) is a \(m \times n\) matrix, and thus \(N(A) \neq \{\boldsymbol{0}\}\), for \(n > m\) (column 數大於 row 數).
\(\Rightarrow\) There exists non-zero \(\boldsymbol{x} \in \mathbb{R}^n\) s.t. \(A\boldsymbol{x} = \boldsymbol{0} \Rightarrow V_0A\boldsymbol{x} = \boldsymbol{0} \Rightarrow W\boldsymbol{x} = \boldsymbol{0}\)
\(\therefore N(W) \neq \{\boldsymbol{0}\}\), but this contradicts the fact that \(\{\boldsymbol{w_1}, ..., \boldsymbol{w_m}\}\) is a basis. (Vectors in a basis must be linearly independent.)
(ii)
Similarly, we can prove that \(n < m\) is impossible either.
\(\therefore n = m\) ∎
\(dim(V) =\) number of vectors in any basis for \(V\) (basis 的 vector 數量) (詳見 [Linear Transformation])