[Engineering Mathematics] 7. Basis

Definition. A subset \(\mathfrak{B}\) of a vector space \(V\) is basis for \(V\) if \(\mathfrak{B}\) spans \(V\) and \(\mathfrak{B}\) is linearly independent.

 

Definition. The coordinate vector of \(\boldsymbol{v}\) with respect to \(\mathfrak{B}=\{\boldsymbol{v}_1, \boldsymbol{v}_2..., \boldsymbol{v}_n\}\) is the vector \([\boldsymbol{v}]_\mathfrak{B}=(c_1, c_2, ..., c_n)^\top \in \mathbb{R}^n\) such that \(c_1\boldsymbol{v}_1+...+c_n\boldsymbol{v}_n=\boldsymbol{v}\).

 

Theorem 1. Let \(\mathfrak{B}\) be a basis for a vector space \(V\). Then \([\boldsymbol{v}]_\mathfrak{B}\) is unique.

 

Proof. 

 Using the proof by contradiction, let \((c_1, ..., c_n)^\top \neq (d_1, ..., d_n)^\top\) and \(c_1\boldsymbol{v}_1+...+c_n\boldsymbol{v}_n=\boldsymbol{v}=d_1\boldsymbol{v}_1+...+d_n\boldsymbol{v}_n\). Then \((c_1-d_1)\boldsymbol{v}_1+...+(c_n-d_n)\boldsymbol{v}_n=\boldsymbol{v}-\boldsymbol{v}=\boldsymbol{0}\). Since a basis is linearly independent, \(c_1-d_1=...=c_n-d_n=0\), that is, \(c_i=d_i\) for any \(1 \leq i \leq n\). It is a contradiction to the assumption. Therefore, \([\boldsymbol{v}]_\mathfrak{B}\) is unique.

\(\square\)

 

Theorem 2. Let \(\mathfrak{B}\) be a basis for a vector space \(V\). Then,

 1) \([\boldsymbol{u}+\boldsymbol{v}]_\mathfrak{B}= [\boldsymbol{u}]_\mathfrak{B}+ [\boldsymbol{v}]_\mathfrak{B}\)

 2) \( [c\boldsymbol{u}]_\mathfrak{B}= c[\boldsymbol{u}]_\mathfrak{B}\)

 

Proof.

 Let \(\boldsymbol{u}=c_1\boldsymbol{v}_1+...+c_n\boldsymbol{v}_n\) and \(\boldsymbol{v}=d_1\boldsymbol{v}_1+...+d_n\boldsymbol{v}_n\).

 1) \(\boldsymbol{u}+\boldsymbol{v}=(c_1+d_1)\boldsymbol{v}_1+...+(c_n+d_n)\boldsymbol{v}_n\). By the uniqueness of the coordinate vector, \([\boldsymbol{u}+\boldsymbol{v}]_\mathfrak{B}=(c_1+d_1, ..., c_n+d_n)^\top=(c_1, ..., c_n)^\top+(d_1, ..., d_n)^\top=[\boldsymbol{u}]_\mathfrak{B}+[\boldsymbol{v}]_\mathfrak{B}\).

 2) \(c\boldsymbol{u}=(cc_1)\boldsymbol{v}_1+...+(cc_n)\boldsymbol{v}_n\). By the uniqueness of the coordinate vector, \([c\boldsymbol{u}]_\mathfrak{B}=(cc_1, ..., cc_n)^\top=c(c_1, ..., c_n)^\top=c[\boldsymbol{u}]_\mathfrak{B}\).

\(\square\)