Question

Prove Theorem 6 for an affinely independent set\(S = \left\{ {{v_1},...,{v_k}} \right\}\)in\({\mathbb{R}^{\bf{n}}}\). [Hint:One method is to mimic the proof of Theorem 7 in Section 4.4.]

Short answer

Theorem 6 is proved for an affinely independent set \(S = \left\{ {{v_1},...,{v_k}} \right\}\) in \({\mathbb{R}^n}\).

Step-by-step solution

Prove theorem 6

Consider the set \(S = \left\{ {{{\bf{v}}_{\bf{1}}},{{\bf{v}}_{\bf{2}}},...,{{\bf{v}}_k}} \right\}\)in\({\mathbb{R}^n}\). Let\({c_1},{c_2},...,{c_k}\), and\({d_1},{d_2},...,{d_k}\)be the scalars such that\({c_1}{{\bf{v}}_1} + {c_2}{{\bf{v}}_2} + ... + {c_k}{{\bf{v}}_k} = {\bf{p}}\), and\({d_1}{{\bf{v}}_1} + {d_2}{{\bf{v}}_2} + ... + {d_k}{{\bf{v}}_k} = {\bf{p}}\), where\({d_1} + {d_2} + ... + {d_k} = 1\).

The systems of equations is shown below:

\({\bf{p}} = {c_1}{{\bf{v}}_1} + {c_2}{{\bf{v}}_2} + ... + {c_k}{{\bf{v}}_k}\)

\({\bf{p}} = {d_1}{{\bf{v}}_1} + {d_2}{{\bf{v}}_2} + ... + {d_k}{{\bf{v}}_k}\)

Prove Theorem 6

Subtract the equations\({\bf{p}} = {c_1}{{\bf{v}}_1} + {c_2}{{\bf{v}}_2} + ... + {c_k}{{\bf{v}}_k}\)and\({\bf{p}} = {d_1}{{\bf{v}}_1} + {d_2}{{\bf{v}}_2} + ... + {d_k}{{\bf{v}}_k}\)as shown below:

\(\begin{array}{l}\left( {{c_1} - {d_1}} \right){{\bf{v}}_1} + \cdots + \left( {{c_k} - {d_k}} \right){{\bf{v}}_k} = {\bf{p}} - {\bf{p}}\\\left( {{c_1} - {d_1}} \right){{\bf{v}}_1} + \cdots + \left( {{c_k} - {d_k}} \right){{\bf{v}}_k} = 0\end{array}\)

It is given that\(S = \left\{ {{{\bf{v}}_{\bf{1}}},{{\bf{v}}_{\bf{2}}},...,{{\bf{v}}_k}} \right\}\)is an independent set. Thus, all the coefficients must be 0 as shown below:

\(\begin{array}{c}{c_1} - {d_1} = 0\\{c_1} = {d_1}\end{array}\)

\(\begin{array}{c}{c_2} - {d_2} = 0\\{c_2} = {d_2}\end{array}\)

And

\(\begin{array}{c}{c_k} - {d_k} = 0\\{c_k} = {d_k}\end{array}\)

Since \({c_1} = {d_1},{c_2} = {d_2},...,{c_k} = {d_k}\), the set of scalars is unique.

Hence, Theorem 6 is proved for an affinely independent set \(S = \left\{ {{v_1},...,{v_k}} \right\}\) in \({\mathbb{R}^n}\).