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Q11E

Expert-verifiedFound in: Page 437

Book edition
5th

Author(s)
David C. Lay, Steven R. Lay and Judi J. McDonald

Pages
483 pages

ISBN
978-03219822384

**Explain why any set of five or more points in \({\mathbb{R}^3}\) must be affinely dependent.**

Any set of five or more points in \({\mathbb{R}^3}\) must be affinely dependent.

The set is said to be affinely dependent if the set \(\left\{ {{{\bf{v}}_{\bf{1}}},{{\bf{v}}_{\bf{2}}},...,{{\bf{v}}_p}} \right\}\) in the dimension\({\mathbb{R}^n}\) exists such that for non-zero scalars\({c_1},{c_2},...,{c_p}\), the sum of scalars is zero i.e.\({c_1} + {c_2} + ... + {c_p} = 0\), and \({c_1}{{\bf{v}}_1} + {c_2}{{\bf{v}}_2} + ... + {c_p}{{\bf{v}}_p} = 0\).

Consider the set of five points \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3},{{\bf{v}}_4},{{\bf{v}}_5}} \right\}\).

Let \(\left\{ {{{\bf{v}}_2} - {{\bf{v}}_1},{{\bf{v}}_3} - {{\bf{v}}_1},{{\bf{v}}_4} - {{\bf{v}}_1},{{\bf{v}}_5} - {{\bf{v}}_1}} \right\}\) be the set of vectors in \({\mathbb{R}^3}\).This new set of four points must be linearly dependent when a set of vectors is translated by eliminating the first point or any other point.

From \({\mathbb{R}^3}\), the number of entries is 3. From the above-considered set of points, the number of vectors is 4. That is, \(n = 3\), and \(p = 4\).

Here, the set contains more vectors or points than the number of entries. So,\(p > n\)and the original set of five points \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3},{{\bf{v}}_4},{{\bf{v}}_5}} \right\}\) is affinely dependent.

Therefore, any set of five or more points in \({\mathbb{R}^3}\) must be affinely dependent.

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