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In Exercises 1-6, determine if the set of points is affinely dependent. (See Practice Problem 2.) If so, construct an affine dependence relation for the points.

4.\(\left[ {\begin{aligned}{{}}{ - 2}\\5\\3\end{aligned}} \right],\left[ {\begin{aligned}{{}}0\\{ - 3}\\7\end{aligned}} \right],\left[ {\begin{aligned}{{}}1\\{ - 2}\\{ - 6}\end{aligned}} \right],\left[ {\begin{aligned}{{}}{ - 2}\\7\\{ - 3}\end{aligned}} \right]\)

Short Answer

Expert verified

The set of points are affinely dependent, and the relation is \( - 6{{\mathop{\rm v}\nolimits} _1} + 3{{\mathop{\rm v}\nolimits} _2} - 2{{\mathop{\rm v}\nolimits} _3} + 5{{\mathop{\rm v}\nolimits} _4} = 0\).

Step by step solution

01

Condition for 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 \({c_1},{c_2},...,{c_p}\) not all zero, and the sum must be zero \({c_1} + {c_2} + ... + {c_p} = 0\), and \({c_1}{{\bf{v}}_1} + {c_2}{{\bf{v}}_2} + ... + {c_p}{{\bf{v}}_p} = 0\).

02

Compute \({{\mathop{\rm v}\nolimits} _2} - {{\mathop{\rm v}\nolimits} _1}\), \({{\mathop{\rm v}\nolimits} _3} - {{\mathop{\rm v}\nolimits} _1}\) and \({{\mathop{\rm v}\nolimits} _4} - {{\mathop{\rm v}\nolimits} _1}\)

Let \({{\mathop{\rm v}\nolimits} _1} = \left( {\begin{aligned}{{}}{ - 2}\\5\\3\end{aligned}} \right),{{\mathop{\rm v}\nolimits} _2} = \left( {\begin{aligned}{{}}0\\{ - 3}\\7\end{aligned}} \right),{{\mathop{\rm v}\nolimits} _3} = \left( {\begin{aligned}{{}}1\\{ - 2}\\{ - 6}\end{aligned}} \right),{{\mathop{\rm v}\nolimits} _4} = \left( {\begin{aligned}{{}}{ - 2}\\7\\{ - 3}\end{aligned}} \right)\).

Compute \({{\mathop{\rm v}\nolimits} _2} - {{\mathop{\rm v}\nolimits} _1}\), \({{\mathop{\rm v}\nolimits} _3} - {{\mathop{\rm v}\nolimits} _1}\), and \({{\mathop{\rm v}\nolimits} _4} - {{\mathop{\rm v}\nolimits} _1}\) as shown below:

\({{\mathop{\rm v}\nolimits} _2} - {{\mathop{\rm v}\nolimits} _1} = \left( {\begin{aligned}{{}}2\\{ - 8}\\4\end{aligned}} \right)\),

\({{\mathop{\rm v}\nolimits} _3} - {{\mathop{\rm v}\nolimits} _1} = \left( {\begin{aligned}{{}}3\\{ - 7}\\{ - 9}\end{aligned}} \right)\),

\({{\mathop{\rm v}\nolimits} _4} - {{\mathop{\rm v}\nolimits} _1} = \left( {\begin{aligned}{{}}0\\2\\{ - 6}\end{aligned}} \right)\)

Write the augmented matrix as shown below:

\(\left( {\begin{aligned}{{}}{{{\mathop{\rm v}\nolimits} _2} - {{\mathop{\rm v}\nolimits} _1}}&{{{\mathop{\rm v}\nolimits} _3} - {{\mathop{\rm v}\nolimits} _1}}&{{{\mathop{\rm v}\nolimits} _4} - {{\mathop{\rm v}\nolimits} _1}}\end{aligned}} \right) \sim \left( {\begin{aligned}{{}}2&3&0\\{ - 8}&{ - 7}&2\\4&{ - 9}&{ - 6}\end{aligned}} \right)\)

03

Apply row operation

At row 1, multiply row 1 by \(\frac{1}{2}\).

\( \sim \left( {\begin{aligned}{{}}1&{\frac{3}{2}}&0\\{ - 8}&{ - 7}&2\\4&{ - 9}&{ - 6}\end{aligned}} \right)\)

At row 2, multiply row 1 by 8 and add it to row 2. At row 3, multiply row 1 by 4 and subtract it from row 3.

\( \sim \left( {\begin{aligned}{{}}1&{\frac{3}{2}}&0\\0&5&2\\0&{ - 15}&{ - 6}\end{aligned}} \right)\)

At row 2, multiply row 2 by \(\frac{1}{5}\).

\( \sim \left( {\begin{aligned}{{}}1&{\frac{3}{2}}&0\\0&1&{\frac{2}{5}}\\0&{ - 15}&{ - 6}\end{aligned}} \right)\)

At row 1, multiply row 2 by \(\frac{3}{2}\) and subtract it from row 1.

\( \sim \left( {\begin{aligned}{{}}1&0&{ - \frac{3}{5}}\\0&1&{\frac{2}{5}}\\0&{ - 15}&{ - 6}\end{aligned}} \right)\)

At row 3, multiply row 2 by \(15\) and add it to row 3.

\( \sim \left( {\begin{aligned}{{}}1&0&{ - \frac{3}{5}}\\0&1&{\frac{2}{5}}\\0&0&0\end{aligned}} \right)\)

The columns are linearly dependent because not every column is a pivot column. So, \({{\mathop{\rm v}\nolimits} _2} - {{\mathop{\rm v}\nolimits} _1},{\rm{ }}{{\mathop{\rm v}\nolimits} _3} - {{\mathop{\rm v}\nolimits} _1},\,\,{\mathop{\rm and}\nolimits} \,\,\,{{\mathop{\rm v}\nolimits} _4} - {{\mathop{\rm v}\nolimits} _1}\) are linearly dependent.

04

Determine whether the set of points is affinely dependent

Theorem 5states that an indexed set \(S = \left\{ {{{\mathop{\rm v}\nolimits} _1},...,{{\mathop{\rm v}\nolimits} _p}} \right\}\) in \({\mathbb{R}^n}\), with \(p \ge 2\), the following statement is equivalent. This means that either all the statements are true or all the statements are false.

  1. The set \(S\) isaffinely dependent.
  2. Each of the points in \(S\)is an affine combination of the other points in \(S\).
  3. In \({\mathbb{R}^n}\), the set \(\left\{ {{{\mathop{\rm v}\nolimits} _2} - {{\mathop{\rm v}\nolimits} _1},...,{{\mathop{\rm v}\nolimits} _p} - {{\mathop{\rm v}\nolimits} _1}} \right\}\)is linearly dependent.
  4. The set \(\left\{ {{{\bar v}_1},...,{{\bar v}_p}} \right\}\) of homogeneous forms in \({\mathbb{R}^{n + 1}}\) is linearly dependent.

By statement (c) in theorem 5, \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\) is affinely dependent. \({{\mathop{\rm v}\nolimits} _4} - {{\mathop{\rm v}\nolimits} _1}\) is a linear combination of \({{\mathop{\rm v}\nolimits} _3} - {{\mathop{\rm v}\nolimits} _1}\) which means that \({{\mathop{\rm v}\nolimits} _4} - {{\mathop{\rm v}\nolimits} _1}\) is in span\(\left\{ {{{\mathop{\rm v}\nolimits} _2} - {{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _3} - {{\mathop{\rm v}\nolimits} _1}} \right\}\). Row reduction of the matrix shows that:

\(\begin{aligned}{}{{\mathop{\rm v}\nolimits} _4} - {{\mathop{\rm v}\nolimits} _1} = - \frac{3}{5}\left( {{{\mathop{\rm v}\nolimits} _2} - {{\mathop{\rm v}\nolimits} _1}} \right) + \frac{2}{5}\left( {{{\mathop{\rm v}\nolimits} _3} - {{\mathop{\rm v}\nolimits} _1}} \right)\\{{\mathop{\rm v}\nolimits} _4} = - \frac{3}{5}{{\mathop{\rm v}\nolimits} _2} + \frac{2}{5}{{\mathop{\rm v}\nolimits} _3} + \frac{6}{5}{{\mathop{\rm v}\nolimits} _1}\\ - \frac{6}{5}{{\mathop{\rm v}\nolimits} _1} + \frac{3}{5}{{\mathop{\rm v}\nolimits} _2} - \frac{2}{5}{{\mathop{\rm v}\nolimits} _3} + {{\mathop{\rm v}\nolimits} _4} = 0\\ - 6{{\mathop{\rm v}\nolimits} _1} + 3{{\mathop{\rm v}\nolimits} _2} - 2{{\mathop{\rm v}\nolimits} _3} + 5{{\mathop{\rm v}\nolimits} _4} = 0\end{aligned}\)

Thus, the set of points are affinely dependent.

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