Question

The conditions for affine dependence are stronger than those for linear dependence, so an affinely dependent set is automatically linearly dependent. Also, a linearly independent set cannot be affinely dependent and therefore must be affinely independent. Construct two linearly dependent indexed sets\({S_{\bf{1}}}\)and\({S_{\bf{2}}}\)in\({\mathbb{R}^2}\)such that\({S_{\bf{1}}}\)is affinely dependent and\({S_{\bf{2}}}\)is affinely independent. In each case, the set should contain either one, two, or three nonzero points.

Short answer

The indexed set \({S_1}\)is affinelyand linearly dependent.

The indexed set \({S_2}\) is affinelyand linearly independent.

Step-by-step solution

State the condition for affinely dependence

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 \({c_1}{{\bf{v}}_1} + {c_2}{{\bf{v}}_2} + ... + {c_p}{{\bf{v}}_p} = 0\).

Show affinely dependence

Let\({S_1} = \left\{ {{{\bf{v}}_{\bf{1}}},{{\bf{v}}_{\bf{2}}},{{\bf{v}}_3}} \right\}\)be the set of vectors, where\({{\bf{v}}_1} = \left[ {\begin{aligned}{{}{}}3\\{ - 3}\end{aligned}} \right]\),\({{\bf{v}}_2} = \left[ {\begin{aligned}{{}{}}0\\6\end{aligned}} \right]\), and\({{\bf{v}}_3} = \left[ {\begin{aligned}{{}{}}2\\0\end{aligned}} \right]\).

Let the newpoints, \({{\bf{v}}_3} - {{\bf{v}}_1}\)and \({{\bf{v}}_2} - {{\bf{v}}_1}\), be obtained by eliminating the first point.

\(\begin{aligned}{}{{\bf{v}}_3} - {{\bf{v}}_1} &= \left[ {\begin{aligned}{{}{}}2\\0\end{aligned}} \right] - \left[ {\begin{aligned}{{}{}}3\\{ - 3}\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{{}{}}{ - 1}\\3\end{aligned}} \right]\end{aligned}\)

And

\(\begin{aligned}{}{{\bf{v}}_2} - {{\bf{v}}_1} &= \left[ {\begin{aligned}{{}{}}0\\6\end{aligned}} \right] - \left[ {\begin{aligned}{{}{}}3\\{ - 3}\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{{}{}}{ - 3}\\9\end{aligned}} \right]\end{aligned}\)

So, \({{\bf{v}}_2} - {{\bf{v}}_1} = 3\left( {{{\bf{v}}_3} - {{\bf{v}}_1}} \right)\).

Simplify the equation \({{\bf{v}}_2} - {{\bf{v}}_1} = 3\left( {{{\bf{v}}_3} - {{\bf{v}}_1}} \right)\).

\[\begin{aligned}{}{{\bf{v}}_2} - {{\bf{v}}_1} &= 3{{\bf{v}}_3} - 3{{\bf{v}}_1}\\2{{\bf{v}}_1} + {{\bf{v}}_2} - 3{{\bf{v}}_3} &= 0\end{aligned}\]

If the vectors are affinely dependent, the sum of weightsis 0. In the equation \(3{{\bf{v}}_1} + {{\bf{v}}_2} - 3{{\bf{v}}_3} = 0\), the weights are 2, 1, and \( - 3\). So,

\(2 + 1 - 3 = 0\).

Thus, the indexed set \({S_1}\) is affinelyand linearly dependent.

 Step 3: Show affinely independence

Let\({S_2} = \left\{ {{{\bf{v}}_{\bf{1}}},{{\bf{v}}_{\bf{2}}}} \right\}\)be the set of vectors, where\({{\bf{v}}_1} = \left[ {\begin{aligned}{{}{}}1\\2\end{aligned}} \right]\)and\({{\bf{v}}_2} = \left[ {\begin{aligned}{{}{}}2\\4\end{aligned}} \right]\).

So, \({{\bf{v}}_2} = 2{{\bf{v}}_1}\).

Simplify the equation \({{\bf{v}}_2} = 2{{\bf{v}}_1}\).

\({{\bf{v}}_2} - 2{{\bf{v}}_1} = 0\)

For thevectors to be affinely dependent, the sum of weights must be 0. In the equation \({{\bf{v}}_2} - 2{{\bf{v}}_1} = 0\), the weights are1 and \( - 2\). So,

\(\begin{aligned}{}1 - 2 &= - 1\\ &\ne 0\end{aligned}\).

Thus, the indexed set \({S_2}\) is affinelyand linearly independent.