Indicate which pairs of the following vectors are linearly independent or linearly dependent.
- $$\vec{u}=(0,2)$$, $$\vec{v}=(1,1)$$
- $$\vec{u}=(2,-2)$$, $$\vec{v}=(1,-1)$$
- $$\vec{u}=(1,-3)$$, $$\vec{v}=(-3,9)$$
Development:
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The vectors $$\vec{u}=(0,2)$$, $$\vec{v}=(1,1)$$ are linearly independent because they do not have the same direction and their coordinates are not proportional $$$\dfrac{2}{1}\neq\dfrac{0}{-2}$$$ Another way of verifying that they are linearly independent is looking to see if any linear combination of these vectors equal to zero implies that the scalars will both be zero: $$$\lambda(2,0)+\mu(1,-2)=(2\lambda+\mu,-2\mu)=(0,0)$$$ $$$\left. \begin{array}{r} 2\lambda+\mu=0 \\ -2\mu=0 \end{array} \right\} \Rightarrow \mu=0, \ \lambda=0$$$
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The vectors $$\vec{u}=(2,-2)$$, $$\vec{v}=(1,-1)$$ are linearly dependent because their coordinates are proportional: $$$2=\dfrac{2}{1}=\dfrac{-2}{-1}=2$$$
- Veiem que en aquest cas the vectors $$\vec{u}=(1,-3)$$, $$\vec{v}=(-3,9)$$ are linearly dependent since: $$$\dfrac{1}{-3}=\dfrac{-3}{9} \Rightarrow -\dfrac{1}{3}=-\dfrac{1}{3}$$$ Another way of verifying that they are linearly dependent is finding the scalar values different from zero in such a way that $$a\vec{u}+b\vec{v}=\vec{0}$$, so: $$$a(1,-3)+b(-3,9)=(a,-3a)+(-3b,9b)=(a-3b,-3a+9b)=(0,0)$$$ $$$\left. \begin{array}{r} a-3b=0 \\ -3a+9b=0 \end{array} \right\} \Rightarrow a=3b \Rightarrow -3(3b)+9b=0 \Rightarrow -9b+9b=0$$$ the second equation does not yield an answer, so it is necessary only to satisfy $$a = 3b$$, for example $$a = 3$$, $$b = 1$$ would be a solution for our system.
Solution:
- Linearly independent.
- Linearly dependent.
- Linearly dependent.