Indicate which pairs of the following vectors form a basis. In the case that they do form one, determine the coordinates of the vector $$\vec{w}=(5,-2)$$ in every basis.
- $$\vec{u}=(2,0)$$, $$\ \vec{v}=(1,1)$$
- $$\vec{u}=(2,1)$$, $$\ \vec{v}=(1,-1)$$
- $$\vec{u}=(1,-3)$$, $$\ \vec{v}=(-3,9)$$
Development:
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The vectors $$\vec{u}=(2,0)$$ and $$\ \vec{v}=(1,1)$$ are linearly independent because they do not have the same direction and their coordinates are not proportional: $$$\dfrac{0}{2}\neq\dfrac{1}{1}$$$ We observe that the second component $$\vec{u}$$ is $$0$$ then we have done the division the other way round because we are interested in verifying if its components are proportional. 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 scalar values of the combination are both zero: $$$\lambda(2,0)+\mu(1,1)=(2\lambda+\mu,\mu)=(0,0)$$$ $$$\left. \begin{array}{r} 2\lambda+\mu=0 \\ \mu=0 \end{array} \right\} \Rightarrow \mu=0, \ \lambda=0$$$ As the vectors are linearly independent they form a basis. The coordinates of $$\vec{w}$$ in this basis are scalars in such a way that $$\vec{w}=a\vec{u}+b\vec{v}$$, so in other words: $$$(5,-2)=a(2,0)+b(1,1)=(2a,0)+(b,b)=(2a+b,b)$$$ $$$\left. \begin{array}{r} 2a+b=5 \\ b=-2 \end{array} \right\} \Rightarrow a=\dfrac{7}{2}, \ b=-2$$$ Therefore $$\vec{w}=(\dfrac{7}{2},-2)$$ in the basis $$B=\{\vec{u},\vec{v}\}$$.
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The vectors $$\vec{u}=(2,1)$$ and $$\ \vec{v}=(1,-1)$$ are linearly independent because its coordinates are not proportional: $$$\dfrac{2}{1}\neq\dfrac{1}{-1}$$$ Therefore they form a basis. The coordinates of $$\vec{w}$$ in this basis fulfill $$\vec{w}=a\vec{u}+b\vec{v}$$, or in other words: $$$(5,-2)=a(2,1)+b(1,-1)=(2a+b,a-b)$$$ $$$\left. \begin{array}{r} 2a+b=5 \\ a-b=-2 \end{array} \right\} \Rightarrow a=1, \ b=3$$$ Therefore $$\vec{w}=(1,3)$$ in the basis $$B=\{\vec{u},\vec{v}\}$$.
- We see, in this case, that the vectors $$\vec{u}=(1,-3)$$ and $$\ \vec{v}=(-3,9)$$ sare linearly dependent since: $$$\dfrac{1}{-3}\neq\dfrac{-3}{9} \Rightarrow -\dfrac{1}{3}=-\dfrac{1}{3}$$$ Another way of evaluating whether they are linearly dependent or not is finding the scalar values when $$a\vec{u}+b\vec{v}=\vec{0}$$ and seeing if they are zero, or in other words: $$$a(1,-3)+b(-3,9)=(a,-3a)+(-3b,9b)=(a-3b,-3a+9b)=(0,0)$$$ $$$\left. \begin{array}{r} a-3b=0 \\ -3a+9b=0 \end{array} \right\} \Rightarrow a=3b \Rightarrow -3(3b)+9b=0 \Rightarrow \Rightarrow -9b+9b=0 $$$ the second equation does not yield information, so it is only necessary to establish that $$a = 3b$$, for example $$a = 3$$, $$b = 1$$ would be a solution for our system. As they are not linearly independent, they cannot form a basis.
Solution:
- The vectors $$\vec{u}=(2,0)$$ and $$\ \vec{v}=(1,1)$$ form a basis. $$\vec{w}=(\dfrac{7}{2},-2)$$ in the basis $$B=\{\vec{u},\vec{v}\}$$.
- The vectors $$\vec{u}=(2,1)$$ and $$\ \vec{v}=(1,-1)$$ form a basis. $$\vec{w}=(1,3)$$ in the basis $$B=\{\vec{u},\vec{v}\}$$.
- The vectors $$\vec{u}=(1,-3)$$ and $$\ \vec{v}=(-3,9)$$ do not form any basis.