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Problem 51

# Let $$L: R^{3} \rightarrow R^{3}$$ be defined by $\mathrm{x}|\quad| \mathrm{x}+1 \mid$ $$\mathrm{L}|\mathrm{y}|=\mid 2 \mathrm{y}$$ Show that $$L$$ is not a linear transformation.

Expert verified
The counterexample provided shows that L does not satisfy the additivity property, as L(x+y) = 0 ≠ L(x) + L(y) = 3. Thus, L is not a linear transformation.
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We need to check if the following equation holds for all x, y in R^3: L(x+y) = L(x) + L(y) To see if this property holds, let's first compute L(x+y) and L(x) + L(y): $$L(x+y) =|(x+y)+1|$$ $$L(x)+L(y) = |x+1|+|y+1|$$ Now, we need to find if there exist any x and y such that L(x+y) is not equal to L(x) + L(y). Consider the following vectors: $$x = \begin{bmatrix} -1 \\ 0 \\ 0 \end{bmatrix}$$ and $$y = \begin{bmatrix} 0 \\ -1 \\ 0 \end{bmatrix}$$ Now, let's compute L(x+y) and L(x)+L(y): $$L(x+y) = L(\begin{bmatrix} -1 \\ -1 \\ 0 \end{bmatrix}) = |(-1) + 1| = |0| = 0$$ $$L(x) = |(-1) + 1| = |0| = 0$$ $$L(y) = |(-1) + 1| = |0| = 0$$ $$L(x) + L(y) = 0 + 0 = 0$$ In this case, L(x+y) = L(x) + L(y). However, we need to check more examples to see if the additivity holds for other vectors as well. Consider the following vectors: $$x = \begin{bmatrix} -2 \\ 0 \\ 0 \end{bmatrix}$$ and $$y = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$$ Now, let's compute L(x+y) and L(x) + L(y): $$L(x+y) = L(\begin{bmatrix} -1 \\ 0 \\ 0 \end{bmatrix}) = |(-1) + 1| = |0| = 0$$ $$L(x) = |(-2) + 1| = |-1| = 1$$ $$L(y) = |1 +1| = |2| = 2$$ $$L(x) + L(y) = 1 + 2 = 3$$ Here, we found a counterexample: L(x+y) = 0 ≠ L(x) + L(y) = 3 Since L doesn't satisfy the additivity property, we can conclude that L is not a linear transformation. There is no need to check the homogeneity property since the violation of any one property is enough to show that L is not linear.

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