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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.

Short Answer

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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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Step 1: Additivity

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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Most popular questions from this chapter

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