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CSS Pure Mathematics QUESTION #4091
Question 1
Consider \(T:\mathbb{R}^3\to\mathbb{R}^2\) defined by \(T(x,y,z)=(|x|,y+z)\). To show \(T\) is NOT linear, the simplest counterexample uses scalar multiplication with \(c=-1\) and vector \((1,0,0)\):
  • Both \(T(c\mathbf{v})\) and \(cT(\mathbf{v})\) equal \((1,0)\)
  • Both equal \((-1,0)\)
  • They differ: \(T(-1,0,0)=(1,0)\) but \(-T(1,0,0)=(-1,0)\)✔️
  • They differ: \(T(-1,0,0)=(0,-1)\) but \(-T(1,0,0)=(0,1)\)
Correct Answer Explanation
For linearity, \(T(c\mathbf{v})=cT(\mathbf{v})\). With \(\mathbf{v}=(1,0,0)\) and \(c=-1\): \(T(-1,0,0)=(|-1|,0)=(1,0)\), but \(-1\cdot T(1,0,0)=-1\cdot(1,0)=(-1,0)\). Since \((1,0)\neq(-1,0)\), \(T\) is NOT linear — homogeneity fails.