Discussion
Let \(T : \mathbb{R}^{3} \to \mathbb{R}^{3}\) be a linear transformation satisfying
T(1, 0, 0) = (0, 1, 1), T(1, 1, 0) = (1, 0, 1) and T(1, 1, 1) = (1, 1, 2).
Then
Answer: C
T is NEITHER one-one NOR onto.
Find the images of the standard basis by linearity:
\[ \begin{aligned} T(1,0,0) &= (0,1,1) \\ T(0,1,0) &= T(1,1,0)-T(1,0,0) = (1,-1,0) \\ T(0,0,1) &= T(1,1,1)-T(1,1,0) = (0,1,1) \end{aligned} \]
The first and third columns are equal, so T(1,0,-1) = 0 and the kernel is non-trivial: T is not one-one. By rank-nullity, the rank is at most 2 (it is exactly 2, since (0,1,1) and (1,-1,0) are independent), so T is not onto \(\mathbb{R}^{3}\). For a linear map from \(\mathbb{R}^{3}\) to itself, one-one and onto stand or fall together, which rules out the two mixed options.
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