Answer:
The answer is that the mapping T: R^2 -> R defined by T(x, y) = (xy, 0) is not a linear mapping. It fails to satisfy either the additivity property or the scalar multiplication property, as shown in both approaches.
Explanation:
To show that the mapping T: R^2 -> R defined by T(x, y) = (xy, 0) is not a linear mapping, we can find specific vectors x, y ∈ R^2 such that T(x + y) ≠ T(x) + T(y) or find a specific scalar r ∈ R and vector x ∈ R^2 such that T(rx) ≠ rT(x).
Let's go with the first approach, finding specific vectors x, y ∈ R^2 such that T(x + y) ≠ T(x) + T(y).
Consider x = (1, 0) and y = (0, 1).
T(x + y) = T(1, 0) + T(0, 1)
= (1 * 0, 0) + (0 * 1, 0)
= (0, 0) + (0, 0)
= (0, 0)
T(x) + T(y) = T(1, 0) + T(0, 1)
= (1 * 0, 0) + (0 * 0, 0)
= (0, 0) + (0, 0)
= (0, 0)
As we can see, T(x + y) = (0, 0) ≠ (0, 0) = T(x) + T(y).
Therefore, T: R^2 -> R defined by T(x, y) = (xy, 0) is not a linear mapping since it fails the additivity property.
Alternatively, we can also show that T: R^2 -> R defined by T(x, y) = (xy, 0) is not a linear mapping by using the second approach. We need to find a specific scalar r ∈ R and vector x ∈ R^2 such that T(rx) ≠ rT(x).
Consider r = 2 and x = (1, 1).
T(rx) = T(2, 2)
= (2 * 2, 0)
= (4, 0)
rT(x) = 2T(1, 1)
= 2(1 * 1, 0)
= 2(1, 0)
= (2, 0)
As we can see, T(rx) = (4, 0) ≠ (2, 0) = rT(x).
Therefore, T: R^2 -> R defined by T(x, y) = (xy, 0) is not a linear mapping since it fails the scalar multiplication property.
In both cases, we have shown that T is not a linear mapping.