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Show that T: R2 -› R defined by T(x,y) = (xy, 0) is not a linear mapping.To do this, either i) Find specific vectors x, y E R2

such that T(x+ y) ‡ T(x) + T(y)

or ii) Find a specific r E R and x E R? such that T(rx) ‡ rT (x)

User Cetin Sert
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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.

User Sanket Phansekar
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