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Let a be a fixed nonzero vector in R2. A mapping of the form L(x)=x+a is called a translation. Show that a translation is not a linear operator. Illustrate geometrically the effect of a translation.

User Kevy
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Final answer:

A translation described by L(x) = x + a does not satisfy the additivity and homogeneity properties required for linear operators, and thus is not linear. Geometrically, a translation slides every point in the plane by the fixed vector 'a' without altering orientations or shapes.

Step-by-step explanation:

A translation in ℝ² is described by the mapping L(x) = x + a, where a is a fixed non-zero vector. To show that a translation is not a linear operator, we must examine whether it satisfies two main properties: additivity and homogeneity of scalar multiplication. For L to be linear, it must hold that L(x+y) = L(x) + L(y) and L(cx) = cL(x) for all vectors x, y and scalar c.

For additivity, if we take any two vectors x and y in ℝ², we can see that:

  • L(x+y) = (x+y) + a
  • L(x) + L(y) = (x + a) + (y + a) = x + y + 2a (since vector addition is commutative)

Clearly, L(x+y) ≠ L(x) + L(y) because of the extra 'a' in the second expression. Therefore, the additivity property is violated.

For homogeneity:

  • L(cx) = cx + a
  • cL(x) = c(x + a) = cx + ca

Here again, L(cx) ≠ cL(x) as 'a' is not multiplied by the scalar 'c' in L(cx), hence violating homogeneity.

Geometrically, a translation moves every point in the plane by the vector 'a', without altering the orientation or shape of the figure being translated. It is like sliding a shape in a spatial direction determined by 'a'. This is fundamentally different from linear transformations such as rotations or dilations, which might change angles or distances between points.

The geometric effect of a translation can be envisioned by considering a figure at one location being picked up and moved to another without any rotation or resizing, creating a congruent figure at the new location.

User Qazi Ammar
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