Question:
The transformation T: ℝ³ → ℝ² defined T(x, y, z) = (x, y, y, z) is:
(a) linear and has zero kernel
(b) linear and has a proper subspace as kernel
(c) neither linear nor one-one
(d) neither linear nor onto?

Answer:
To determine whether the given transformation T is linear and has zero kernel, we need to check two conditions: linearity and the kernel.

Linearity:
A transformation T: ℝ³ → ℝ² is linear if it satisfies the following properties:
1. T(u + v) = T(u) + T(v) for all vectors u and v in ℝ³.
2. T(cu) = cT(u) for all vectors u in ℝ³ and scalar c.

Let's check these properties for the given transformation T(x, y, z) = (x, y, y, z):

1. T(u + v) = T(x₁ + x₂, y₁ + y₂, z₁ + z₂) = (x₁ + x₂, y₁ + y₂, y₁ + y₂, z₁ + z₂)
T(u) + T(v) = (x₁, y₁, y₁, z₁) + (x₂, y₂, y₂, z₂) = (x₁ + x₂, y₁ + y₂, y₁ + y₂, z₁ + z₂)

The property T(u + v) = T(u) + T(v) holds true.

2. T(cu) = T(cx, cy, cz) = (cx, cy, cy, cz)
cT(u) = c(x, y, y, z) = (cx, cy, cy, cz)

The property T(cu) = cT(u) holds true.

Therefore, the given transformation T is linear.

Kernel:
The kernel of a linear transformation T is the set of all vectors u in the domain of T such that T(u) = 0.

Let's find the kernel of the given transformation T(x, y, z) = (x, y, y, z):

T(x, y, y, z) = (0, 0)

This implies x = 0, y = 0, and z = 0.

Therefore, the kernel of the given transformation is { (0, 0, 0) }.

Conclusion:
Based on the analysis above, we can conclude that the given transformation T: ℝ³ → ℝ² defined T(x, y, z) = (x, y, y, z) is linear and has zero kernel.

Therefore, the correct answer is (a) linear and has zero kernel.