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Transformation
(x, y, z) → (x + y, y + z) : R3 → R2 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
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Transformation(x, y, z) → (x + y, y + z) : R3→R2 isa)linear ...
We are given that a linear transformation T : R3 → R2 defined by T(x, y, z) = (x + y, y + z)
Here, we need to check its linearity and also need to find ker T.
Let (x1, y1, z1) and (x2, y2, z2) be any two vector of R3, Then
T[(x1, y1, z1) + (x2, y2, z2)]

Again let α be a scalar number then,
T[α(x,y, z)]
Hence, T is a linear.
Let (x, y, z) ∈ ker T, then T(x, y, z) = (0,0)
using the definition of ker T, we get
(x+y,y + z) = (0, 0)
Implies x + y - 0 and
            y + z = 0
Implies x = - y and y = - z
Therefore, ker T = {(x, -x, x ) : x e R}
Therefore, dim ker T = 1
Hence, ker T is a proper subspace of R3
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Most Upvoted Answer
Transformation(x, y, z) → (x + y, y + z) : R3→R2 isa)linear ...
The given transformation can be represented as (x', y', z'), where:

x' = x
y' = y
z' = z
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Transformation(x, y, z) → (x + y, y + z) : R3→R2 isa)linear and has zero kernelb)linear and has a proper subspace as kernelc)neither linear nor one-oned)neither linear nor ontoCorrect answer is option 'B'. Can you explain this answer?
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