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Transformation
(x, y, z) → (x + y, y + z) : R3 → R2 is
(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?
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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
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
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
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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