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Problem 46

# Show that the zero mapping and the identity transformation are linear transformations.

Expert verified
The zero mapping, T₀, and the identity transformation, I, are both linear transformations. This is because they satisfy the two properties of linear transformations: For the zero mapping, T₀(u+v) = T₀(u) + T₀(v) and T₀(cu) = cT₀(u), as T₀ maps all vectors to the zero vector. For the identity transformation, I(u+v) = I(u) + I(v) and I(cu) = cI(u), since I maps each vector to itself.
See the step by step solution

## Step 1: Zero Mapping

Let T₀ : V → W be the zero mapping. We want to show that it satisfies the properties of a linear transformation. 1. Let u, v be arbitrary vectors in V. Then: T₀(u+v) = 0 (since T₀ maps all vectors to the zero vector) T₀(u) + T₀(v) = 0 + 0 = 0 (since T₀ also maps u and v to the zero vector) So, T₀(u+v) = T₀(u) + T₀(v). 2. Let u be an arbitrary vector in V and c be an arbitrary scalar. Then: T₀(cu) = 0 (since T₀ maps all vectors to the zero vector) cT₀(u) = c(0) = 0 (since T₀ maps u to the zero vector) So, T₀(cu) = cT₀(u). Since the zero-mapping satisfies both properties of a linear transformation, it is a linear transformation.

## Step 2: Identity Transformation

Let I : V → W be the identity transformation. We want to show that it satisfies the properties of a linear transformation. 1. Let u, v be arbitrary vectors in V. Then: I(u+v) = u+v (since I maps each vector to itself) I(u) + I(v) = u + v (since I maps u to u and v to v) So, I(u+v) = I(u) + I(v). 2. Let u be an arbitrary vector in V and c be an arbitrary scalar. Then: I(cu) = cu (since I maps each vector to itself) cI(u) = cu (since I maps u to u) So, I(cu) = cI(u). Since the identity transformation satisfies both properties of a linear transformation, it is a linear transformation.

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