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

Short Answer

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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.
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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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Most popular questions from this chapter

Chapter 2

1 ) Let \(\mathrm{V}\) be a vector space over a field \(\mathrm{F}\) and let \(\alpha \varepsilon \mathrm{F}\) be any nonzero scalar. Show that multiplication by \(\alpha\), $\mathrm{M}_{\alpha}: \mathrm{V} \rightarrow \mathrm{V}$ is a vector-space isomorphism. 2) Let \(\mathrm{V}\) and \(\mathrm{w}\) be any vector spaces over \(\mathrm{F}\) and \(\mathrm{V} \varepsilon \mathrm{W}\) their Cartesian product. Show that \(\mathrm{V} \cong \mathrm{V} \times\\{0\\}\) where \(0 \varepsilon \mathrm{W}\).

Chapter 2

Let \(\mathrm{C}[0,1]\) be the vector space of bounded continuous functions defined on the interval \(0 \leq \mathrm{x} \leq 1 .\) Let \(\mathrm{M}\) be the subspace of \(\mathrm{C}[0,1]\) consisting of functions \(\mathrm{f}\) such that both \(\mathrm{f}\) and its derivative \(\mathrm{f}\) are defined and continuous for \(0 \leq \mathrm{x} \leq 1\). Show that the operations of differentiation and integration are linear transformation.

Chapter 2

Let \(\mathrm{V}\) and \(\mathrm{W}\) be vector spaces over the field \(\mathrm{K}\), and \(\mathrm{T}: \mathrm{V} \rightarrow \mathrm{W}\) a linear transformation. Find the kernel and range of \(\mathrm{T}\) if \(\mathrm{T}\) takes the following forms: 1) \(\mathrm{T}\) is the scalar operator \(\alpha\) I where \(\mathrm{I}\) is the identity operator and \(\alpha \neq 0\). 2) \(\mathrm{T}\) is a rotation of elements in \(\mathrm{R}^{2}\).

Chapter 2

Let \(\mathrm{T}\) be a linear operator on \(\mathrm{R}^{2}\) defined by $\mathrm{T}\left(\mathrm{x}_{1}, \mathrm{x}_{2}\right)=\left(\mathrm{x}_{1}+\mathrm{x}_{2}, \mathrm{x}_{1}\right)$ Show that \(\mathrm{T}\) is non-singular.

Chapter 2

Let \(\mathrm{X}=[1,0,5,-2]\) and \(\mathrm{Y}=[3,5,7,1]\) be points in \(\mathrm{R}^{4}\). Find proj \(\mathrm{X}\) and the cosine of the angle between \(\mathrm{X}\) and \(\mathrm{Y}\).

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