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Chapter 6: Problem 36

Let \(V\) be a vector space with basis \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{k}\right\\}\) and suppose \(T: V \rightarrow W\) is a linear transformation such that \(T\left(\mathbf{v}_{i}\right)=\mathbf{0}\) for each \(i=1,2, \ldots, k .\) Prove that \(T\) is the zero transformation; that is, \(T(\mathbf{v})=\mathbf{0}\) for each \(\mathbf{v} \in V.\)

Short Answer

Expert verified
Let \(\mathbf{u}\) be an arbitrary vector in V. We express \(\mathbf{u}\) as a linear combination of basis vectors and then apply the linear transformation T to \(\mathbf{u}\). Using the properties of linear transformations, we distribute and apply the given information about the transformation of basis vectors. Finally, we combine the terms and simplify the equation, proving that the transformation of any arbitrary vector \(\mathbf{u} \in V\) under T results in the zero vector in W. Therefore, T is the zero transformation.

Step by step solution

01

Choose an arbitrary vector in V.

Let \(\mathbf{u}\) be an arbitrary vector in V. Since \(\{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\}\) is a basis for V, we can express \(\mathbf{u}\) as a linear combination of basis vectors: \[\mathbf{u} = c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \cdots + c_k \mathbf{v}_k\]
02

Apply T to the arbitrary vector.

Now, we find the transformation of \(\mathbf{u}\) under T: \[T(\mathbf{u}) = T(c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \cdots + c_k \mathbf{v}_k)\]
03

Use the properties of linear transformations.

Because T is a linear transformation, we can apply the properties of linear transformations to split and distribute: \[ T(\mathbf{u}) = c_1 T(\mathbf{v}_1) + c_2 T(\mathbf{v}_2) + \cdots + c_k T(\mathbf{v}_k) \]
04

Use the given information about the transformation of basis vectors.

Now, we replace each \(T(\mathbf{v}_i)\) with \(\mathbf{0}\), as given in the problem: \[T(\mathbf{u}) = c_1 \mathbf{0} + c_2 \mathbf{0} + \cdots + c_k \mathbf{0}\]
05

Combine the terms and simplify the equation.

Finally, combine the terms and simplify the equation: \[T(\mathbf{u}) = (c_1 + c_2 + \cdots + c_k) \mathbf{0} = \mathbf{0}\] Since we proved that the transformation of any arbitrary vector \(\mathbf{u} \in V\) under T results in the zero vector in W, we can conclude that T is the zero transformation.

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Most popular questions from this chapter

Prove that if \(T: V \rightarrow W\) is a linear transformation with \(\operatorname{dim}[W]=n=\operatorname{dim}[\operatorname{Rng}(T)],\) then \(T\) is onto.

Suppose \(T: V \rightarrow W\) is a linear transformation and \(\left\\{\mathbf{w}_{1}, \mathbf{w}_{2}, \ldots, \mathbf{w}_{m}\right\\}\) spans \(W .\) If there exist vectors \(\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{m}\) in \(V\) such that \(T\left(\mathbf{v}_{i}\right)=\mathbf{w}_{i}\) for each \(i=1,2, \dots, m,\) prove that \(T\) is onto.

Let \(V\) be a vector space with basis \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{k}\right\\}\) and suppose \(T: V \rightarrow W\) is a linear transformation such that \(T\left(\mathbf{v}_{i}\right)=\mathbf{0}\) for each \(i=1,2, \ldots, k .\) Prove that \(T\) is the zero transformation; that is, \(T(\mathbf{v})=\mathbf{0}\) for each \(\mathbf{v} \in V.\)

Prove that if \(T: V \rightarrow V\) is a linear transformation such that \(T^{2}=0\) (that is, \(T(T(\mathbf{v}))=\mathbf{0}\) for all \(\mathbf{v} \in V\) ) then \(\operatorname{Rng}(T)\) is a subspace of \(\operatorname{Ker}(T)\).

Let \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\) be a basis for the vector space \(V,\) and suppose that \(T_{1}: V \rightarrow V\) and \(T_{2}: V \rightarrow V\) are the linear transformations satisfying $$\begin{array}{ll} T_{1}\left(\mathbf{v}_{1}\right)=\mathbf{v}_{1}-\mathbf{v}_{2}, & T_{1}\left(\mathbf{v}_{2}\right)=2 \mathbf{v}_{1}+\mathbf{v}_{2} \\ T_{2}\left(\mathbf{v}_{1}\right)=\mathbf{v}_{1}+2 \mathbf{v}_{2}, & T_{2}\left(\mathbf{v}_{2}\right)=3 \mathbf{v}_{1}-\mathbf{v}_{2} \end{array}$$ Determine \(\left(T_{2} T_{1}\right)(\mathbf{v})\) for an arbitrary vector \(\mathbf{v}\) in \(V\)
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