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Given \(T:V \to W\) as in Exercise 35, and given a subspace \(Z\) of \(W\), let \(U\) be the set of all \({\mathop{\rm x}\nolimits} \) in \(V\) such that \(T\left( {\mathop{\rm x}\nolimits} \right)\) is in \(Z\). Show that \(U\) is a subspace of \(V\).

Short Answer

Expert verified

It is proved that \(U\) is a subspace of \(V\).

Step by step solution

01

Linear transformation

Alinear transformation \(T\)from a vector space \(V\) into a vector space \(W\) is a rule that assigns to each vector x in \(V\) a unique vector \(T\left( {\mathop{\rm x}\nolimits} \right)\) in \(W\), such that

1.\(T\left( {{\mathop{\rm u}\nolimits} + {\mathop{\rm v}\nolimits} } \right) = T\left( {\mathop{\rm u}\nolimits} \right) + T\left( {\mathop{\rm v}\nolimits} \right)\) for all \({\mathop{\rm u}\nolimits} ,{\mathop{\rm v}\nolimits} \,\,{\mathop{\rm in}\nolimits} \,\,V\), and

2 \(T\left( {c{\mathop{\rm u}\nolimits} } \right) = cT\left( {\mathop{\rm u}\nolimits} \right)\) for all \({\mathop{\rm u}\nolimits} \,\,\,{\mathop{\rm in}\nolimits} \,\,V\) and all scalar \(c\).

02

Step 2: \(U\) is closed under vector addition

Consider \(c\) is any vector. Then, \(c{\mathop{\rm x}\nolimits} \) is in \(U\) because \({\mathop{\rm x}\nolimits} \) is in \(U\), and \(U\) is a subspace of \(V\). \(cT\left( {\mathop{\rm x}\nolimits} \right)\) is in \(T\left( U \right)\) and \(T\left( {c{\mathop{\rm x}\nolimits} } \right) = cT\left( {\mathop{\rm x}\nolimits} \right)\) because \(T\) is a linear transformation. Therefore, \(T\left( U \right)\) is closed under multiplication by scalar and \(T\left( U \right)\) is a subspace of \(W\).

Thus, it is proved that \(T\left( U \right)\) is a subspace of \(W\).

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