Question

Let \(V\) and \(W\) be vector spaces, and let \(T:V \to W\) be a linear transformation. Given a subspace \(U\) of \(V\), let \(T\left( U \right)\) denote the set of all images of the form \(T\left( {\mathop{\rm x}\nolimits} \right)\), where x is in \(U\). Show that \(T\left( U \right)\) is a subspace of \(W\).

Short answer

It is proved that \(T\left( U \right)\) is a subspace of \(W\).

Step-by-step solution

Define linear transformation

The conditions forlinear transformation T are as follows:

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 \(c\).

Show that \(T\left( U \right)\) is closed under vector addition

\({0_V}\)is in \(U\) because \(U\) is a subspace of \(V\). \(T\left( {{0_V}} \right) = {0_W}\) since \(T\) is linear. Then, \({0_W}\) is in \(T\left( U \right)\).

Consider \(T\left( {\mathop{\rm x}\nolimits} \right)\) and \(T\left( {\mathop{\rm y}\nolimits} \right)\) are typical elements in \(T\left( U \right)\). Then, \({\mathop{\rm x}\nolimits} \) and \(y\) are in \(U\)and \({\mathop{\rm x}\nolimits} + y\) is also in \(U\) because \(U\) is in the subspace of \(V\). \(T\left( {\mathop{\rm x}\nolimits} \right) + T\left( {\mathop{\rm y}\nolimits} \right) = T\left( {{\mathop{\rm x}\nolimits} + y} \right)\) because \(T\) is a linear transformation. Therefore, \(T\left( {\mathop{\rm x}\nolimits} \right) + T\left( {\mathop{\rm y}\nolimits} \right)\) is in \(T\left( U \right)\). Thus, \(T\left( U \right)\) is closed under vector addition.

Show that \(T\left( U \right)\)  is a subspace of \(W\)

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 \(W\).