91Ó°ÊÓ

Given that\(f:{\mathbb{R}^n} \to {\mathbb{R}^m}\)is a linear transformation,\(T\)be an affine subset of\({\mathbb{R}^m}\), and\(S = \left\{ {x \in {\mathbb{R}^n}\,:\,f\left( {\bf{x}} \right) \in T} \right\}\).

Assume a subspace for \(S\) and \(\mathbb{R}\), that is, \({\bf{x,y}} \in S\) and \(t \in \mathbb{R}\). To show that \(S\) is affine, it suffices to show that for any pair \({\bf{x}}\) and \({\bf{y}}\) of points in \(S\), the line through \({\bf{x}}\) and \({\bf{y}}\)lies in \(S\).

As \(S = \left\{ {x \in {R^n}\,:\,f\left( x \right) \in T} \right\}\). So, for each real \(t\), \(f\left( {\left( {1 - t} \right){\rm{x}} + t{\rm{y}}} \right) = \left( {1 - t} \right)f\left( {\rm{x}} \right) + tf\left( {\rm{y}} \right)\).

Since \(T\) is an affine subspace of \({\mathbb{R}^n}\), \(\left( {1 - t} \right)f\left( {\rm{x}} \right) + tf\left( {\rm{y}} \right) \in T\). Moreover, \(\left( {1 - t} \right){\rm{x}} + t{\rm{y}} \in S\), as \({\rm{x,y}} \in S\), and \(f\left( {\rm{x}} \right) \in T\).

The statement \(\left( {1 - t} \right){\rm{x}} + t{\rm{y}} \in S\) is satisfied by all the points in the subspace of \(S\), and \(\mathbb{R}\).

Therefore, \(S\) is an affine subset of \({\mathbb{R}^m}\).