Exercises 31 and 32 concern finite-dimensional vector spaces V and W and a linear transformation \(T:V \to W\).

Let H be a nonzero subspace of V, and suppose T is a one-to-one (linear) mapping ofV and W.Prove that \({\bf{dim}}T\left( H \right) = {\bf{dim}}H\). If T happens to be one-to-one mapping of V onto W, then \({\bf{dim}}V = {\bf{dim}}W\). Isomorphic finite-dimensional vector spaces have the same dimensions.

Let the set \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},....{{\bf{v}}_p}} \right\}\) be a basis for H,i.e. \(\dim H = p\). For any \({\bf{v}}\) in the subspace H:

\(\begin{array}{c}{\bf{v}} = {c_1}{{\bf{v}}_1} + .... + {c_p}{{\bf{v}}_p}\\T\left( {\bf{v}} \right) = T\left( {{c_1}{{\bf{v}}_1} + .... + {c_p}{{\bf{v}}_p}} \right)\\ = {c_1}T\left( {{{\bf{v}}_1}} \right) + {c_2}T\left( {{{\bf{v}}_2}} \right) + ..... + {c_p}T\left( {{{\bf{v}}_p}} \right)\end{array}\)

Consider that the vectors are linearly independent:

\[\begin{array}{c}{c_1}T\left( {{{\bf{v}}_1}} \right) + {c_2}T\left( {{{\bf{v}}_2}} \right) + ..... + {c_p}T\left( {{{\bf{v}}_p}} \right) = 0\\T\left( {{c_1}{{\bf{v}}_1}} \right) + T\left( {{c_2}{{\bf{v}}_2}} \right) + .... + T\left( {{c_p}{{\bf{v}}_p}} \right) = 0\\T\left( {{c_1}{{\bf{v}}_1} + {c_2}{{\bf{v}}_2} + .... + {c_p}{{\bf{v}}_p}} \right) = 0\end{array}\]

As \(T\left( 0 \right) = 0\) and T is one-to-one, so

\[{c_1}{{\bf{v}}_1} + {c_2}{{\bf{v}}_2} + .... + {c_p}{{\bf{v}}_p} = 0\]

Vectors \({{\bf{v}}_1}\), \({{\bf{v}}_2}\),…., \({{\bf{v}}_p}\) arelinearly independent, which is contrary to the assumed statement.

Therefore, \(\dim T\left( H \right) = \dim H\)