Question

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 let \(T\left( H \right)\) be the set of images of vectors in H. Then \(T\left( H \right)\) is a subspace of W, by Exercise 35 in section 4.2. Prove that \({\bf{dim}}T\left( H \right) \le {\bf{dim}}\left( H \right)\).

Short answer

\(\dim T\left( H \right) \le p = \dim H\)

Step-by-step solution

Write the transformation vector in subspace H

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:

\(T\left( {\bf{v}} \right) = T\left( {{c_1}{{\bf{v}}_1} + .... + {c_p}{{\bf{v}}_p}} \right)\)

Check for statement (b)

Any vector \(T\left( {\bf{v}} \right) \in T\left( H \right)\) is a linear combination of \(T\left( {{{\bf{v}}_1}} \right)\),….,\(T\left( {{{\bf{v}}_p}} \right)\), i.e.

\(T\left( H \right) = {\rm{span}}\left\{ {T\left( {{{\bf{v}}_1}} \right),.....,T\left( {{{\bf{v}}_p}} \right)} \right\}\)

As \(T\left( {{{\bf{v}}_1}} \right)\),….,\(T\left( {{{\bf{v}}_p}} \right)\) are not linearly independent, \(\dim T\left( H \right) \le p = \dim H\).