37.3: One-to-one and Inverse transform
- Page ID
- 70390
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Read Section 4.9 of the textbook if you are not familiar with this part.
A transformation \(T:U\mapsto V\) is said to be one-to-one if each element in the range is the image of just one element in the domain. That is, for two elements (\(x\) and \(y\)) in \(U\). \(T(x)=T(y)\) happens only when \(x=y\).
Let \(T:U\mapsto V\) be a one-to-one linear transformation. If \(\{u_1,\dots,u_n\}\) is linearly independent in \(U\), then \(\{T(u_1),\dots,T(u_n)\}\) is linearly independent in \(V\).
A linear transformation \(T:U\mapsto V\) is said to be invertible if there exists a transformation \(S:V\mapsto U\), such that \(S(T(u))=u,\qquad T(S(v))=v,\) for any \(v\) in \(V\) and any \(u\) in \(U\).
If linear transformation \(T:U\mapsto V\) is invertible, and the dimension of \(U\) is 2, what is the dimension of \(V\)? Why?


