Let T be a linear transformation that maps \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\). Is \({T^{ - 1}}\) also one-to-one?

Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^m}\) be a linear transformation, and Abe the standard matrix for T. Then, according to theorem 12,

1. Tmaps \({\mathbb{R}^n}\) onto \({\mathbb{R}^m}\) if and only if the columns of Aspan \({\mathbb{R}^m}\);
2. T is one-to-one if and only if the columns of Aare linearly independent.

When T maps \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\), the columns of standard matrix Aspan \({\mathbb{R}^n}\), according to theorem 12.

A is invertible according to the invertible matrix theorem.

Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) be a linear transformation and Abe the standard matrix for T. Then, according to theorem 9, Tis invertible if and only if Ais an invertible matrix. The linear transformation S, given by \[S\left( x \right) = {A^{ - 1}}{\mathop{\rm x}\nolimits} \], is a unique function satisfying the equations

1. \(S\left( {T\left( {\mathop{\rm x}\nolimits} \right)} \right) = {\mathop{\rm x}\nolimits} \)for all x in \({\mathbb{R}^n}\), and
2. \(T\left( {S\left( {\mathop{\rm x}\nolimits} \right)} \right) = {\mathop{\rm x}\nolimits} \)for all x in \({\mathbb{R}^n}\).

Tis invertible and \({A^{ - 1}}\) is the standard matrix of \({T^{ - 1}}\), according to theorem 9.

By the invertible matrix theorem, the columns of \({A^{ - 1}}\) are linearly independent, and they span \({\mathbb{R}^n}\) since \({A^{ - 1}}\) is invertible. Thus, the transformation \({T^{ - 1}}\) is one-to-one mapping of \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\), according to theorem 12.

Thus, the transformation \({T^{ - 1}}\) is one-to-one mapping.