Question

Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) be an invertible linear transformation. Explain why T is both one-to-one and onto \({\mathbb{R}^n}\). Use equations (1) and (2). Then give a second explanation using one or more theorems.

Short answer

It is proved that Tis both one-to-one and onto.

Step-by-step solution

Show that T is one-to-one

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 \({\mathbb{R}^n}\).

Let \[T\left( {\mathop{\rm u}\nolimits} \right) = T\left( {\mathop{\rm v}\nolimits} \right)\] for vectors u and v in \({\mathbb{R}^n}\). Then, \(S\left( {T\left( {\mathop{\rm u}\nolimits} \right)} \right) = S\left( {T\left( {\mathop{\rm v}\nolimits} \right)} \right)\), where Srepresents the inverse of T.

By equation (1), \({\mathop{\rm u}\nolimits} = S\left( {T\left( {\mathop{\rm u}\nolimits} \right)} \right)\) and \(S\left( {T\left( {\mathop{\rm v}\nolimits} \right)} \right) = {\mathop{\rm v}\nolimits} \). Thus, \[{\mathop{\rm u}\nolimits} = v\] and T is one-to-one.

Show that T is onto

Let \(y\) be an arbitrary vector in \({\mathbb{R}^n}\) and \({\mathop{\rm x}\nolimits} = S\left( y \right)\). Equation (2) shows that \(T\left( {\mathop{\rm x}\nolimits} \right) = T\left( {S\left( {\mathop{\rm y}\nolimits} \right)} \right) = {\mathop{\rm y}\nolimits} \), which indicates that Tmaps \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\).

Thus, it is proved that Tmaps \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\).

Show that T is both one-to-one and onto

Let\(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) be a linear transformation and Abe the standard matrix for T. Then, according totheorem 9, Tis invertibleif 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 \({\mathbb{R}^n}\).

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.

The standard matrix Afor T is invertible, according to theorem 9.

Following the invertible matrix theorem, the columns of Aare linearly independent, and they span \({\mathbb{R}^n}\). Therefore, T is one-to-one and it maps \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\), based on theorem 12.

Thus, it is proved that Tis both one-to-one and onto.