Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) be an invertible linear transformation, and let Sand U be functions from \({\mathbb{R}^n}\) into \({\mathbb{R}^n}\) such that \(S\left( {T\left( {\mathop{\rm x}\nolimits} \right)} \right) = {\mathop{\rm x}\nolimits} \) and \(\)\(U\left( {T\left( {\mathop{\rm x}\nolimits} \right)} \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\). Show that \(U\left( v \right) = S\left( v \right)\) for all v in \({\mathbb{R}^n}\). This will show that Thas a unique inverse, as asserted in theorem 9. [Hint: Given any v in \({\mathbb{R}^n}\), we can write \({\mathop{\rm v}\nolimits} = T\left( {\mathop{\rm x}\nolimits} \right)\) for some x. Why? Compute \(S\left( {\mathop{\rm v}\nolimits} \right)\) and \(U\left( {\mathop{\rm v}\nolimits} \right)\)].

Step by step solution

01

Show that T is onto mapping

For any v in \({\mathbb{R}^n}\), you can write \({\mathop{\rm v}\nolimits} = T\left( x \right)\) for some x (since \(T\) is onto mapping).

02

Show that \(U\left( v \right) = S\left( v \right)\) for all v in \({\mathbb{R}^n}\)

According to the assumed properties of Sand U, \(S\left( v \right) = S\left( {T\left( x \right)} \right) = x\) and \(U\left( v \right) = U\left( {T\left( x \right)} \right) = x\). Therefore, \[S\left( v \right)\] and \[U\left( v \right)\] are equal for any v.

This means that Sand U are the same functions from \({\mathbb{R}^n}\) into \({\mathbb{R}^n}\).

Thus, it is proved that \(U\left( v \right) = S\left( v \right)\).

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