Question

Suppose Tand U are linear transformations from \({\mathbb{R}^n}\) to \({\mathbb{R}^n}\) such that \(T\left( {U{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\) . Is it true that \(U\left( {T{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\)? Why or why not?

Short answer

It is proved that \(U\left( {T{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \), for all x in \({\mathbb{R}^n}\), is true.

Step-by-step solution

Show that the mapping is identity mapping

Consider Aand Barethe standard matrices of Tand U. Then, by matrix multiplication, ABis the standard matrix of the mapping \(x \mapsto T\left( {U\left( x \right)} \right)\). The mapping is identity mapping. Thus, according to the hypothesis, \(AB = I\).

Determine whether \(U\left( {T{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \), for all x, is true

The matrices are invertible because, according to the invertible matrix theorem, A and B are square matrices and \(B = {A^{ - 1}}\). Therefore, \(BA = I\). It follows that the mapping \(x \mapsto T\left( {U\left( x \right)} \right)\) is identity mapping. This indicates that \(U\left( {T{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\).

Thus, it is proved that \(U\left( {T{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\).