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### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

# Suppose T and 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?

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.

See the step by step solution

## Step 1: Show that the mapping is identity mapping

Consider A and B are the standard matrices of T and U. Then, by matrix multiplication, AB is 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, $$AB = I$$.

## Step 2: 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 $${\mathbb{R}^n}$$.