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Q2.3-39Q

Expert-verified
Found in: Page 93

### 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

# Let $$T:{\mathbb{R}^n} \to {\mathbb{R}^n}$$ be an invertible linear transformation, and let S and 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 T has 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 $$U\left( {\mathop{\rm v}\nolimits} \right)$$].

It is proved that $$U\left( v \right) = S\left( v \right)$$.

See the step by step solution

## Step 1: 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).

## Step 2: Show that $$U\left( v \right) = S\left( v \right)$$ for all v in $${\mathbb{R}^n}$$

According to the assumed properties of S and 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 S and U are the same functions from $${\mathbb{R}^n}$$ into $${\mathbb{R}^n}$$.

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