Concept check: Let V and W be subspaces, and $T: V \rightarrow W$ be linear.

Question

Concept check: Let V and W be vector spaces, and $T: V \rightarrow W$ be linear. My question is if we find that $T: V \rightarrow W$ is invertible, can we, without proving, say that there exists a $U: W \rightarrow V$ such that U is linear and invertible?

Answer 1

It ultimately depends on the context how much you can take as given. In a first course on linear algebra, if this fact has not been demonstrated in class, I would say it cannot be taken as given for instance. That said, the fact is true: invertible linear transforms have an inverse which is itself a linear transform, i.e. $U=T^{-1}$ exists in your case, and is in fact invertible. Outside of first courses, I would ultimately say that this is almost so obvious that it can be taken as a given without a problem.

Of course this ultimately would come down to context as mentioned. If this is for a class, talk to your professor - they will ultimately know far better than we would on the matter.

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That said...

Proving that $T^{-1}$ is linear and invertible is easy enough to do from the definitions. I'll show one property for you, leaving the rest for you - they have a similar "flavor". Linear transforms should be additive, and thus we want $T^{-1}(x+y) = T^{-1}(x) + T^{-1}(y)$. Also note that $x = T(T^{-1}(x))$ since $T^{-1}$ is the inverse. The same holds for $y$ obviously. Then

$$T^{-1}(x+y) = T^{-1}(T(T^{-1}(x)) + T(T^{-1}(y))) \tag 1$$

Since $T$ is known to be linear, $T(T^{-1}(x)) + T(T^{-1}(y)) = T(T^{-1}(x) + T^{-1}(y))$ - it's the definition of linearity, but in reverse. Thus, $(1)$ implies

$$T^{-1}(x+y) = T^{-1}(T(T^{-1}(x) + T^{-1}(y))) \tag 2$$

The outermost $T^{-1},T$ negate owing to being inverses and thus

$$T^{-1}(x+y) = T^{-1}(x) + T^{-1}(y)$$

giving linearity.

What remains to be shown:

• $T^{-1}(ax) = aT^{-1}(x)$ for scalars $a$
• $(T^{-1})^{-1}$ exists, i.e. show $T^{-1}$ is invertible (show $T$ is the inverse)

Answer 2

Whether or not you can say such a thing without giving a proof depends on what you have agreed with the person who reads and evaluates. It is universally known that $T^{-1}$ is linear and bijective if $T$ is linear and bijective.