Suppose is finite-dimensional, is a subspace of , and . Prove there exists an invertible operator such that for every if and only if is injective.
Suppose is injective. Let be a basis for then extend to a basis of . since is injective is independent, meaning we can extend it to a basis of .
Finally define and for all . is obviously surjective, implying is invertible by 3.69. (You can also show injectivity directly by using the independence of .)
Now suppose is not injective on . This would imply is not injective on hence cannot be invertible.