Halmos Section 36: Inverses (Pt. I)
Point of post: In this post we complete the problems at the end of section 36 in Halmos.
Problem: Which of the following linear transformations are invertible?
a) , where is considered a real vector space.
b) where .
c) , where we clearly this is meant to mean that is the unique element of for which
a) Since it suffices to show that is injective. But, this is clear since
and so injectivity follows. In fact, a quick computation shows that .
c) We merely recall that it’s sufficient to check that injects a basis into a basis. Recall though that
is a basis for and evidently is an injection and so is a monomorphism. But, since is finite dimensional the conclusion follows.
d) Suppose that and let be arbitrary. We then see that
But, since was arbitrary it follows that . Thus, and so is a monomorphism. Since is finite dimensional we may conclude that is invertible.
e) This is not invertible because it isn’t injective. Recall that for a fixed is a bijection. Thus,
Thus, choosing a non-symmetric and using the above shows that is not injective and thus not invertible.
f) The same concept applies
so choosing an even and a suitable the above violates injectivity. Thus, is not injective thus not invertible.
Problem: Consider given by
Prove that if and only if .
Proof: Suppose that . If we’re done since the equation is trivially satisfied and , and thus is not injective. So, assume not. Note then that
and since it follows that and so is not injective and thus not invertible. Thus, if then .
Conversely, suppose that . Now, clearly we can’t have and otherwise contradictory to our assumption. So, suppose that but . Then, we may assume without loss of generality that . So, by assumption
We see from that
But, note that
and so plugging into this gives
contradictory to our assumption. Thus, it follows that and thus is injective. Therefore, since is finite dimensional we may conclude that .
1. Halmos, Paul R. Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. Print