## Halmos Section 36: Inverses (Pt. III)

**Point of post: **This is a continuation of this post.

8.

**Problem: ** If is an -space and . Prove that if has precisely one right inverse then is invertible.

**Proof:**** **We merely note that if (taking to be the unique right inverse) then

so that is a right inverse for , thus by assumption and so solving this for gives . Thus and .

9.

**Problem: **If is a finite dimensional -space and then there exists a polynomial such that .

**Proof: **It’s true in general that given that there exists a minimal polynomial which annihilates this, but for the invertible case its’s easier to show that there exists just a polynomial with non-zero constant coefficient which annihilates . Intuitively, since we know that there exists some non-zero polynomial

which annihilates . If then we may factor this as

but since is invertible and the above expression identically zero we may conclude that

continuing with this process (realizing that it must clearly end otherwise we arrive at is identically zero, which implies that which contradicts that ) we arrive at a polynomial

for which and . Thus, we see that

multiplying both sides by proves that

from where the conclusion follows.

R*emark: *Too the interested/precocious reader: you can attempt to prove the fact stated in the first sentence of the above paragraph by first considering the polynomial ring and and showing that is a proper ideal of . Prove then that is generated by some monic polynomial, and this polynomial is the one we seek).

10.

**Problem: **Formulate a sensible definition of invertibility for linear transformations from one -space to another -space . Use that definition to determine which of the following linear transformations are invertible:

**a)**

**b) **

**c) **Assuming that we may define

**d) **Fix and define

where .

**Proof: **If we merely call invertible if it’s bijective.

**a) **Recall from an earlier post that if and is such that then .In other words, . Thus, recalling that will be a monomorphism if and only if which is true if and only if we may conclude that is an isomorphism (recalling that a non-zero linear functional is always surjective) if and only if , in which case in which case, by an appropriate composition of maps, we can interpret as being an endomorphism.

**b) **This is evidently not a monomorphism (in most cases) since . Thus, is invertible (since it’s evidently surjective) if and only if in which case and the same idea of being able to view as an endomorphism discussed in **a) **still applies.

**c) **Note that if were an isomorphism then and thus . Thus, (as one can easily check the other direction) is an isomorphism if and only if .

**d) **Once again using a dimension argument we may conclude that the only time this could possibly be an isomorphism is when (using a dimension argument). Moreover, this is highly dependent upon the identity of . For example, if then this is evidently not an isomorphism. In fact, we can say this mapping is an isomorphism precisely when and is *non-degenerate in the second column*. In the sense that for all implies that . To see this we first note that if then for all , which by assumption implies that . Thus, is injective. Recalling though that since we may conclude that is an isomorphism (since is a subspace of with equal dimension).

**References:**

1. Halmos, Paul R. *Finite-dimensional Vector Spaces,*. New York: Springer-Verlag, 1974. Print.

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