Inverses of linear operators

References: edX online course MIT 8.05.1x Week 3.

Sheldon Axler (2015), Linear Algebra Done Right, 3rd edition, Springer. Chapter 3.

On a vector space {V}, a linear operator {T} has an inverse {S} if {TSv=STv=v} for all {v\in V}. Here, we’re restricting the arguments given in Axler’s section 3.D by assuming that all linear operators act from {V} back onto {V}. (Axler’s arguments allow {T} to map vectors from {V} into another vector space {W}.) We can show first that {S} is unique:

Theorem 1 If a linear operator {T} has an inverse, the inverse is unique.

Proof: Suppose that there are two distinct inverses {S_{1}} and {S_{2}}. Then

\displaystyle S_{1}=S_{1}I=S_{1}\left(TS_{2}\right)=\left(S_{1}T\right)S_{2}=IS_{2}=S_{2}

We’ve made use of the fact that {TS_{2}=S_{1}T=I}, the identity operator, by the definition of the inverse, and we’ve also used the associativity property of a vector space to go from the third to the fourth term. \Box

Because the inverse is unique, we can refer to it by the notation {T^{-1}}.

There is an important general result about operator inverses:

Theorem 2 A linear operator is invertible if and only if it is both injective and surjective.

Proof: We first recall the definitions of injective and surjective. An injective operator is one-to-one, so that if {Tv_{1}=Tv_{2}}, then {v_{1}=v_{2}}. A surjective operator has the entire vector space {V} as its range.

As this is an ‘if and only if’ proof, we need to prove the theorem in both directions. First, assume that {T^{-1}} exists, so that the operator is invertible. Then for {u,v\in V} suppose that {Tu=Tv}. Applying the inverse, we get

\displaystyle T^{-1}Tu=u=T^{-1}Tv=v \ \ \ \ \ (1)

so we must have {u=v}, making {T} injective.

Next, to prove that the existence of an inverse implies surjectivity, we note that we can write any vector {v\in V} as

\displaystyle v=T\left(T^{-1}v\right) \ \ \ \ \ (2)

That is, there is a vector {T^{-1}v} which, when operated on by {T}, gives any vector {v\in V}. Thus every vector {v\in V} is in the range of {T}, making {T} surjective.

Now we need to prove that if {T} is both injective and surjective, it has an inverse. Let the operator {S} be defined by the property that for any vector {w\in V}, {v=Sw} gives a vector {v\in V} such that {Tv=w}. Because {T} is injective, we know that {v=Sw} must be unique, since only one vector {v} can satisfy {Tv=w}. We also know that {v} must exist for every {w\in V} because {T} is surjective, and must produce every vector in {V} in its range. Since {Tv=T\left(Sw\right)=\left(TS\right)w=w}, we must have {TS=I}, so {S} is an inverse on the RHS side of {T}. To prove that {ST=I} as well, we have

\displaystyle T\left(STv\right)=\left(TS\right)Tv=ITv=Tv \ \ \ \ \ (3)

Comparing first and last terms, we see that {ST=I}. Thus {TS=ST=I} and {S=T^{-1}} so {T} has an inverse.

[The full proof also requires that we show that {S} is linear; the details are done in Axler’s theorem 3.56 if you’re interested.] \Box

In the above proof, we did not need to assume anything about the dimension of the vector space {V}, so that the result is valid for both finite and infinite-dimensional vector spaces. For a finite vector space, the result can be made even stricter. The fundamental theorem of linear maps states that for a finite vector space {V} and linear operator {T}

\displaystyle \mbox{dim }V=\mbox{dim null }T+\mbox{dim range }T \ \ \ \ \ (4)

If {T} is injective, then {\mbox{dim null }T=0}, since 0 is the only vector in the null space. Thus {\mbox{dim range }T=\mbox{dim }V} and {T} is surjective. Conversely, if we know that {\mbox{dim range }T=\mbox{dim }V}, then the theorem tells us that the null space has 0 dimensions. In other words, for a finite-dimensional vector space, injectivity implies surjectivity and vice versa. Thus, in this case, if we know that {T} is either injective or surjective, we know it has an inverse.

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