# 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.