# Proving that an injective and surjective linear map is invertible

I'm trying to understand how a surjective and injective linear map is invertible. I'm using Axler's book but I found the proof there hard to follow (in one of the directions only; I can see why an invertible linear map is surjective and injective).

Say $$T: V \to W$$ is a linear map from vector space $$V$$ to vector space $$W$$. It is injective and surjective. I want to prove that there exists a linear map $$S:W \to V$$ such that $$TS = I, ST = I$$ where each $$I$$ is the identity map in $$W$$ and $$V$$ respectively.

Axler's proof introduces $$S$$ as:

For each $$w \in W$$, define $$Sw$$ to be the unique element of $$V$$ such that $$TSw$$ = $$w$$ (the existence and uniqueness of such an element follow from the surjectivity and injectivity of $$T$$).

First of all, what is $$S$$ here? Is it a function or some sort of operator? Or is it a linear map $$S:W \to V$$?

But the proof ends with a final section:

To complete the proof, we need to show that $$S$$ is linear

So $$S$$ was not a linear map, then what was it?

### If $$S$$ can be taken as a linear map

• If I assume $$S:W \to V$$, I can use the surjective $$T$$ to say that $$\exists v \in V$$ such that $$Tv = w \forall w \in W$$.
• Additionally with injective $$T$$, $$\forall w_1, w_2 \in W$$ such that $$w_1 \neq w_2$$ we know $$v_1, v_2 \in V$$ exists such that $$Tv_1 = w_1, Tv_2 = w_2$$. If $$v_1 = v_2$$ then $$Tv_1 = Tv_2 \implies w_1 = w_2$$ which contradicts our assumption that $$w_1 \neq w_2$$. So $$v_1 \neq v_2$$. So $$w_1 \neq w_2 \implies v_1 \neq v_2$$.
• Define $$S$$ as a map from each $$w \in W$$ to a unique (proved to exist in previous point) $$v \in V$$ such that $$Sw = v$$ where $$Tv = w$$. So $$TSw = Tv= w$$.

Is this a valid way to interpret Axler's bracketed text?

What is happening is the following.

Let $$T: V \to W$$ be a linear map that is both injective and surjective. We want to construct an inverse $$S: W \to V$$. (

Given $$w \in W$$, how would we define $$Sw \in V?$$ Well, since $$T$$ is injective and surjective, there exists a unique element $$v \in V$$ with $$Tv = w$$. This uniqueness allows us to define $$Sw:= v$$.

Axler used suggestive notation and immediately wrote $$v = Sw.$$

Note that you only need to check that $$S$$ is an inverse of $$T$$ and not that it is linear. The inverse of a linear map is always automatically linear (exercise!).

• But what is $Sw$? I assume it cannot be a linear map since "To complete the proof, we need to show that S is linear" is the last section of the proof. Commented Dec 19, 2020 at 10:40
• I told you what $Sw$ is: it is the unique element $v \in V$ satisfying $Tv = w$. Once you have shown that $S$ is inverse to $T$, linearity comes for free. Indeed, $S( v+w) = S(TSv + TSw) = S(T(Sv+Sw)) = Sv + Sw$ and similarly for scalar multiplication. Commented Dec 19, 2020 at 10:42
• Thanks for the useful note about the inverse. So I guess $S$ at first is simply a map, not necessarily linear? Commented Dec 19, 2020 at 10:45
• @sprajagopal Yes, you first define $S$ as a map. A priori, you don't know it is linear. Then, you notice that $S$ is inverse to $T$, and then by my previous comment $S$ will be linear. Commented Dec 19, 2020 at 10:46

Here, $$S$$ is a map from $$W$$ to $$V$$. Asserting that $$T\colon V\longrightarrow W$$ is invertible means that there is a linear map from $$W$$ to $$V$$ such that:

• $$(\forall v\in V):S(T(v))=v$$;
• $$(\forall w\in W):T(S(w))=w$$.
• To complete the proof, we need to show that S is linear" is the last section of the proof. So at the beginning there is no assumption that $S$ is a linear map. Is it possible it's a different kind of map? Commented Dec 19, 2020 at 10:40
• Axler first defines a map $S\colon W\longrightarrow V$ and then he proves that $S$ is linear. Commented Dec 19, 2020 at 10:42

If $$T:V\to W$$ is a linear map then also it is a function.

It is well known that a function is bijective if and only if it has an inverse.

Also it is well known that this inverse is unique.

Denoting this inverse as $$S:W\to V$$ we have $$T\circ S=\mathsf{id}_W$$ and $$S\circ T=\mathsf{id}_V$$.

What remains now is proving that this $$S$$ is a linear map.