Invertible Linear Map

Question

Suppose $V$ is finite-dimensional and $S,T\in \mathcal{L}(V)$. Prove that if $ST$ is invertible, then both $S$ and $T$ are invertible.

This is the partial question retrieved from Linear Algebra Done Right, and I have come out with different solution from the solution guide.

My solution: Since $ST$ is invertible, then the only possibility for $(ST)u=0$ is when $u=0$. Therefore I have $(ST)(0)=0=S(T(0))=0$. Since $T(0)=0$ (as proven in book), I must have $S(0)=0$. Since $T(0)=0$ and $S(0)=0$ comes directly from $(ST)(0)=0$, both $S$ and $T$ are injective. Since both of them are linear operator, we can deduce they are both invertible (by equivalece statement.)

Is my proof valid?

Answer 1

No. For any linear map $f$ it is true that $f(0)=0$. Therefore, nothing can be deduced from the fact that that happens for a specific linear map.

Answer 2

Note that since ST is invertible

$$\forall u\neq 0\iff (ST)u\neq0$$

then since

$$(ST)u=(S)((T)u)\neq0 \implies v=(T)u\neq0$$

thus T is invertible.

Since T is surjective, we have that $\forall v\neq0 \quad\exists u\neq0 \quad v=(T)u$ and $S(v) \neq 0$, therefore also S is invertible.

Answer 3

As José Carlos Santos stated, your proof is not correct. You have to show two things: injectivity and surjectivity of both $S$ and $T$. If you want to use zero element in your proof then injectivity can be shown by proving the following implication $$ f(u) = 0 \Rightarrow u = 0 \ ,$$ that is, you do not assume in the beginning that $u$ is zero.

Other hint that might be useful in the proof: If $ST$ is injective then $T$ must be injective. If $ST$ is surjective then $S$ must be surjective.