Show that $L$ is an isomorhism iff $L\circ L$ is an isomorphism too.

Question

Let $V$ be a vector space. Show that a linear operator $L:V\rightarrow$ $V$ is an isomorphism if and only if $L\circ L$ is an isomorphism.

My attempt at a solution:

Since $L\circ L$ is an isomorphism then $L^{-1}$ exists and thus we set $L(V)=0$ $$L^{-1}(L(V) \Longrightarrow V=0$$ Thus $L(V)$ is one-to-one. Then since $L(V)$ is one-to-one this implies that $dim(ker(L(V)))=0$ so by the dimension theorem we have that L(V) is onto.

Im aware that I must prove the proposition in the other direction but is this a correct proof for the forward direction? Any feedback is apprecited!

Answer 1

One direction is obvious since composition of isomorphism is an isomorphism again.

To prove the other direction: we know that $L\circ L$ is injective and surjective, so: let $v\in V$, and suppose $L(v)=0$, then $L(L(v))=0$, that is $L\circ L(v)=0 \Longrightarrow v=0,$ thus $L$ is injective. Now $L\in \textsf{End}(V)$, thus $\textsf{dim}(Ker L)=\textsf{dim} (V)-\textsf{dim}(Im L)=0$ then $\textsf{dim} (V)=\textsf{dim}(Im L)$ so $L$ is also surjective. Thus $L$ is an isomorphism.

Answer 2

If $L\circ L$ is an isomorphism, then $(L\circ L)^{-1}$ exists. Claiming the existence of $L^{-1}$ is exactly what you need to prove.

Suppose $L\circ L$ is an isomorphism. Then there exists $M$ which is a left and right inverse of $L\circ L$: $$ M\circ (L\circ L)=1_V,\qquad (L\circ L)\circ M=1_V $$ Can you tell whether $L$ has a left and right inverse?

Answer 3

This is a purely algebraic statement. If $L\circ L$ is an isomorphism, then there is an inverse $A$ such that $L\circ L\circ A=I$. Then $L\circ (L\circ A)=I$, so $L\circ A$ is a right inverse for $L$ and hence $L$ is surjective. The dual of this argument shows that $L$ is injective as well, hence an isomorphism.

We did not use any properties of vector spaces here in particular.