How to prove $\operatorname{rank}(M \circ L) <= rank(M)$ if $L$ and $M$ is inverse mapping of $L$

Question

Let $L:V\to W$ and $M:W\to V$ be linear maps, where $V, B$ are finite dimensional vector spaces.

Note: $L, V$ are not necessary one-to-one and onto, they are general linear mappings only.

I can prove $\DeclareMathOperator{rank}{rank} \DeclareMathOperator{nullity}{nullity} \rank(M \circ L) \leq \rank(L)$ like below:

1. Based on rank-nullity theorem, From linear map $L:\dim(V)=\nullity(L)+\rank(L)$ From linear map $M \circ L: dim(V)=\nullity(M \circ L)+\rank(M \circ L)$

2. I can prove $\nullity(L) \leq \nullity(M \circ L)$ like below: If any $x\in \ker L$, then we have $L(x)=0$, as $M$ is linear so $M(0)=0$, so combined together we have $M(0)=M(L(x))=(M \circ L)(x)=0$, means $x$ belong to $\ker(M \circ L)$, then $\ker(L)$ is subset of $\ker(M \circ L)$, hence $\nullity(L) \leq \nullity(M \circ L)$.

3. Based on 1 and 2: we can see $\nullity(L)+\rank(L) = \nullity(M \circ L)+\rank(M \circ L)$, as $\nullity(L) \leq \nullity(M \circ L)$, so $\rank(M \circ L) \leq \rank(L)$.

However, to prove $\rank(M \circ L) <= \rank(M)$, I'm blocked, not able to use above strategy, could you please help? Thanks a lot!

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I have another question, I'm working on it right now. Could you give me any hint? Still the same condition, if $L:V\to W$ and $M:W\to V$ are linear maps, and $V, B$ are finite dimensional vector spaces. $L, M$ are not necessary one-to-one and onto, they are general linear mappings only. Now, assume $0<\dim(W)<\dim(V)$, we need check if $M \circ L = Iv$, and can $L \circ M = Iw$, with example if can, and explain why if no. $I$ is the identity map here.

I'm thinking we still need rank-nullity theorem, but do not have a way to solve it.

Please advise. Thanks!

Answer 1

Think I found the solution to prove rank(M∘L)≤rank(M). Thanks the hint from @red_trumpet and @P.Lawrence

Step 1: prove Range(M∘L) ⊆ Range(M) Let y∈Range(M∘L), then means there's a x∈V, such that: y∈{M(L(x))∈V : x∈V} by definition of range As L(x)∈W, we can write L(x)=z, such that z∈W, then y∈{M(L(x))∈V : x∈V} can write to: y∈{M(z)∈V : z∈W} which is the Range(M) Hence Range(M∘L) ⊆ Range(M)

Step 2: As we proved Range(M∘L) ⊆ Range(M) then dim(Range(M∘L)) <= dim(Range(M)) hence rank(M∘L) <= rank(M) by definition of rank.