Isomorphisms between finite vector spaces

Question

Let $U,V$ and $W$ be finite dimensional real vector spaces. Let $L:U\rightarrow V$ and $M:V\rightarrow W$ be one-to-one linear mappings. If $U$ and $W$ are isomorphic, prove that $V$ is isomorphic to $U$.

I've been thinking about this question for awhile now and I don't see how the existence of the one-to-one linear mappings $L$ and $M$ help me prove that $V$ and $U$ are isomorphic. Thanks for your help.

Answer 1

Hints.

1. Vector spaces are isomorphic iff they have the same dimension.

2. Use the rank-nullity theorem on (the matrix representations of) $L$ and $M$.

Answer 2

The condition "finite-dimensional" suggests that we look at bases and/or dimensions. Since $L,M$ are one-to-one we have $\dim V\ge \dim U$ and $\dim W\ge \dim V$. As $\dim W=\dim U$, we conclude $\dim V=\dim U =\dim W)$.

Answer 3

observe that $dim W \geqslant dim V \geqslant dim U$ and $dim U= dim W$

Answer 4

the linear map $T=M \circ L$ is one-to-one, so the dimension of the $Ker(T)$ is zero, apply the rank–nullity theorem

$dim(W) = dim(U) = dim(Ker(T)) + dim(Im(T)) = dim(Im(T))$

therefore, T is an isomorphism between U and W.

if L is not isomorphism we get that there is $v \in V$ that is not $L(u)$ for all $u \in U$.

from that we get, that $M(v) \neq M(L(u))$ for all vectors $u \in U$ because M is one to one, and if there is such vector $v=L(u) $ for some $u \in U$ contradicts our assumption that L is not isomorphism.

but if there is such $v$ we get that $T$ is not an isomorphism, because there is $w \in W$ such that $w = M(v) \neq T(u)$ for all $u \in U$, and that is contradiction, so we conclude that L is an isomorphism, therefore V is isomorphic to U.