Let $T: V \rightarrow W$ be a vector space isomorphism. Prove that the $T$ maps proper subspaces to proper subspaces.

Question

Let $T: V \rightarrow W$ be a vector space isomorphism. Prove that the $T$ maps proper subspaces to proper subspaces.

Let $T:V_1 \rightarrow W_1.$ Prove that if $V_1$ is a proper subspace of $V,$ then $W_1$ is a proper subspace of $W.$

Consider a vector $\vec{0} \neq \vec{v} \notin V_1.$ For the sake of contradiction, assume $ T(\vec{v}) \in W_1. $ Then $ T(\vec{v}) = \vec{w} $ for some $ \vec{w} \in W, $ but since $T$ is an isomorphism, $T^{-1}:W_1 \rightarrow V_1.$ Therefore,$ T^{-1}(\vec{w}) = \vec{v} \in V_1, $ a contradiction.

Is this a sufficient proof?

Answer 1

You proof is not correct. $W_1$ is not any subspace of $W$. It's precisely $T(V_1)$. So let $V_1$ a proper subspace of $V$. If $T(V_1)=W$, then $V_1=V$ (recall that $T(V)=W$ and $T$ is injective). Therefore we get a contradiction.

Answer 2

That $T$ maps supspaces to supspaces should already be known for arbitrary linear transformations.

Being an isomorphism, $T$ is in particular an injective map, hence maps distinct subsets to distincts sets. In particular, $V_1\ne V$ implies $T(V_1)\ne T(V)=W$