Isomorphism and dimension

The problem: Let $V$ and $W$ be finite-dimensional vector spaces and $T:V \rightarrow W$ be an isomorphism. Let $V_o$ be a subspace of $V$. Prove that $T(V_o)$ is a subspace of $W$ and that $dim(V_o) = dim(T(V_o))$.

Please let me know if my solutions are correct.

My solution to the first claim: Because $V_o$ is a subspace of $V$, $0_v \in V_o$. Because $T$ is an isomorphism, $T$ is linear, and $T(0_v) = 0_w$, so $T(O_v) \in W$. Let $u,v \in V_o$ and let $c \in F$. Then $T(cu + v) = cT(u) + T(v) \in W$ because $T$ is linear. Therefore $T(V_o)$ is a subspace of W.

My solution to the second claim: Because $T:V \rightarrow W$ is an isomorphism, T is both one-to-one and onto. Because $V_o$ is a subspace of V, $V_o \subseteq V$ and for all $v_1,v_2 \in V_o$, $T(v_1) \neq T(v_2)$, so $T:V_o \rightarrow T(V_o)$ is one-to-one. Furthermore, because $T(V_o)$ is a subspace of $W$, $T(V_o) \subseteq W$ and for every $w \in T(V_o)$, there exists a $v \in V$ such that $T(v) = w$. Assume that $v \in V - V_o$. Then $T(v) \in T(V-V_o) = T(V) - T(V_o)$, which is a contradiction. Therefore $v \in V_o$, and $T:V_o \rightarrow T(V_o)$ is onto.

A linear map is an isomorphism if it is invertible. I know that isomorphic vector spaces should have the same dimension and I am trying to use that theorem in the proof, but I need to show first that $T:V_o \rightarrow T(V_o)$ is an isomorphism (invertible) and I'm not sure how to do that. – user90593 Oct 15 '13 at 2:39
Hint: a linear map is $\;1-1\;$ iff it maps linearly dependent sets to linearly dependent sets... – DonAntonio Oct 15 '13 at 4:02