How do I prove that something is a subspace AND an isomorphic? Let φ : V → W be an isomorphism of vector spaces V and W. Let U ⊆ V be a subspace of V. Let φ(U) = { φ(u)| u∈U } ⊆ W. Prove that φ(U) is a subspace of W and that U is isomorphic to φ(U).
I know that in an isomorphism, there is an invertible linear transformation the two vector spaces. Therefore, if V U ⊆ V is a subspace of V, then there is an invertible linear transformation in that subspace. Furthermore, since φ is already an isomorphism, U is isomorphic to φ(U) since U is an element of V. I don't know if this is even close to the right answer.
 A: To prove that $\varphi(U)$ is a subspace of $W$, we simply need to verify that it is closed under addition and scalar multiplication. Take $w,z \in \varphi(U)$, $\alpha \in F$ (where $F$ is the field that $V$ and $W$ are defined over. Since $w,z \in \varphi(U)$, there are $x, y \in U$, such that $\varphi(x) = w, \varphi(y) = z$. Since $U$ is a subspace of $V$, it is closed under addition and scalar multiplication. Then $x+y \in U$ and $\alpha x \in U$. By linearity of $\varphi$, we have $\varphi(x+y) = \varphi(x) + \varphi(y) = w + z$ which shows that $w+z \in \varphi(U)$. Similarly, $\varphi(\alpha x) = \alpha\varphi (x) = \alpha w$ so $\alpha w \in \varphi(U)$. Thus $\varphi(U)$ is closed under addition and scalar multiplication so it is a subspace of $W$.
The proof that $\varphi$ is an isomorphism from $U \to \varphi(U)$ is almost trivial. We have that $\varphi$ is a surjective map from $U \to \varphi(U)$ by definition of $\varphi(U)$. Also, $\phi$ is an injective map from $V \to W$ so in particular it is injective on $U$. Hence it is bijective and so it is a isomorphism.
