Show that $\{a+H\,|\,a \in B_Q\}$ is a basis for $Q$ when $B_Q$ is an extension in $B_H \cup B_Q$ of $B_H$ to a basis for $V$

Question

Let $H$ be a subspace of $V$.

For $z \in V$, define $E(z) = \{z + h\,|\,h\in H\}$

Let $Q = \{E(v)\,|\,v\in V\}$.

Define Addition in $Q$: for $v, w\in V, E(v) \oplus E(w) = E(v+w)$
Define Scalar Multiplication in $Q$: for $v\in V$ and $\rho\in\mathbb{R},\, E(\rho) = E(\rho v)$

Let $B_H$ be a basis for $H$ where $B_H \cup B_Q$ is an extension of $B_H$ to a basis for $V$.

Show that $\{E(a)\,|\,a\in B_Q\}$ is a basis for $Q$.

Answer 1

Consider $T:V \rightarrow Q$ defined by $v \mapsto E(v)$. Clearly $T$ is surjective with kernel $H$ (since $z \in H \iff E(z) = H)$. So $T(v_i)$ must span $\text{im }T = Q$ for $v_i \in B_Q$. Now by rank nullity, we see that $|B_Q| = \dim \text{im }T = \dim Q$, and so $T(v_i) = E(v_i)$ must form a basis for the image, or $Q$.