A basis of $V$ is linearly independent in $H$ where $V \subset H$ are Hilbert spaces?

Question

Let $V \subset H$ both be separable Hilbert spaces with continuous and dense embedding.

Let $\{v_j\}$ be a basis for $V$, so every $v \in V$ can be written as $v = \sum_{j=1}^\infty a_jv_j$ with $a_j \in \mathbb{R}$.

Why is it true that $\{v_1, ..., v_n\}$ is a linearly independent set in $H$ for each fixed $n$?

I see no mention of linear independency for the infinite-dimensional basis $\{v_j\}$.

Answer 1

By definition, an embedding $V \subset H$ (usually) means that the inclusion map $\iota : V \to H$ is injective.

It is an easy exercise to show that an injective linear map maps linear independent sets onto linear independent sets.

By definition of a basis (in what sense whatsoever), $(v_j)$ is linearly independent in $V$, hence in $H$.