A basic question on orthogonal vector

Question

Let $V$ be a finite dimensional vector space and $X$ be a subspace. Let $$\langle u,y\rangle=0 \forall u $$ with the property that $$\langle u,x\rangle =0 \;\forall x \in X$$ where $u,y \in V$. Then I need to prove that $y \in X$.

This looks trivial intuitively. But finding it difficult to prove it. I guess I need to write $y$ as a combination of elements of $X$.

Actually I am trying to prove that orthogonal complement of an orthogonal complement of a subspace is a subset of that subspace (actually it is equal, but I have proved the other part)

Answer 1

We have $y\in X \Rightarrow \langle u,y \rangle=0 $, need to show that $\langle u,y \rangle=0 \Rightarrow y\in X $. I think this depends on $X$, if e.g. $X$ is a 2D subspace in 3D $V$, then I guess it's true, while it is not true if $X$ is a 1D subspace in 3D $V$.

Answer 2

Let $w \in (V^{\bot})^{\bot}$. I need to prove that $w \in V$.

Now, $w$ can be expressed uniquely as $w= v + v'$ where $v \in V$ and $v' \in V^{\bot}$. Now, because $V^{\bot}$ and $(V^{\bot})^{\bot}$ are complement of each other $$\langle w, v' \rangle=0$$ which implies $v'=\theta_{v}$ using the fact that $\langle v, v' \rangle$=0. This implies $w=v$ thus $w \in V$

Answer 3

If the kernel of $V$ has one than more dimension (lets say $n > 1$), $X$ could be a subspace with dimension $1$, and $y$ could belong to $Y$, where $\operatorname{Ker}\{V\} = X \oplus Y$. Therefore $y \notin X$.

On the other hand, if the kernel of $V$ has dimension one or less, then it must be true that $y \in X$, because $\operatorname{Ker}\{V\} = X$.