I'm studying a problem from linear algebra:
For a bilinear form $b: V \times V$ and a subvector space $U $ in $ V $ we set
$ U^{\perp}:=\{v \in V \mid b(u, v)=0 \text { for all } u \in U\} . $
Now let $ A=\left(\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right)$. We consider the bilinear form
$ b: \mathbb{R}^{2} \times \mathbb{R}^{2} \longrightarrow \mathbb{R}, \quad(x, y) \mapsto x^{t} A y $
Prove or disprove (by giving a counterexample): For every $ \mathbb{R} $ subvector space $U $ in $ \mathbb{R}^{2}$ we have $ \left(U^{\perp}\right )^{\perp}=U$.
This should be correct because we just deal with finite dimensional vector spaces.
Here is my proof:
Let $ u \in U $ with $\langle u, w\rangle =0 \quad \forall w \in U^{\perp},$ so $ u \in U^{\perp \perp} $.
Consequently $ U \subseteq U^{\perp \perp} $.
Since we know that V is the direct sum of $U$ and $U^{\perp}$:
$\operatorname{dim}(U) = \operatorname{dim}(V)-\operatorname{dim}\left(U^{\perp}\right) =\operatorname{dim}(V)-\left(\operatorname{dim}(V)-\operatorname{dim}\left(U^{\perp \perp}\right)\right)=\operatorname{dim} \left(U^{\perp \perp}\right),$
it follows $U=U^{\perp \perp}$.
Best Answer
$\qquad$Disproof !
The specific bilinear form you propose allows for the canonical counter-example $\,U = \{0\}\,$ since the defining matrix $A$ is singular, which corresponds to the associated bilinear form being degenerate.
We clearly & always have $\,U^\perp = \{0\}^\perp = V$, but then & here $$\left(U^{\perp}\right )^{\perp}\;=\; V^\perp\;=\; \langle{1\choose 1}\rangle \;\neq\;\{0\}=U.$$ Going into more detail:
The defining matrix $A$ has the eigenvalues $\{0,2\}$ with respective (non-normalised) eigenvectors ${1\choose 1}$ and ${1\choose -1}\,$. As a statement about diagonalisation of $A$ this may be written as $$\begin{pmatrix}1 & -1 \\ -1 & 1\end{pmatrix} \;\frac1{\sqrt 2}\begin{pmatrix}1& 1\\ 1& -1\end{pmatrix} \;=\; \frac1{\sqrt 2}\begin{pmatrix}1& 1\\ 1& -1\end{pmatrix} \begin{pmatrix} 0& 0\\ 0& 2\end{pmatrix}$$
Regarding your proof attempt:
Your first conclusion $U\subseteq U^{\perp \perp}$ is true.
In case of degeneracy, $V$ is in general not the direct sum, e.g., it does not fit for $U^{\perp}$ and $U^{\perp\perp}\,$.