Subspace and orthogonal complement span whole space even if the form is degenerate? Let $V$ be a finite dimensional vector space over a field $\mathbb{R}$. Let $B\colon V\times V\rightarrow \mathbb{R}$ be a bilinear form. 
Assume that $B$ is degenerate.
Q.1 Does $V$ has orthogonal / orthonormal basis?
Q.2 If $U$ is a subspace of $V$, then is it always true that $U$ and $U^{c}$ (the orthogonal complement of $U$) span whole $V$?
 A: By definition of a degenerate form, there exists $x\in V$, $x\neq0$, such that $B(x,y)=0$ for all $y\in V$. In particular, if $\{v_1,\ldots,v_n\}$ is a basis of $V$ then $B(x,v_j)=0$ for all $j$. If we assume $B(v_i,v_j)=0$ for $i\neq j$ then this would imply $B(v_j,v_j)=0$ for some $j$. I suppose you could in principle define an orthogonal basis to be one where $B(v_i,v_j)=0$ for $i\neq j$, regardless of whether or not $B(v_j,v_j)=0$, in which case the Gram-Schmidt process would still work.
The second question is true although the sum is not necessarily direct. Take a basis $\{v_1,\ldots,v_k\}$ of $U$, extend to a basis $\{v_1,\ldots,v_n\}$ of $v$, apply Gram-Schmidt to get a basis $\{e_1,\ldots,e_n\}$ of $V$ such that $\{e_1,\ldots,e_k\}$ spans $U$ and $B(e_i,e_j)=0$ for $i\neq j$. If $x=\sum_{j=1}^na_je_j\in V$, then $x=y+z$ where $y=\sum_{j=1}^ka_je_j\in U$ and $z=\sum_{j=k+1}^na_je_j\in U^c$, this last statement following since $B(e_i,e_j)=0$ for all $1\leq i\leq k,k+1\leq j\leq n$. If $x\in U$ is such that $B(x,y)=0$ for all $y\in V$ then $x\in U\cap U^c$.
