Consider $\operatorname{Bil}(V)=\{\text{bilinear maps}\ V\times V\rightarrow F\}$. Let $\dim V<\infty$, $\operatorname{char}F\neq 2$, and let $w\in\operatorname{Sym}(V)$, ie. $w$ is symmetric ($w(x,y)=w(y,x)$ for all $x,y\in V$). Then there exists a basis for $V$ which is orthogonal w.r.t. $w$.
A proof is given in the textbook, which I'm having trouble understanding.
Proof: By induction on $n=\dim V$. If $V=\{0\}$, there's nothing to prove. Also, if $w=0$, every basis is orthogonal, so assume $w\neq 0$. Then $w$ is not skew and hence not alternating.
Let $x_1\in V$ be such that $w(x_1,x_1)\neq0$. Consider the linear form $y(x)=w(x_1,x)$ for $x\in V$. Here $y\in V'\setminus\{0\}$ and hence its nullspace $N(y)=\{x_1\}^\bot$ has dimension $n-1$. By induction it has an orthogonal basis $\{x_2,\ldots,x_n\}$ and then $\{x_1,x_2,\ldots,x_n\}$ is an orthogonal basis for $V$.
We say $x$ and $y$ are orthogonal if $w(x,y)=0$.
For $U\subset V$ then $U^\bot=\{y\in V \mid \forall x\in U: w(x,y)=0$}.
$V'$ denotes the dual space of $V$.
Now here are my questions.
- In the case where $n=0$, then $V=\{0\}$ since $V$ must have a neutral additive element. $0$ can't be a basis element or else we wounldn't have unique ways to write elements in $V$ as linear combinations, right? But how does there "exists a basis for $V$ which is orthogonal w.r.t. $w$"? What basis is that?
- Next as we define $y(x)$, why do we mention that $y\in V'\setminus\{0\}$? By writing out the sets, isn't it clear that $N(y)=\{x_1\}^\bot$?
- Why does $\{x_1\}^\bot$ have dimension $n-1$?
- At last, we get that $\{x_2,\ldots,x_n\}$ is an orthogonal basis for $\{x_1\}^\bot$, but why does that imply that $\{x_1,x_2,\ldots,x_n\}$ is a basis for $V$? I do however see why it's orthogonal.
Here are some answers.
The basis of the zero vector space $V=\{0\}$ is the empty set $\varnothing$. This may seem a bit counterintuitive, but if we check we can see that it formally satisfies the definition of a basis.
To emphasize that $y$ is NOT the identically zero functional. That's important because if $y=0$ (in $V'$), i.e if $y$ is the identically zero functional, then its kernel (or null-space) is $\ker(y)=N(y)=\{x_1\}^{\bot}=V$, the entire space $V$, and we can't proceed in any useful way. However, since $w(x_1,x_1)\ne0$, we know that $y(x_1)\neq0$, so $y$ is not the zero functional. Therefore, $\ker(y)=N(y)=\{x_1\}^{\bot}\subsetneq V$, a proper subset of $V$.
And yes, of course, $N(y)=\{x_1\}^{\bot}$. But once again, the real question is whether it is all of $V$ or not.
By the Rank-Nullity Theorem.
Because they are $n$ linearly independent vectors in an $n$-dimensional vector space.