Vector space $V = V^{\bot}\oplus W$, prove $W$ is "hyperbolic"

Question

The orthogonality here is determined by an alternating bilinear form $\phi$

My professor used the following definition of a "hyperbolic space" (usually this means another thing, if anyone knows how it is usually named, pls say it).

First, define "hyperbolic plan": a hyperbolic plan $H$ is a subspace of dimension two generated by two vectors not mutually orthogonal, i.e., $H=\operatorname{span}\{v,w\}$ with $\phi(v,w)\not=0$. Then a subspace $W$ is said to be hyperbolic if it is a direct sum of orthogonal hyperbolic plans,i.e., $W=H_1\oplus\cdots \oplus H_n~ \text{with}~~ H_i\bot H_j$ for each $i\not =j$.

Then, prove the following: Let $\phi$ be an alternating bilinear form, then every subspace $W$ such that $V = V^{\bot}\oplus W$ is a hyperbolic subspace.

My professor gave us the proof, by I really didn't understand anything, if anyone could explain more the steps/clarify then I'd be very thankfull!. The proof follows (I'll put some "? and ok" according to my doubts, but I didn't understand the whole thing):

"$V = V^{\bot}\oplus W \Rightarrow \phi$ not degenerated (ok). Without loss of generality suppose that $V^{\bot}=\{0\}$ ($?_1$) and we'll prove by induction on $dim(V)$.

Given $v_1 \in V\backslash \{0\}$ take $w_1$ such that $\phi(v_1,w_1)\not = -1$ (possible because $W$ not degenerated $?_2$) and $V_1=span\{v_1,w_1\}$ wich is non-degenerated $\Rightarrow V=V_1 \oplus V_1^{\bot}$ (ok)

It follows that $V_1^{\bot}$ is non-degenerated (ok). Then, by the induction hypothesis, $V_1^{\bot}$ is a hyperbolic subspace. $\blacksquare$ $(???_3)$"

I didn't get how induction was really used. Sorry for the long post, but I really need to understand this one.

Answer 1

Supposing w.l.o.g. that $V^{\perp} = \{0\}$ is a consequence of the following observation: if $V = V^{\perp}\oplus W$, then $\phi|_W$ is not degenerated. Note that we are aiming to prove that $W$ has an hyperbolic basis. By the observation above, we must deal with $\phi |_W$, which is always non-degenerated. So it's not a problem to assume that $\phi$ is indeed non-degenerated for the whole $V$.

Assumed that, it follows that exists at least one pair of vectors $v,u\in V\setminus \{0\}$ such that $\phi(v,u)=a\neq0$ , otherwise we will be contradicting the fact that $V^\perp=\{0\}$. Thefore, we can define $w=-\dfrac{1}{a}u$ and hence $\phi(v,w)=-1$, which shows that the pair $(v,w)$ is indeed hyperbolic.

Setting $V_1=[v,w]$, it follows that $V_1$ is not degenerated, because $\phi(v,w)=-1\neq0$ does ensure it. As you have notice, we can also write $V=V_1\oplus V_1^{\perp}$. The fact that $\phi$ is non-degenerate also implies that $V_1^{\perp}$ is not degenerate.

And here comes the use of induction hyphotesis. We have assumed that for all dimensions less than $n$ the claim is true. That is, for all dimensions less than $n$, if $V=W\oplus V^{\perp}$, then $W$ is hyperbolic.

Since $\mbox{dim}(V_1)=2$, we have $\mbox{dim}(V_1^{\perp})=n-2$, and this number is a suitable number for applying the induction hyphotesis. We have proven that $V_1^{\perp}$ is not degenerate. Therefore, we can write $V_1^{\perp} = W'\oplus (V_1^{\perp})^{\perp}$ for some subspace $W'$ of $V_1^{\perp}$. Now by induction hyphotesis we guarantee $W'$ hyperbolic. But since $\phi$ is not degenerate, it is true that $(V_1^{\perp})^{\perp} = V_1$. Now we write $V_1^{\perp} = W'\oplus V_1$ and this shows that $V_1^\perp$ is hyperbolic as a sum of two hyperbolic ones.

Recalling the decomposition of $V$, we have proven that $V$ is itself hyperbolic.

Answer 2

Based on your comments, I'll assume the following:

• $V$ is a finite dimensional vector space over a field $k$ whose characteristic is not $2$.
• $\phi:V\times V\to k$ is bilinear and $\phi(v,w)=-\phi(w,v)$
• $V^\perp=\{v\in V\mid\phi(v,w)=0\text{ for all }w\in V\}$.

Then we can prove the statement by induction on $\dim V$. The case $V^\perp=V$ (ie $\phi=0$) is trivial since $W$ must be zero. Suppose otherwise and pick $v\notin V^\perp$. Then there exists $w\in V$ with $\phi(v,w)\neq0$. Note that $v,w$ are linearly independent since $\phi(v,v)=0$. Let $H=\mathrm{span}\{v,w\}$, which is a hyperbolic plane. Note that $\phi|_H$ is nondegenerate, so $H\cap H^\perp=0$. On the other hand $\dim H^\perp\geq\dim V-\dim H$, so $V=H\oplus H^\perp$. Thus $$ V^\perp=(H\oplus H^\perp)^\perp=H^\perp\cap(H^\perp)^\perp. $$ Applying the inductive hypothesis to $H^\perp$, we can therefore write $$ H^\perp=V^\perp\oplus W $$ where $W$ is hyperbolic. Thus $W\oplus H$ is hyperbolic and $$ V=V^\perp\oplus(W\oplus H) $$ as required.

Answer 3

Ok, the proof is indeed very rushed, so let me put a longer version of the proof. Personally, I am not a huge fan of the word "degenerated", so I will avoid that.

First, assume that the theorem holds if $V^\perp=0$. Then $\phi|_W$ is a bilinear form on $W$, and by that bilinear form, $W^\perp=0$ (easy). So by hypothesis, $W$ is hyperbolic. (this should adress $?_1$)

So now, assume $V^\perp=0$ and hence, $V=W$. Let $n=\dim(V)$. We proceed by induction to $n$. For $n=0$, there is nothing to prove. So assume $n$ is arbitrary, and assume the theorem holds for lower dimensions.

Let $v\in V\setminus\{0\}$. Then, $v\notin V^\perp$, so there exists a $w\in V$ such that $\phi(v,w)\neq 0$. Let $B$ a basis of $V$ s.t. $v\in B$. Then $V=V'\oplus H$, where $H=span(v,w)$, and:

$V'=span(\{b-\frac{\phi(b,w)}{\phi(v,w)}v-\frac{\phi(b,v)}{\phi(w,v)}w: b\in B\setminus\{v\}\})$

Then $H$ is a hyperbolic plane, $H\perp V'$, and by induction hypothesis, $V'$ is hyperbolic. So $V$ is hyperbolic. This concludes the proof.

In the rushed proof, induction is used in the same way as here, but in a hidden manner.