Let \(V\) be a finite-dimensional vector space and let \(f\) be a symmetric bilinear form on \(V\). If \(W\) is a subspace of \(V\), define \[ W^{\perp}=\{\alpha\in V:f(\alpha,\beta)=0\text{ for every }\beta\in W\}. \] Prove the following.
\(W^{\perp}\) is a subspace of \(V\).
\(\{0\}^{\perp}=V\).
\(V^{\perp}=\{0\}\) if and only if \(f\) is non-degenerate.
\(\operatorname{rank}f=\dim V-\dim V^{\perp}\).
If \(\dim V=n\) and \(\dim W=m\), then \[ \dim W^{\perp}\ge n-m. \]
The restriction of \(f\) to \(W\) is non-degenerate if and only if \[ W\cap W^{\perp}=\{0\}. \]
One has \[ V=W\oplus W^{\perp} \] if and only if the restriction of \(f\) to \(W\) is non-degenerate.
Solution
1. Prove that \(W^\perp\) is a subspace and compute \(\{0\}^\perp\).
The zero vector lies in \(W^\perp\). If \(x,y\in W^\perp\) and \(a,b\) are scalars, then for every \(w\in W\), \[ f(ax+by,w)=a f(x,w)+b f(y,w)=0. \] Thus \[ \boxed{W^\perp\le V,} \] which proves (a).
For \(W=\{0\}\), every vector is orthogonal to its only element, so \[ \boxed{\{0\}^\perp=V.} \] This proves (b).
2. Identify \(V^\perp\) with the radical and prove (c)–(d).
Define \[ L_f:V\longrightarrow V^*, \qquad L_f(x)(y)=f(x,y). \] Since \(f\) is symmetric, \[ \ker L_f =\{x:f(x,y)=0\text{ for all }y\in V\} =V^\perp. \] Hence \(f\) is non-degenerate exactly when \(L_f\) is injective, i.e. exactly when \[ \boxed{V^\perp=\{0\}.} \] This is (c).
By definition \(\operatorname{rank}f=\operatorname{rank}L_f\). Rank-nullity therefore gives \[ \boxed{ \operatorname{rank}f =\dim V-\dim V^\perp.} \] This proves (d).
3. Prove the lower bound on \(\dim W^\perp\).
Let \[ \Phi:V\longrightarrow W^*, \qquad \Phi(x)(w)=f(x,w). \] Then \[ \ker\Phi=W^\perp. \] Since \(\dim W^*=m\), \[ \operatorname{rank}\Phi\le m. \] Rank-nullity yields \[ \dim W^\perp =n-\operatorname{rank}\Phi \ge n-m. \] Thus \[ \boxed{\dim W^\perp\ge n-m,} \] proving (e).
4. Characterize non-degeneracy of the restriction to \(W\).
Let \(f_W\) denote the restriction of \(f\) to \(W\times W\). Its radical is \[ \begin{aligned} \operatorname{rad}(f_W) &=\{w\in W:f(w,u)=0\text{ for every }u\in W\}\\ &=W\cap W^\perp. \end{aligned} \] Therefore \(f_W\) is non-degenerate exactly when its radical is zero, i.e. exactly when \[ \boxed{W\cap W^\perp=\{0\}.} \] This proves (f).
5. Prove the direct-sum criterion.
Assume first that \(f_W\) is non-degenerate. By step 4, \[ W\cap W^\perp=\{0\}. \] Thus \[ \dim(W+W^\perp)=\dim W+\dim W^\perp. \] Step 3 gives \[ \dim W^\perp\ge n-m, \] so \[ \dim(W+W^\perp)\ge m+(n-m)=n. \] But \(W+W^\perp\le V\) and \(\dim V=n\), hence equality holds and \[ V=W+W^\perp. \] Together with the zero intersection this gives \[ V=W\oplus W^\perp. \]
Conversely, if \[ V=W\oplus W^\perp, \] then necessarily \[ W\cap W^\perp=\{0\}. \] Step 4 then shows that the restriction \(f_W\) is non-degenerate. Therefore \[ \boxed{ V=W\oplus W^\perp \iff f|_{W\times W}\text{ is non-degenerate}.} \] This proves (g).