Question about degenerate inner product subspaces

Question

I can't seem to prove that, in a vector space V with a degenerate inner product <,>, the set N=$\{v:$<$v,v$>$=0\}$ is a linear subspace of V. It seems like I need to prove that $Re$(<$v,w$>)$=0$ $\forall v,w\in N$, but I don't see a way to prove it.

Answer 1

If you mean a degenerate bilinear form, it is false.

Let $\mathbb R^3$ have the degenerate metric $B(x,y)=x_1y_1-x_2y_2$.

Then $v=(1,1,0)$ and $w=(-1,1,0)$ both satisfy $B(v,v)=0$ but their sum does not. So, the subset of isotropic vectors is not closed under addition.

---

If you mean a positive semidefinite Hermitian inner product on a $\mathbb C$ vector space, then a version of the Cauchy–Bunyakovsky–Schwarz inequality$^{[1]}$ will allow you to show the isotropic vectors form a subspace.

If both $v$ and $w$ are isotropic, then the CBS inequality gives

$$ 0\leq |\langle v,w\rangle|^2\leq \langle v,v\rangle\langle w,w\rangle=0\cdot 0 $$

This shows both $\langle v,w\rangle = \langle w,v\rangle=0$

Then after expansion one finds $\langle v+w,v+w\rangle=0$. Easiest of all is $\langle \lambda v, \lambda v\rangle=\lambda \bar\lambda\cdot 0=0$. This establishes the isotropic vectors form a subspace.

---

$^{[1]}$ Thanks to Sha Vuklia for this reference on the topic: I.1.4 'The Cauchy-Bunyakowsky-Schwarz inequality' in Conway's Functional Analysis.

Answer 2

If $V$ is a complex vector space and $\left< \cdot, \cdot \right>$ is a sesquilinear Hermitian bilinear form that satisfies $\left< v, v \right> \geq 0$ for all $v \in V$ then the set $N = \left \{ v \in V \, | \left< v , v \right> \right \}$ does form a vector subspace of $V$ and $\left< \cdot, \cdot \right>$ induces an honest complex inner product on the quotient $V / N$.

To see this, set $||v|| := \left< v, v \right>^{\frac{1}{2}}$ and note that the standard Cauchy-Schwartz proof carries verbatim to such bilinear forms (strict positivity is not used in the proof) and the triangle inequality also holds for $||\cdot||$ is it follows from the Cauchy-Schwartz inequality. Then, if $v, w \in N$ we have $\|v\| = \|w\| = 0$ but then $\|av + bw\| \leq |a|\|v\| + |b|\|w\|$ which shows that $av + bw \in N$.