Index of nullity of orthogonal complement of vector space

Question

From Serge Lang's Linear Algebra, I've been just introduced to the concept of index of nullity and Sylvester's theorem based on non positive-definite scalar products:

Let $V$ be a finite dimensional vector space over $\mathbb{R}$, with a scalar product. Assume $\textrm{dim} \, V > 0$. Let $V_0$ be the subspace of $V$ consisting of all vectors $v \in V$ such that $\langle{v}, w \rangle = 0$ for all $w \in V$. Let $\{v_1, ... , v_n\}$ be an orthogonal basis for $V$. Then the number of integers $i$ such that $\langle{v_i} , v_i \rangle$ is equal to the dimension of $V_0$.

The proof is fairly simple, suppose $\{v_1, ... , v_n\}$ is ordered so that:

$\langle v_1, v_1 \rangle \neq 0, ... ,\langle v_s, v_s \rangle \neq 0$ but $\langle v_i, v_i \rangle = 0$ for all $i > s$.

Considering that $\{v_1, ... , v_n\}$ is orthogonal basis, it is obvious that $\{v_{s+1}, ... , v_n\}$ lies in $V_0$. Any element $v \in V_0$ can be thus written as:

$$v = x_1v_1 + ... + x_sv_s + ... + x_nv_n$$

with $x_i \in X \in \mathbb{R}^n$. Taking scalar product of $v$ with any $v_j$ such that $j \leq s$, it can be seen by bilinearity that:

$$0=\langle v, v_{j} \rangle = x_j \langle v_j, v_j \rangle$$

Considering that $\langle v_j, v_j \rangle \neq 0$, by trivial factor rule $x_j = 0$. Hence $\{v_{s+1}, ... , v_n\}$ forms an orthogonal basis for $V_0$.

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Contradiction by orthogonal complement:

I've studied much before a concept of orthogonal complement in positive-definite cases, such that:

$$\textrm{dim} \, W + \textrm{dim} \, W^{\perp} = \textrm{dim} \, V$$

if $W$ is a subspace of $V$ and $W^{\perp}$ is its orthogonal complement.

But in this case, $V_0$ is an orthogonal complement of $V$, and thus:

$$\textrm{dim} \, V + \textrm{dim} \, V^{\perp} = \textrm{dim} \, V$$ $$\textrm{dim} \, V^{\perp} = \textrm{dim} \, V - \textrm{dim} \, V$$ $$\textrm{dim} \, V^{\perp} = 0$$

Thus this contradicts the proof above, because instead of $\{v_{s+1}, ... ,v_n\}$, basis of trivial vector space must be $\{0\}$.

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Am I missing something? The note on index of nullity does not mention whether or not scalar product is positive-definite. Perhaps basis of $V_0$ is $\{0\}$ iff $V$ has positive-definite scalar product?

Thank you!

Answer 1

By definition $V_0^\perp = V$, so $\dim V_0 + \dim V_0^\perp = \dim V_0 + \dim W$. In particular, if $V_0 \neq \{ 0 \}$ then $\dim V_0 + \dim V_0^{\perp} > \dim W$, and so the formula you recall from the case of a positive definite scalar product does not apply.

Remark Any scalar product $\langle\,\cdot\,,\,\cdot\,\rangle$ on $V$ determines a nondegenerate scalar product $\langle\!\langle\,\cdot\,,\,\cdot\,\rangle\!\rangle$ on $V / V_0$ by setting $\langle\!\langle v + V_0, w + V_0 \rangle\!\rangle = \langle v, w \rangle$. Then, replacing $V$ and $W$ in the identity $\dim W + \dim W^\perp = \dim V$ (which applies just as well to general nondegenerate scalar products as to positive definite ones) respectively with $V / V_0$ and $W / (W \cap V_0)$ gives the identity $$\dim W + \dim W^\perp = \dim V + \dim (W \cap V_0) $$ which holds even for degenerate scalar products on $V$.