# Index of nullity of orthogonal complement of vector space

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 $$V_0$$.

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 $$\{0\}$$.

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!

• You're correct: the scalar product is not supposed to be positive definite here. However, we also get $V_0=\{0\}$ in the negative-definite case, and probably in certain indefinite cases, too. Jul 30, 2019 at 13:28
• @Berci I thought about making mistake when I mentioned "iff" on my last statement, but now it makes sense, thank you! Jul 30, 2019 at 14:32

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 $$V$$.
• Thank you for the answer and general identity. I just have a question about generalization, you mentioned in the Remark that $\textrm{dim} \, W + \textrm{dim} \, W^{\perp} = \textrm{dim} \, V$ applies to general nondegenerate scalar products as well as positive definite ones, but when $\textrm{dim} \, V = \textrm{dim} \, W$ doesn't this contradict the statement that this identity does not hold in indefinite cases? I apologize if I misunderstood anything. Jul 30, 2019 at 14:46
• You're welcome, I'm glad you found it useful. I'm not sure I parsed your follow-up question correctly, but if $\dim V = \dim W$ then (for finite-dimensional vector spaces) $V = W$, and if the scalar product on $V$ is nondegenerate, by definition there is no nonzero vector $v_0$ such that $\langle v, v_0 \rangle = 0$ for all $v \in V$ (this is precisely the statement that $V_0 = 0$), so $V^{\perp} = \{ 0 \}$ and $\dim V^{\perp} = 0$ as expected. Jul 30, 2019 at 18:09
• Yes, that's what I was asking, I was little confused but then I remembered that nondegenerate scalar product has a trivial kernel, Serge Lang even mentioned later in the section that index of nullity is $0$ iff scalar product is nondegenerate, so now it makes sense. Thanks again! Jul 31, 2019 at 8:45