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 $V_0$.

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 $\{0\}$.

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!

  • 1
    $\begingroup$ 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. $\endgroup$
    – Berci
    Jul 30, 2019 at 13:28
  • $\begingroup$ @Berci I thought about making mistake when I mentioned "iff" on my last statement, but now it makes sense, thank you! $\endgroup$
    – ShellRox
    Jul 30, 2019 at 14:32

1 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 $V$.

  • $\begingroup$ 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. $\endgroup$
    – ShellRox
    Jul 30, 2019 at 14:46
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    $\begingroup$ 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. $\endgroup$ Jul 30, 2019 at 18:09
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    $\begingroup$ To summarize: For a nondegenerate scalar product there are no nonzero vectors orthogonal to everything, for a degenerate scalar product there are. For a definite scalar product there are no nonzero vectors orthogonal to themselves, for an indefinite scalar product there are. $\endgroup$ Jul 30, 2019 at 18:12
  • $\begingroup$ 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! $\endgroup$
    – ShellRox
    Jul 31, 2019 at 8:45

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