Gram-Schmidt for "indefinite inner product"

Question

Define an "inner product" on $\mathbb{C}^3$ as $\langle x,y \rangle = y^* A x$ where $A$ is Hermitian and indefinite (and its entries are given). This "inner product" satisfies all properties of an inner product except for the positive-definiteness property (as $A$ is indefinite, not even positive semidefinite).

I am asked to

find an orthogonal basis $v_1, v_2, v_3$ from the basis of unit vectors $e_1, e_2, e_3$ using the Gram-Schmidt method.

In every source I have consulted so far, the Gram-Schmidt process assumes an underlying inner product space, but here we do not have a true inner product.

Are there spaces more general than inner product spaces for which the Gram-Schmidt process is guaranteed to yield an orthogonal basis?

Answer 1

Naively applying Gram-Schmidt with an indefinite not-inner-product is the sort of thing that could work out just fine, unless it doesn't.

The main danger is that one of the vectors $v_i$ you end up computing satisfies $v_i^*A v_i = 0$. If this happens, then:

• You would not be able to normalize $v_i$, though that's not a big deal compared to the other issues.
• You cannot transform another vector $x$ into a vector $A$-orthogonal to $v_i$, because the operation $x \mapsto x - \frac{v_i^* A x}{v_i^* A v_i} v_i$ requires us to divide by $0$.
• Even if the previous bullet point somehow does not prevent you from carrying out Gram-Schmidt (say, because everything is already $A$-orthogonal to $v_i$ by coincidence), it would not be guaranteed that a set of $n$ $A$-orthogonal vectors containing $v_i$ is actually linearly independent. (For one thing, $v_i$ is $A$-orthogonal to itself.) The usual proof does not work.

If you happen not to get any such vectors, then everything goes just fine.

Answer 2

Without focusing on the Gram-Schmidt procedure in particular, orthonormal bases (for a reasonable generalisation of "orthonormal") are possible in pseudo-Euclidean spaces, which are finite-dimensional real vector spaces equipped with a non-degenerate billinear form (and are important in special relativity). In finite-dimensions, this means there is no non-zero vector orthogonal to every vector, i.e. $$\forall y : \langle x, y \rangle = 0 \implies x = 0.$$ In terms of $A$, this is equivalent to $A$ being invertible.

Non-degenerate forms do have some differences to inner products: we can have $\langle x, x \rangle$ be negative, or $0$ even when $x \neq 0$ (i.e. $x$ is self-orthogonal). The "norm" associated with the non-degenerate form is $\sqrt{|\langle x, x \rangle|}$, and isn't a norm at all, not least because it may send non-zero vectors $x$ to $0$.

Because of this, Gram-Schmidt is not really possible. However, building an orthonormal basis is possible, where orthonormal means, as usual, distinct vectors are orthogonal, and the "norm" of these vectors are all $1$ (meaning $\langle x, x \rangle = \pm 1$).

The procedure is simple. Let us call our pseudo-Euclidean space $V$.

1. Choose any vector $x_1 \in V$ such that $\langle x_1, x_1 \rangle \neq 0$. This is always possible, because otherwise $$\langle x, y \rangle = \frac{1}{4}(\langle x + y, x + y \rangle - \langle x - y, x - y \rangle) = 0$$ for all $x, y$, contradicting non-degeneracy. Note: $x_1 \neq 0$.
2. Divide $x_1$ by $\|x_1\|$ to get your first orthonormal vector $e_1 \neq 0$.
3. Restricting $\langle \cdot, \cdot \rangle$ to the subspace $\{e_1\}^\perp \subseteq V$ produces another non-degenerate bilinear form. We can first note that $$V = \operatorname{span}\{e_1\} \oplus \{e_1\}^\perp$$ in the usual way: we can write, for any $x \in V$, $$x = \frac{\langle x, e_1 \rangle}{\langle e_1, e_1\rangle} e_1 + \left(x - \frac{\langle x, e_1\rangle}{\langle e_1, e_1 \rangle} e_1\right),$$ and because $\langle e_1, e_1 \rangle \neq 0$, the sum is direct. Hence, if we have $x \in \{e_1\}^\perp$ such that $\langle x, y \rangle = 0$ for all $y \in \{e^\perp\}$, then $\langle x, y \rangle = 0$ for all $y \in V$. As such, we can simply repeat steps $1$ and $2$ until we run out of dimensions, calling the resulting vectors $e_2, e_3, \ldots, e_n$. It's not difficult to see that the resulting list of vectors is orthonormal.

It's not really Gram-Schmidt because we can't really transform a basis to an orthonormal basis in the same way. Indeed, we might find that the first vector in the given basis has length $0$, or that subsequent vectors have orthogonal components that have length $0$. So, it's not quite an answer to your question, but it seemed relevant.