Extension to the complex numbers for ex. 12 in ch. 6 of Axler's "Linear Algebra Done Right"

Question

I'm wondering how the answer to Sheldon Axler's exercise 12 of chapter 6 "Linear Algebra Done Right" changes when the underlying field is extended from the reals to the complex numbers.

The exercise asks:

Suppose $V$ is a real inner product space and $(v_1,\ldots,v_m)$ is a linearly independent list of vectors in $V$. Prove there exist exactly $2^m$ orthonormal lists $(e_1, ..., e_m)$ such that for $j = 1 \ldots n$, $$\text{span}(v_1,\ldots,v_j) = \text{span}(e_1,\ldots,e_j).$$

The answer to the original question over the reals is obvious, but thinking about the complex case, it seems that there at least $4^m$ possible lists since $$\text{span}(e_1,\ldots,e_j) = \text{span}(e_1,\ldots, i e_j) = \text{span}(e_1,\ldots, -i e_j)$$ and both $(e_1,\ldots, i e_j)$ and $(e_1,\ldots, -i e_j)$ are orthonormal if $(e_1,\ldots, e_j)$ is.

Is this correct? And is $4^m$ an upper as well as lower bound on the number of such lists?

Answer 1

anomaly answered this question in the comment above:

Your argument is correct as far as it goes, but $4^m$ is not an upper bound. What if you replace $i$ or $−i$ with an arbitary complex scalar λ with |λ|=1? Consider the case $\text{dim} V=1$, for example.