I have a set of vectors defined as
$[\mathbf{v}(x)]_n = e^{jn\pi x}; \quad n = 0 ~\text{to}~ (N-1)$
where $\mathbf{v}$ is an $N \times 1$ vector, $j$ is $\sqrt{-1}$, and $-1 \leq x < 1$. For a choice $\mathbf{v}(x_0)$ it is possible to show
$ \mathbf{v}^H(x_0)\mathbf{v}(x) = 0$
for $N-1$ different values of $x$ i.e. there are a set of $N-1$ orthogonal vectors among $\mathbf{v}(x)$ for any choice one particular vector.
Now consider the quadratic form
$\mathbf{v}^H(x_0)\mathbf{R}\mathbf{v}(x)$
where $\mathbf{R}$ is a hermitian matrix.
Is it possible to show that there exist another set of $N-1$ vectors among $\mathbf{v}(x)$ such that above quadratic form still equals to zero?
Since $\mathbf{R}$ is Hermitian, it admits a spectral decomposition $$\mathbf{R} = \sum\limits_i {{\lambda _i}{{\mathbf r}_i}{\mathbf r}_i^{\text{H}}}$$ with $\lambda _i$ real.
Letting $\mathbf{v}_0$ denote $\mathbf{v}^{\rm H}\left( x_0\right)$, we have then that $\mathbf{v}^{\rm H}_0\mathbf{R}$ is $${\mathbf{v}}_0^{\text{H}}{\mathbf{R}} = \sum\limits_i {{\lambda _i}{\mathbf{v}}_0^{\text{H}}{{\mathbf{r}}_i}{\mathbf{r}}_i^{\text{H}}}$$
Denote the $N-1$ "twist" vectors orthogonal to $\mathbf{v}_0$ as $\mathbf{v}_1 ,\ldots , \mathbf{v}_{N-1}$. Since these vectors must collectively span ${\mathbb C}^{N}$ along with $\mathbf{v}_0$, we can write $${{\mathbf{r}}_i} = {\iota _{i0}}{{\mathbf{v}}_0} + \sum\limits_j {{\iota _{ij}}{{\mathbf{v}}_j}} $$ for scalars $\iota _{ij}$.
Plugging this into the equation above, we get $${\mathbf{v}}_0^{\text{H}}{\mathbf{R}} = \sum\limits_i {{\lambda _i}{\mathbf{v}}_0^{\text{H}}\left( {{\iota _{i0}}{{\mathbf{v}}_0} + \sum\limits_j {{\iota _{ij}}{{\mathbf{v}}_j}} } \right){\mathbf{r}}_i^{\text{H}}}$$ $$ = \sum\limits_i {{\lambda _i}\left( {{\iota _{i0}}{\mathbf{v}}_0^{\text{H}}{{\mathbf{v}}_0} + \sum\limits_j {{\iota _{ij}}{\mathbf{v}}_0^{\text{H}}{{\mathbf{v}}_j}} } \right){\mathbf{r}}_i^{\text{H}}}$$ $$ = \sum\limits_i {{\lambda _i}{\iota _{i0}}{{\left\| {{{\mathbf{v}}_0}} \right\|}^2}{\mathbf{r}}_i^{\text{H}}}$$ Since we want the proof to hold for all Hermitian $\mathbf{R}$, we can choose the values of $\lambda _i$ and $\mathbf{r}_i$ as broadly as we want for all $i$. Clearly then, we can choose our values so that the sum evaluates to any desired complex-valued row vector $\mathbf{w}^{\rm H}$ we want, either by choosing $${{\mathbf{r}}_1} = {\mathbf{w}}{e^{\phi /2}}$$ $${\lambda _1} = {\left| {{{\mathbf{w}}^{\rm H}}{{\mathbf{v}}_0}} \right|^{ - 1}}$$ $$\phi = \arg {{\mathbf{w}}^{\text{H}}}{{\mathbf{v}}_0}$$ and setting all other $\lambda _i$'s to zero, if $\mathbf{w}$ isn't orthogonal to $\mathbf{v}_0$; or by using two $\mathbf{r}_i$'s and $\lambda _i$'s to sum to $\mathbf{w}$ if $\mathbf{w}$ is orthogonal to $\mathbf{v}_0$.
Hence for any Hermitian $\mathbf{R}$, we can find a set mutually-orthogonal "twist vectors", $\mathbf{v'} _{j}$, that satisfies $${\mathbf{v}}_0^{\text{H}}{\mathbf{R}}{{{\mathbf{v'}}}_j} = 0$$ if and only if for any vector $\mathbf{w} \in {\mathbb C}^N$ we can find $$\mathbf{w}^{\rm H}\mathbf{v'}_{j} = 0$$ That is, if we can find a set of twist vectors orthogonal to $\mathbf{w}$.
Since we know a twist vector exists that is orthogonal to $\mathbf{v'} _{j}$, and we know that all vectors orthogonal to $\mathbf{v'} _{j}$ must lie in this one-dimensional subspace, it follows by contradiction that no solution for $\mathbf{v'} _{j}$ exists for any $\mathbf{w}$ not equal to a scalar multiple of a twist matrix.
Hence the answer is 'No'. It is not always possible to find a set of twist vectors such that the quadratic form is zero.