Let $\lambda,\mu \in \mathbb{C}$ be two distinct roots of unity.
Let $v \in \mathbb{C}^{n-1}$ be the vector given by
$$v_i = \sum_{j=0}^{n-1-i} \lambda^j, \ i \in \{1,...,n-1\}$$
that is $$v = \begin{pmatrix} 1 + \lambda + \lambda^2 + \cdot\cdot\cdot + \lambda^{n-3}+ \lambda^{n-2} \\ 1 + \lambda + \lambda^2 + \cdot\cdot\cdot + \lambda^{n-3} \\ \cdot\cdot\cdot \\ 1 \end{pmatrix}.$$
Another way to write this would be that $v$ is given by the product $$ M \tilde{v}$$ where $M$ is the matrix $$M = \begin{pmatrix} 1 & 1 & 1& \cdot \cdot \cdot & 1 & 1 \\ 1 & 1 & 1& \cdot \cdot \cdot & 1 & 0 \\ & & & \cdot \cdot \cdot & & \\ 1 & 0 & 0 & \cdot \cdot \cdot & 0 & 0 \end{pmatrix}$$ and $\tilde{v}_i = \lambda^i$.
Let $w \in \mathbb{C}^{n-1}$ be the vector given by $$w_i = \overline{\mu^i}.$$ That is, $$\overline{w} = (\mu,\mu^2,\mu^3,...,\mu^{n-1}).$$
The inner product of $v,w$ is then given by
$$\langle v, w \rangle = \sum_{i=1}^{n-1} \sum_{j=0}^{n-1-i} \lambda^j \mu^i.$$
I want to check whether it always holds that $\langle v,w \rangle \neq 0$.
Is there a neat way to proof/disprove this?
Edit: According to the comment, geometric sums might be of help.
Indeed we have (for $\lambda \neq 1$)
$$\sum_{j=0}^{n-1-i} \lambda^j = \frac{1-\lambda^{n-i}}{1-\lambda}$$
Thus we get $$\langle v, w \rangle = \sum_{i=1}^{n-1} \frac{1-\lambda^{n-i}}{1-\lambda} \mu^i = \frac{\sum_{i=1}^{n-1} \mu^i - \sum_{i=1}^{n-1} (\mu\lambda^{-1})^i}{1-\lambda}=\frac{-1 -(-1)}{1-\lambda} = 0$$