What am I missing in this proof? Unbiased vectors I have a set of vectors $\{P_{1},P_2,...,P_{d+1}\}$ in a $d$-dimensional Hilbert space, which satisfy the following two conditions on the inner product
$$|P_{i}|^2=d, \quad \forall i=1,...,d+1,  \qquad \qquad P_i^*\cdot P_j =-1, \quad i,j=1,...,d+1 \text{ and } i\neq j.$$
I want to prove that this set is not linearly independent but if you remove any vector, it becomes linearly independent.
The first part of the proof is easy if one notices that any of the vectors can be written as
\begin{equation}
P_i=-\sum_{\substack{j=1\\ i\neq j}}^{d+1}P_j,
\tag{1}
\label{eq:pi}
\end{equation}
because this satisfies both conditions on the inner products.
Therefore, let us now remove $P_1$ from the set, for example. Is the set $\{P_2,P_3,...,P_{d+1}\}$ linearly independent? I would say that it is. Let us take $P_2$ for example and see if it can be written as a linear combination of $\{P_3,...,P_{d+1}\}$. From \eqref{eq:pi} we have that
$$P_2=-P_1-P_3-P_4-...-P_{d+1},$$
and this doesn't seem like it can be written as a linear combination of $\{P_3,...,P_{d+1}\}$. However, I feel that this proof is missing rigour somewhere.
What am I missing?

The vectors $P_1 - P_{d+1}, \dots , P_d - P_{d+1}$ are all non-zero and orthogonal to each other, hence linearly independent:
Let $i,j<d+1$, $i\ne j$ then:
$$
\| P_i - P_{d+1}\|^2 = d + 2 + d = 2d+2>0
$$
$$
(P_i- P_{d+1},  P_j- P_{d+1}) = 0.
$$
