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?
 A: Consider the $d+1\times d+1$ Gram matrix $G$ where $g_{i,j} := \langle P_i, P_j\rangle$
(or $g_{i,j} := \langle P_j, P_i\rangle$ depending on inner product convention)
And let $G'$ be $G$ with column $k$ and row $k$ deleted.
(1) Then $G= (d+1) \cdot I_{d+1} - \mathbf 1_{d+1}\mathbf 1_{d+1}^T$
$G\mathbf 1_{d+1}= \mathbf 0$ so $G$ is singular.
(2) $G'= (d+1) \cdot I_{d} - \mathbf 1_{d}\mathbf 1_{d}^T$
and $G'$ is nonsingular because either  (i) strict diagonal dominance, (ii) Gershgorin discs, (iii) rank one update determinant formula, (iv) spectral decomposition, etc.
(3)  Finally use the correspondence between Gram matrix (non)singularity and vector linear (in)dependence, i.e. consider arbitrary $n\times n$ Gram matrix $G^{''}$ with $g^{''}_{i,j}:=\langle \mathbf v_i,\mathbf v_j\rangle$
$\mathbf x^* G^{''}\mathbf x = \sum_{i=1}^n \bar{x_{i}} \sum_{j=1}^n x_{j}\langle \mathbf v_i, \mathbf v_j \rangle = \langle\big(\sum_{j=1}^n x_{j} \mathbf v_j\big),  \big(\sum_{j=1}^n x_{j} \mathbf v_j \big)\rangle\geq 0 $
with equality iff $\sum_{j=1}^n x_{j} \mathbf v_j  = \mathbf 0$
A: 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:
$$
\| P_i - P_{d+1}\|^2 = d + 2 + d = 2d+2>0
$$
$$
(P_i- P_{d+1},  P_j- P_{d+1}) = 0.
$$
