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After working on this Some kind of projection in a non-orthogonal basis , it turns out it suffices (ask why if you can't figure it) to solve the following problem:

Suppose the existence of $a_1,...a_d$ be $d$ real numbers, and $\mathbf{e_1,\ldots,e_d}$ be a basis of $\mathbb R^d$ such that $$\forall (x_1,...,x_d) \in \mathbb R^d, \sum_{i=1}^{d} \sum_{k=1}^{d} x_k \langle \mathbf{e_k,e_i} \rangle a_i=0 $$ Prove that $a_1=\ldots=a_d=0$.

Equivalently, and more simply,

Let $\mathbf{e_1,\ldots,e_d}$ be a basis of $\mathbb R^d$ and $G$ the associated Gramian matrix.

If there exists $A \in \mathcal{M}_{d,1}(\mathbb R)$ such that $\forall X \in \mathcal{M}_{d,1}(\mathbb R), X^TGA=0$, prove that $A=0$

How do I tackle this ? (I don't know much about Gramian matrices).

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Gabriel Romon
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Let $A$ have the desired property.

The Gram matrix $G$ is invertible if, and only if, $\{e_1,\ldots,e_d\}$ is a linearly indepedent set.

One can prove that $\forall X \in \mathcal{M}_{d,1}(\mathbb R)\left(X^TGA=\mathbf 0\right)$ implies $GA=\mathbf 0$ and since $G$ is invertible, the desired result follows.

To prove that $\forall X \in \mathcal{M}_{d,1}(\mathbb R)\left(X^TGA=\mathbf 0\right)\implies GA=\mathbf 0$, note that $\forall X \in \mathcal{M}_{d,1}(\mathbb R)\left(X^TGA=\mathbf 0\right)$ entails in particular $u_iGA=\mathbf 0$, for all $i\in \{1, \ldots ,d\}$, where each $u_i$ is the $i^{\text{th}}$ vector of the standard basis.
This in turn implies that a certain $d\times d$ matrix $B$ has the property $BGA=\mathbf 0$. This $B$ is very special. Can you find it?

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Git Gud
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