Some kind of projection in a non-orthogonal basis

Question

Sorry if the title sounds convoluted, I couldn't find any better.

In $R^d$, let $(e_1,\ldots, e_d)$ be a basis. Show there exists $(a_1,\ldots, a_d)$ d vectors of $R^d$ such that $$\forall x \in R^d, x=\sum_{i=1}^{d} \langle x, e_i \rangle a_i $$

I tried to mimic the proof of Gram-Schmidt orthogonalization, but failed. After that, I introduced a linear mapping of which it suffices to prove the injectivity but it yields awful computations...

Thanks for your help.

I think I might see how a linear mapping from $\Bbb R^d$ to $\Bbb R^d$ might prove to be the right path. Maybe you can prove injectivity by writing down the inverse of the mapping? — Greg Martin, Mar 20 '14 at 17:48
@GregMartin Well since it's linear it might be more simple to prove that the Ker is $0$. This led to this question http://math.stackexchange.com/questions/720086/proving-that-exists-a-forall-x-xtga-0-rightarrow-a-0?lq=1 . Can you write the inverse mapping explicitly ? — Gabriel Romon, Mar 20 '14 at 17:51
Aren't we just looking for a dual basis, where $a_i\cdot e_i=1$, $a_i\cdot e_j=0$ for $i\ne j$? — Gerry Myerson, Mar 21 '14 at 12:22
@GarryMyerson I see what you mean. But how do you prove that the linear forms we're dealing with are linearly independent? — Gabriel Romon, Mar 21 '14 at 18:22
@GarryMyerson I found it. I'll write it as an answer tomorrow — Gabriel Romon, Mar 21 '14 at 20:53

score 0 · Accepted Answer · edited Apr 13 '17 at 12:20

Solution 1:

Consider the $d$ following linear forms $\begin{array}{ccccc} f_i & : & E & \to & \mathbb R \\ & & x & \mapsto & \langle x|e_i\rangle \\ \end{array}$

We first prove that $(f_1,...,f_d)$ is a linearly independent family (hence a basis) of the dual space $E^*$: it suffices to prove that $\operatorname{rank}(f_1,...,f_d)=d $.

But $$\operatorname{rank}(f_1,...,f_d)=d-\dim(\cap_{i=1}^d \ker f_i)$$ $$\operatorname{rank}(f_1,...,f_d)=d-\dim(\cap_{i=1}^d \{e_i\}^\bot)$$

Since $(e_1,...,e_d)$ is a basis, $$\operatorname{rank}(f_1,...,f_d)=d-0=d$$

Therefore $(f_1,...,f_d)$ is basis of $E^*$, which has a unique "prebasis" $(a_1,...,a_d)$ where $(a_1,...,a_d)$ is a basis of $E$ and $a_i^*=f_i$.

Now, $$\forall x \in E, x=\sum_{i=1}^{d} a_i^*(x) a_i=\sum_{i=1}^{d} f_i(x) a_i=\sum_{i=1}^{d} \langle x|e_i\rangle a_i$$

This achieves the proof.

Solution 2

Here : Proving that $\exists A, \forall X, X^TGA=0 \Rightarrow A=0 $

Some kind of projection in a non-orthogonal basis

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