If $\{u_1,\ldots,u_n\}$ is a complete orthonormal set for the inner product induced by $G$, then $\sum_{j=1}^n u_ju_j^T=G^{-1}$

Question

For $G$, an $n\times n$ and symmetric positive definite matrix, the $G$-inner product on $\mathbb{R}^{n}$ is given by $$(x,y)_G=x^{T}Gy.$$ A complete $G$-orthonormal set $\{u_1,u_2,\ldots,u_{n}\}$ satisfies $(u_i,u_j)_G=\delta_{ij}$. Prove that $\sum_{j=1}^{n}u_{j}u_{j}^{T}=G^{-1}$ for any complete $G$-orthonormal set.

(Hint: expand an arbitrary vector in terms of this set).

I have proved that $\{u_1,u_2,\ldots,u_{n}\}$ is a basis. But I do not know how to continue.

Answer 1

First, since $G$ is symmetric positive definite, you know in particular that it is indeed invertible.

Now, let $T := \sum_{j=1}^n u_j u_j^T$. The question is, what is $TGe_k$ for each $k$?

Answer 2

We know $\displaystyle u_i^T G u_j = \begin{cases} 1 & \text{if }i=j, \\ 0 & \text{if }i\ne j. \end{cases}$

How to prove that $$ u^T G\left(\sum_{j=1}^n u_j u_j^T\right) = u^T, $$ so that $$ G\left(\sum_{j=1}^n u_j u_j^T\right) = I\text{ ?} $$

It is enough to show this for $u=u_i$, $i=1,\ldots,n$, since these elements form a basis. $$ \begin{align} u_i^T G\left(\sum_{j=1}^n u_j u_j^T\right) & = \cdots+u_i^T (Gu_i u_i^T) + \cdots \\[12pt] & = \cdots+(u_i^T G u_i) u_i^T+\cdots. \end{align} $$ The other terms are $0$, and this term is $u_i^T$.

A similar argument works if you commute the two factors.