Equivalence between two matrix expressions Let $G$ be a $n \times n$ symmetric and invertible matrix, and $b \in \mathbb{R}^n$, and let $V$ be a full-rank $n \times (n-1) $ matrix, such that $b^T V = 0 $, i.e. the vector $b$ is orthogonal to all the columns of $V$.  The first expression is
$ \theta_1 = \dfrac{1}{b^T b}  \bigg( I_n - V (V^T G V)^{-1} V^T G \bigg) b  $
The second expression is
$ \theta_2 = \dfrac{1}{b^T G^{-1} b}  G^{-1} b $
I want to show that the two expressions are equivalent.  I have no idea how to proceed.  I have verified numerically that they are equivalent.
 A: You have $b^\top\theta_1=b^\top\theta_2=1$ and $V^\top G\theta_1=V^\top G\theta_2=0$, so the expressions are equivalent if  $b^\top$ and the rows of $V^\top G$ together span the entire space. If not, then $b^\top$ is in the span of the rows of $V^\top G$, and since $V^\top G G^{-1}b=V^\top b=0$, that implies that the denominator $b^\top G^{-1}b$ in $\theta_2$ is zero. Thus, the expressions are equivalent whenever they’re both well-defined.
A: Let us re-write the equations defining $\theta_1$ and $\theta_2$ :
$$\theta_1 := \dfrac{1}{b^T b}( I_n - V (V^T G V)^{-1} V^T G ) b\tag{1}$$
$$\theta_2 := \dfrac{1}{b^T G^{-1} b}  G^{-1} b\tag{2}$$
with constraint :
$$b^TV=0 \ \ \iff \ \ V^Tb=0\tag{3}$$
A preliminary remark is that matrix $V^TG$ which has the same dimensions $(n-1) \times n$ as matrix $V^T$ has as well the same rank $n-1$.
Therefore the kernel $\mathcal{K}$ of $V^TG$ is 1-dimensional.
It is easy to see that $\theta_1, \theta_2$ belong to $\mathcal{K}$ by left-multiplying them by $V^TG$, taking into account property (3) for the case of $\theta_2$.
Therefore $\theta_1=k\theta_2$.
In order to show that $k=1$ it suffices to prove that both $\theta_1$ and $\theta_2$ have the same dot product with a same vector. Let us take the dot product of (1) with vector $b$ by left-muliplication by $b^T$ :
$$b^T \theta_1= \dfrac{1}{b^T b}( b^T - \underbrace{b^TV}_{= 0} (V^T G V)^{-1} V^T G ) b= \dfrac{1}{b^T b}(b^Tb)=1 $$
Same operation with expression (2) : left-multiplying it by $b^T$ gives the same result, i.e., $1$.
