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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.

Hosam Hajeer
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2 Answers2

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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.

joriki
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  • Matrix $V$ is not square. – Hosam Hajeer Feb 05 '23 at 16:46
  • @GloriousErin: Sorry, I overlooked that. I edited the answer – I hope it answers the question now. – joriki Feb 05 '23 at 17:22
  • Thank you. That was elegant and to the point. – Hosam Hajeer Feb 05 '23 at 17:33
  • I hadn't seen your answer. Mine is formulated along the very same ideas. Have you noticed that the symmetry condition is not necessary. – Jean Marie Feb 05 '23 at 19:16
  • One more "afterthought". I have spend time, with no avail, trying to apply the fact that $P=I-M(M^TM)^{-1}M^T$ is the projector onto the orthogonal of the column space of $M$. I am still convinced that one could find a proof using this property. – Jean Marie Feb 05 '23 at 19:29
  • @JeanMarie: No, I hadn't noticed that there was a symmetry condition I hadn't used. – joriki Feb 05 '23 at 19:30
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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., $1$.

Jean Marie
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    The fact that $G$ is symmetric has not been needed ; I have checked numerically on several examples that property $\theta_1=\theta_2$ indeed still holds without this hypothesis. – Jean Marie Feb 05 '23 at 19:13