Let $p \in \mathbb{R}^{n}$ and $p=\lambda\_1 e\_1+...+\lambda\_n e\_n$ where $e\_i$ are standard basis vectors then if I want to find the component along which I can get closest to the point $p$ then it will just be $e\_j$ with $j \in \{1,...,n\}$ such that $\lambda\_j$ satisfies $|\lambda\_j| = max\_{1 \leq i\ \leq n} \{|\lambda\_1|,...,|\lambda\_n|\}$ and the closest point to $p$ along this direction is $ \lambda\_j e\_j$. As $\lambda\_j$ was maximum, we have
\begin{equation\*}|\lambda\_j| \geq \frac{|p|}{\sqrt{n}} \end{equation\*} and
\begin{equation} |\lambda\_j e\_j - p | \leq \sqrt{\frac {n-1} {n} } |p|. \end{equation}


I am looking for similar estimates when $p$ is represented by a different basis. What can I say if the point $p= \alpha\_1 x\_1 +...+\alpha\_n x\_n$ where $\{x\_1,...,x\_n\}$ are such that $ \angle(x\_i, \text{span}(x\_1,...,x\_{i-1})) \geq \theta$ for $i=2,...,n$ and $\theta >0$. Can I choose a direction $x\_j$, $j \in \{1,...,n\}$ so that $|\alpha x\_j - p| \leq C |p|$ where $\alpha>0$, $C<1$ and may depend on $n$ and $\theta$?


This is a repost from stackexchange (<https://math.stackexchange.com/questions/4041205/the-direction-that-gets-me-closest-to-a-given-point-in-mathbbrn>) where I got no answers. Hopefully it's OK to repeat the question here.



The direction that gets me closest to a given point in $\mathbb{R}^n$
For $p, x\_i\in \mathbb{R}^n$, with $1\le i\le n$, define
$$\theta\_i=\angle(x\_i,\text{span}(\{x\_j \;|\; j\ne i\})),\quad \theta=\min(\{\theta\_i\})$$


$$\beta\_i=\frac{\pi}{2}-\angle(p,x\_i),\quad\beta=\max(\{\beta\_i\})$$


**CLAIM.**
$$\sin(\beta)\ge\frac{sin(\theta)}{\sqrt{n(1+(n-1)\cos(\theta))}}$$


**PROOF.**
Trivial if $\theta=0$, so assume $\theta>0$. Write $p$ and the $x\_i$'s as $1\times n$ matrices and rescale them so that $pp^T=x\_i x\_i^T=1$. Let $X$ be the matrix with rows $x\_1,\dots x\_n$. Then $\theta>0$ implies that $X$ is invertible, therefore


$$(pX^T)(XX^T)^{-1}(pX^T)^T=pp^T=1$$


But $pX^T=(\sin(\beta\_1),\dots \sin(\beta\_n))$ and the inverse correlation matrix $(XX^T)^{-1}=(c\_{ij})$ satisfies


$$\displaystyle c\_{ij}=\frac{1}{\sin(\theta\_i)\sin(\theta\_j)}\cdot \frac{x'\_i {x'\_j}^T}{|x'\_i||x'\_j|}$$


where $x'\_i$ is the component of $x\_i$ orthogonal to $\text{span}(\{x\_j \;|\; j\ne i\})$. (Different proofs can be found [here](https://stats.stackexchange.com/questions/140080/why-does-inversion-of-a-covariance-matrix-yield-partial-correlations-between-ran) and [here](https://stats.stackexchange.com/questions/73463/what-does-the-inverse-of-covariance-matrix-say-about-data-intuitively), but a standard linear algebra text reference would be welcome.) Moreover $x'\_j \perp x\_i$ for $j\ne i$ also implies $\text{span}(\{x'\_j \;|\; j\ne i\}$ is the hyperplane orthogonal to $x\_i$, and thus $\angle(x'\_i,\text{span}(\{x'\_j\;|\; j\ne i\}))=\angle(\text{span}(\{x\_j \;|\; j\ne i\}),x\_i)=\theta\_i$ and then $\angle(x'\_i,x'\_j)\ge \max(\theta\_i,\theta\_j)\ge \theta$. In conclusion


$$\displaystyle |c\_{ij}|\le \frac{\cos(\theta)}{\sin(\theta)^2}\quad\text{if } i\ne j$$
$$\displaystyle |c\_{ii}|\le \frac{1}{\sin(\theta)^2}$$


and therefore
$$1=\sum\_{i,j}c\_{ij}\sin(\beta\_i)\sin(\beta\_j)\le \sin(\beta)^2\sum\_{i,j}|c\_{ij}|\le \sin(\beta)^2\frac{n+(n^2-n)\cos(\theta)}{\sin(\theta)^2}$$
$$\sin(\beta)\ge\frac{sin(\theta)}{\sqrt{n(1+(n-1)\cos(\theta))}}$$


This also implies the good (as $\theta\rightarrow 0$) first order approximation $\sin(\beta)\ge \sin(\theta)/n$. $\quad\blacksquare$


I verified empirically that $\frac{sin(\theta)}{\sqrt{n(1+(n-1)\cos(\theta))}}\ge \sin(\frac{\theta}{n})$, which would imply $\beta\ge\frac{\theta}{n}$ too, but I'll leave the proof of that as an exercise, if true.


The example in my other answer led me to **conjecture** that the best possible result should be
$$\sin(\beta)\ge \sin(\theta)\sqrt{\frac{(n-1)\tan(\theta)^2+n}{n((n-1)\tan(\theta)^2+n^2)}}$$


which is not much stonger than the claim, but probably much harder to prove.


