Let $v\in\mathbb{R}^N$, and $H$ a non-trivial subspace (i.e. $H\neq\{0\}$ nor $H\neq\mathbb{R}^N$), and denote by $H^\perp$ its orthogonal complement.

Furthermore denote by $p_{H}(v)$ ($p_{H^\perp}(v)$) the orthogonal projection of $v$ onto $H$ ($H^\perp$).

Is it always true that the sum of the angles between $v$ and each projection is always $\frac{\pi}{2}$? In other words is $$\theta_{vp_{H^\perp}(v)} + \theta_{vp_{H}(v)} = \frac{\pi}{2}$$

where $\theta_{ab}$ denotes the angle between vector $a$ and vector $b$.


The answer depend on which scalar product you have on $\mathbb{R}^n$ and which vector you are considering. If it is the standard scalar product the answer is yes if $v\notin H$ and $v\notin H^\perp$ because the projection to the other factor is the $0-$vector , but your statement is not true in general.

  • $\begingroup$ Is there an elegant proof for this? $\endgroup$ – gota Oct 27 '16 at 16:02
  • $\begingroup$ You don't need proof but counterexample. Try do some counterexample to understand why your claim is false. $\endgroup$ – InsideOut Oct 27 '16 at 16:06
  • $\begingroup$ I meant when all your conditions are satisfied and when the scalar product is the usual one $\endgroup$ – gota Oct 27 '16 at 22:35

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