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Let's define $x$ as a vector in $\mathbb R^n$

Let's define $V$ as the set of all vectors orthogonal to $x$, i.e $V$={$y$ in $\mathbb R^n$|$x·y=0$}

Let's define $z$ as another vector in $\mathbb R^n$

Calculate the distance between $z$ and the nearest point to $z$ in $V$, i.e min||z-y|| for a vector $y$ in $V$.

After thinking about this, would the answer be $0$? For example, let's say x is the z-axis (0,0,1). So the vectors in V would be the ones around it of any length in the x-y axis. If z is any other vector in $\mathbb R^n$, wouldn't the euclidean distance between z and a vector in V be $0$? Because you could find any vector in V that would intersect z or be infinitesimally close.

If my thinking process is wrong, any help would be great! I'm looking for a way to formalize my thoughts better :)

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Are you familiar with orthogonal decomposition? If we take a vector x as you have done, then we could normalize it and obtain an orthonormal basis with $\frac{x}{||x||}$ as a first vector. Your V would then be the space created by all the other vectors in the orthonormal basis. Also, what do u really mean by "intersection"? If what you defined as $z$ has a non-zero component in the $\frac{x}{||x||}$ axis, then your minimum will be bigger than zero. If you have any questions, feel free to ask

Edit: The correct generalization of the distance from a given vector to the given subspace is the projection of the given vector onto the subspace. The projection can be found by, for example, switching your vector into a basis where the some vectors are the ones that create the subspace and thus looking at the components of the vector in the remaining subspaces.

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  • $\begingroup$ Hi, thanks for the answer! I think I remember orthogonal decomposition (it's been a while) and your explanation makes sense. To make sure I'm understanding this correctly, the other vectors in the orthonormal basis would be (1,0,0) and (0,1,0) and (x,y,0) where root(x^2+y^2)=1, right? And can you explain a bit more about the minimum being greater than zero? Thanks! $\endgroup$
    – Anthony
    Jan 25 '19 at 18:42
  • $\begingroup$ Yes, for your example basis you can complete it with (1,0,0) and (0,1,0).For the minimum, the key thing to realize is that the distance between a vector and some subspace can only be zero when the vector itself is part of the subspace. In your example, if you have take a vector where the z component is non zero, it cannot be part of that subspace because it wouldn't necessarily be orthogonal. $\endgroup$ Jan 25 '19 at 18:45
  • $\begingroup$ Ah, that totally makes sense to me. If I try to explain this more, my thoughts were that since the set V contains all vectors orthogonal to the z-axis, V contains all vectors of the form (x,y,0), where x,y∈ℝ. So if z was (0,0,1) and a vector in V was (x,0,0) where x was infinitesimally small, wouldn't the distance between the vector and z be almost 0? $\endgroup$
    – Anthony
    Jan 25 '19 at 18:56
  • $\begingroup$ Ah, well i understand your confusion. This isn't what people commonly mean by distance when vectors are involved. You seem to be thinking that the distance is 0 when vectors intersect when taken as lines from the origin. But in that case, since every vector is given relative to the origin, wouldnt all vectors intersect(at the origin)? $\endgroup$ Jan 25 '19 at 18:57
  • $\begingroup$ You're right. I understand that now (my bad!). But am I still correct that V contains all vectors of the form (x,y,0), where x,y∈ℝ? Because if that's the case, I still find it difficult to generalize this answer without using this specific example :( $\endgroup$
    – Anthony
    Jan 25 '19 at 19:07
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In your example you chose that $x=(0,0,1)\Rightarrow y=(x,y,0)$ where $x,y\in\mathbb{R}$. If you choose that $z=(1,1,1)$ in your example, you will have that $\min|z-y|=\min|(1-x,1-y,1)|=1$.

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  • $\begingroup$ That makes sense. Do you have any idea on how to generalize that example? or would the minimum always be 1 $\endgroup$
    – Anthony
    Jan 25 '19 at 18:45
  • $\begingroup$ The general idea is that the minimum distance between z and y is how much z "points" in the x-direction. If we let $z=(a,b,c)$ we have $\min|(a-x,b-y,c)|=|c|$, since $x$ and $y$ are just arbitrary numbers we choose $x=a$ and $y=b$. $\endgroup$
    – kvang
    Jan 25 '19 at 18:54

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