Distance between a vector and a subspace












1














Let $V$ be a vector space of finite dimension, and let $0 neq u in V$. Denote $U=(span text{{u}})^bot$.



Prove that for every $v in V$



$$dist(v,U)=frac {lvert leftlangle v,u rightranglervert}{lVert urVert}$$






linear-algebra vector-spaces inner-product-space orthogonality







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  • 1




    Think in terms of projections.
    – Anurag A
    Jun 10 '17 at 17:07
















1














Let $V$ be a vector space of finite dimension, and let $0 neq u in V$. Denote $U=(span text{{u}})^bot$.



Prove that for every $v in V$



$$dist(v,U)=frac {lvert leftlangle v,u rightranglervert}{lVert urVert}$$






linear-algebra vector-spaces inner-product-space orthogonality







share|cite|improve this question




















  • 1




    Think in terms of projections.
    – Anurag A
    Jun 10 '17 at 17:07














1












1








1


1





Let $V$ be a vector space of finite dimension, and let $0 neq u in V$. Denote $U=(span text{{u}})^bot$.



Prove that for every $v in V$



$$dist(v,U)=frac {lvert leftlangle v,u rightranglervert}{lVert urVert}$$






linear-algebra vector-spaces inner-product-space orthogonality







share|cite|improve this question















Let $V$ be a vector space of finite dimension, and let $0 neq u in V$. Denote $U=(span text{{u}})^bot$.



Prove that for every $v in V$



$$dist(v,U)=frac {lvert leftlangle v,u rightranglervert}{lVert urVert}$$






linear-algebra vector-spaces inner-product-space orthogonality




linear-algebra vector-spaces inner-product-space orthogonality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 20 '18 at 11:49









José Carlos Santos

150k22122222




150k22122222










asked Jun 10 '17 at 17:01









user401516

91539




91539








  • 1




    Think in terms of projections.
    – Anurag A
    Jun 10 '17 at 17:07














  • 1




    Think in terms of projections.
    – Anurag A
    Jun 10 '17 at 17:07








1




1




Think in terms of projections.
– Anurag A
Jun 10 '17 at 17:07




Think in terms of projections.
– Anurag A
Jun 10 '17 at 17:07










1 Answer
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The distance form $v$ to $U$ is the distance from $v$ to the only $win U$ such that $v-w$ is orthogonal to $U$, that is, such that $v-w=lambda u$, for some $lambdainmathbb R$. Since $v-frac{langle v,urangle}{|u|^2}uin U$, you just have to take $lambda=frac{langle v,urangle}{|u|^2}$. So, $w=v-lambda u=v-frac{langle v,urangle}{|u|^2}u$ and the distance from $v$ to $w$ is equal to $frac{bigl|langle v,uranglebigr|}{|u|}$.






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    The distance form $v$ to $U$ is the distance from $v$ to the only $win U$ such that $v-w$ is orthogonal to $U$, that is, such that $v-w=lambda u$, for some $lambdainmathbb R$. Since $v-frac{langle v,urangle}{|u|^2}uin U$, you just have to take $lambda=frac{langle v,urangle}{|u|^2}$. So, $w=v-lambda u=v-frac{langle v,urangle}{|u|^2}u$ and the distance from $v$ to $w$ is equal to $frac{bigl|langle v,uranglebigr|}{|u|}$.






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      The distance form $v$ to $U$ is the distance from $v$ to the only $win U$ such that $v-w$ is orthogonal to $U$, that is, such that $v-w=lambda u$, for some $lambdainmathbb R$. Since $v-frac{langle v,urangle}{|u|^2}uin U$, you just have to take $lambda=frac{langle v,urangle}{|u|^2}$. So, $w=v-lambda u=v-frac{langle v,urangle}{|u|^2}u$ and the distance from $v$ to $w$ is equal to $frac{bigl|langle v,uranglebigr|}{|u|}$.






      share|cite|improve this answer
























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        The distance form $v$ to $U$ is the distance from $v$ to the only $win U$ such that $v-w$ is orthogonal to $U$, that is, such that $v-w=lambda u$, for some $lambdainmathbb R$. Since $v-frac{langle v,urangle}{|u|^2}uin U$, you just have to take $lambda=frac{langle v,urangle}{|u|^2}$. So, $w=v-lambda u=v-frac{langle v,urangle}{|u|^2}u$ and the distance from $v$ to $w$ is equal to $frac{bigl|langle v,uranglebigr|}{|u|}$.






        share|cite|improve this answer












        The distance form $v$ to $U$ is the distance from $v$ to the only $win U$ such that $v-w$ is orthogonal to $U$, that is, such that $v-w=lambda u$, for some $lambdainmathbb R$. Since $v-frac{langle v,urangle}{|u|^2}uin U$, you just have to take $lambda=frac{langle v,urangle}{|u|^2}$. So, $w=v-lambda u=v-frac{langle v,urangle}{|u|^2}u$ and the distance from $v$ to $w$ is equal to $frac{bigl|langle v,uranglebigr|}{|u|}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jun 10 '17 at 17:13









        José Carlos Santos

        150k22122222




        150k22122222






























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