Mixing
Distance between a vector and a subspace
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
edited Nov 20 '18 at 11:49
José Carlos Santos
150k22122222
asked Jun 10 '17 at 17:01
user401516
91539
add a comment |
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
edited Nov 20 '18 at 11:49
José Carlos Santos
150k22122222
asked Jun 10 '17 at 17:01
user401516
91539
1
Think in terms of projections.
– Anurag A
Jun 10 '17 at 17:07
add a comment |
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
edited Nov 20 '18 at 11:49
José Carlos Santos
150k22122222
asked Jun 10 '17 at 17:01
user401516
91539
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
edited Nov 20 '18 at 11:49
José Carlos Santos
150k22122222
asked Jun 10 '17 at 17:01
user401516
91539
edited Nov 20 '18 at 11:49
José Carlos Santos
150k22122222
asked Jun 10 '17 at 17:01
user401516
91539
edited Nov 20 '18 at 11:49
José Carlos Santos
150k22122222
edited Nov 20 '18 at 11:49
José Carlos Santos
150k22122222
edited Nov 20 '18 at 11:49
José Carlos Santos
150k22122222
150k22122222
asked Jun 10 '17 at 17:01
user401516
91539
asked Jun 10 '17 at 17:01
user401516
91539
asked Jun 10 '17 at 17:01
user401516
91539
91539
1
Think in terms of projections.
– Anurag A
Jun 10 '17 at 17:07
add a comment |
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
add a comment |
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 $frac{bigl|langle v,uranglebigr|}{|u|}$.
answered Jun 10 '17 at 17:13
José Carlos Santos
150k22122222
add a comment |
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1 Answer
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1 Answer
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active
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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 $frac{bigl|langle v,uranglebigr|}{|u|}$.
answered Jun 10 '17 at 17:13
José Carlos Santos
150k22122222
add a comment |
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 $frac{bigl|langle v,uranglebigr|}{|u|}$.
answered Jun 10 '17 at 17:13
José Carlos Santos
150k22122222
add a comment |
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 $frac{bigl|langle v,uranglebigr|}{|u|}$.
answered Jun 10 '17 at 17:13
José Carlos Santos
150k22122222
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 $frac{bigl|langle v,uranglebigr|}{|u|}$.
answered Jun 10 '17 at 17:13
José Carlos Santos
150k22122222
answered Jun 10 '17 at 17:13
José Carlos Santos
150k22122222
answered Jun 10 '17 at 17:13
José Carlos Santos
150k22122222
answered Jun 10 '17 at 17:13
José Carlos Santos
150k22122222
150k22122222
add a comment |
add a comment |
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1
Think in terms of projections.
– Anurag A
Jun 10 '17 at 17:07