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How is $I-vv^T$ the projection onto the orthogonal complement of the line through the origin in the direction $v$?
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Given a vector $vin mathbbR^n$, I have seen from several sources that $I-vv^T$ is the projection onto the orthogonal complement of the line through the origin in the direction $v$?
I have unfortunately memorized this expression instead of figuring out what it actually means. I know that the orthogonal complement of a line is a plane with the line being its normal vector. And from what I understand this means that $P=I-vv^T$ is a projection matrix which places vectors into that aforementioned plane. But beyond that, I'm having difficulty connecting $I-vv^T$ to all vectors which are orthogonal to $v$ (which is what an orthogonal complement is, after all).
Can someone help me understand how this is true?
vector-spaces vectors orthogonality plane-geometry projection-matrices
asked Jul 28 at 6:27
Hunle
10311
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up vote
0
down vote
favorite
Given a vector $vin mathbbR^n$, I have seen from several sources that $I-vv^T$ is the projection onto the orthogonal complement of the line through the origin in the direction $v$?
I have unfortunately memorized this expression instead of figuring out what it actually means. I know that the orthogonal complement of a line is a plane with the line being its normal vector. And from what I understand this means that $P=I-vv^T$ is a projection matrix which places vectors into that aforementioned plane. But beyond that, I'm having difficulty connecting $I-vv^T$ to all vectors which are orthogonal to $v$ (which is what an orthogonal complement is, after all).
Can someone help me understand how this is true?
vector-spaces vectors orthogonality plane-geometry projection-matrices
asked Jul 28 at 6:27
Hunle
10311
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given a vector $vin mathbbR^n$, I have seen from several sources that $I-vv^T$ is the projection onto the orthogonal complement of the line through the origin in the direction $v$?
I have unfortunately memorized this expression instead of figuring out what it actually means. I know that the orthogonal complement of a line is a plane with the line being its normal vector. And from what I understand this means that $P=I-vv^T$ is a projection matrix which places vectors into that aforementioned plane. But beyond that, I'm having difficulty connecting $I-vv^T$ to all vectors which are orthogonal to $v$ (which is what an orthogonal complement is, after all).
Can someone help me understand how this is true?
vector-spaces vectors orthogonality plane-geometry projection-matrices
asked Jul 28 at 6:27
Hunle
10311
Given a vector $vin mathbbR^n$, I have seen from several sources that $I-vv^T$ is the projection onto the orthogonal complement of the line through the origin in the direction $v$?
I have unfortunately memorized this expression instead of figuring out what it actually means. I know that the orthogonal complement of a line is a plane with the line being its normal vector. And from what I understand this means that $P=I-vv^T$ is a projection matrix which places vectors into that aforementioned plane. But beyond that, I'm having difficulty connecting $I-vv^T$ to all vectors which are orthogonal to $v$ (which is what an orthogonal complement is, after all).
Can someone help me understand how this is true?
vector-spaces vectors orthogonality plane-geometry projection-matrices
asked Jul 28 at 6:27
Hunle
10311
asked Jul 28 at 6:27
Hunle
10311
asked Jul 28 at 6:27
Hunle
10311
asked Jul 28 at 6:27
Hunle
10311
10311
add a comment |Â
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
2
down vote
Given a vector $x$, and a unit vector $v$, the projection of $x$ onto $v$ is going to be $(v^Tx)v=v(v^Tx)=(vv^T)x$.
Hence the orthogonal complement is $x-(v^Tx)v=x-(vv^T)x=(I-vv^T)x.$
answered Jul 28 at 6:36
Siong Thye Goh
77k134794
I see, so $v$ must be a unit vector - not just any vector.
– Hunle
Jul 28 at 19:36
Yup, suppose not, divide by the norm to make it a unit vector.
– Siong Thye Goh
Jul 29 at 1:57
add a comment |Â
up vote
1
down vote
We have that
$$(vv^T)b=(v^Tb)v$$
is the othogonal projection of $b$ onto $v$ then
$$b-(vv^T)b=(I-vv^T)b$$
is the projection onto the orthogonal complement.
answered Jul 28 at 6:41
gimusi
64.8k73483
only if $v$ has norm 1, it seems to me.
– Hunle
Jul 28 at 19:38
@Hunle Yes of course, we are assuming that $v$ is a direction vector with $|n|=1$.
– gimusi
Jul 28 at 19:39
add a comment |Â
up vote
1
down vote
Let $v$ an unit vector and $(v,u_1,dots,u_n-1)$ an orthonormal basis of $mathbbR^n$ (Constructed by Grahm-Schmitt for example).
Then $x$ can be decomposed in that basis, i.e. : $$x=lambda v+sum_i=1^n-1 lambda_i u_i.$$
Then $v^Tx$ is the inner product and its values is $lambda$. Thus $vv^Tx= lambda v$. In other words, $vv^Tx$ is the orthogonal projection into the vector space spanned by $v$.
Then $$(I-vv^T)x=sum_i=1^n-1 lambda_i u_i.$$
What we have done is removing the $v$ component of $x$. Since our basis is orthogonal, $$v^T(I-vv^T)x=sum_i=1^n-1lambda_i v^Tu_i=0.$$
$(I-vv^T)$ fails to be a projection into $ Span(u_1,..,u_n-1)$ if $v$ is not a unit vector, as $vv^Tx=v|v|_2lambda$, hence :
$$(I-vv^T)x=(1-|v|_2)lambda v +sum_i=1^n-1 lambda_i u_i.$$
which still have a component in $v$.
answered Jul 28 at 6:42
nicomezi
3,4121819
1
Can someone explain the downvote ? Did I say something wrong ?
– nicomezi
Jul 28 at 10:40
add a comment |Â
up vote
1
down vote
Let $E$ be the orthogonal space to the line defined by $v$. The orthogonal projection $P: mathbb R^n to E$ is the unique operator such that the following conditions are satisfied:
- The image of $P$ is $E$
- $langle x - Px, E rangle = 0$
- $E$ is $P$-invariant
To see uniqueness, suppose that $P'$ was another orthogonal projection, then
$$
0 = langle Px - P'Px, E rangle = langle Px - P'x, E rangle,
$$
and since $P, P'$ both map to $E$, they are equal. (Existence is typically shown by using an orthonormal basis of $E$ that is extended to an orthonormal basis of $mathbb R^n$, using e.g. basis extension and Gram--Schmidt.) Hence, it is sufficient to show that your projection satisfies these properties.
So, let $x in mathbb R^n$.
- $I - v v^t$ maps to $E$, since $langle (I - vv^t)x, v rangle = 0$ (here we need $v$ to be a unit vector and use $langle v v^t x, v rangle = langle x, v v^t v rangle$. For reasons of dimension ($vv^t$ is a matrix of rank $1$), the image equals $E$.
- This is because $langle -v v^t x, y rangle = 0$ for $y in E$.
- This is again by orthogonality, namely $v v^t y = v langle v, yrangle = 0$ for $y in E$.
answered Jul 28 at 6:48
AlgebraicsAnonymous
66611
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Given a vector $x$, and a unit vector $v$, the projection of $x$ onto $v$ is going to be $(v^Tx)v=v(v^Tx)=(vv^T)x$.
Hence the orthogonal complement is $x-(v^Tx)v=x-(vv^T)x=(I-vv^T)x.$
answered Jul 28 at 6:36
Siong Thye Goh
77k134794
I see, so $v$ must be a unit vector - not just any vector.
– Hunle
Jul 28 at 19:36
Yup, suppose not, divide by the norm to make it a unit vector.
– Siong Thye Goh
Jul 29 at 1:57
add a comment |Â
up vote
2
down vote
Given a vector $x$, and a unit vector $v$, the projection of $x$ onto $v$ is going to be $(v^Tx)v=v(v^Tx)=(vv^T)x$.
Hence the orthogonal complement is $x-(v^Tx)v=x-(vv^T)x=(I-vv^T)x.$
answered Jul 28 at 6:36
Siong Thye Goh
77k134794
I see, so $v$ must be a unit vector - not just any vector.
– Hunle
Jul 28 at 19:36
Yup, suppose not, divide by the norm to make it a unit vector.
– Siong Thye Goh
Jul 29 at 1:57
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Given a vector $x$, and a unit vector $v$, the projection of $x$ onto $v$ is going to be $(v^Tx)v=v(v^Tx)=(vv^T)x$.
Hence the orthogonal complement is $x-(v^Tx)v=x-(vv^T)x=(I-vv^T)x.$
answered Jul 28 at 6:36
Siong Thye Goh
77k134794
Given a vector $x$, and a unit vector $v$, the projection of $x$ onto $v$ is going to be $(v^Tx)v=v(v^Tx)=(vv^T)x$.
Hence the orthogonal complement is $x-(v^Tx)v=x-(vv^T)x=(I-vv^T)x.$
answered Jul 28 at 6:36
Siong Thye Goh
77k134794
answered Jul 28 at 6:36
Siong Thye Goh
77k134794
answered Jul 28 at 6:36
Siong Thye Goh
77k134794
answered Jul 28 at 6:36
Siong Thye Goh
77k134794
77k134794
I see, so $v$ must be a unit vector - not just any vector.
– Hunle
Jul 28 at 19:36
Yup, suppose not, divide by the norm to make it a unit vector.
– Siong Thye Goh
Jul 29 at 1:57
add a comment |Â
I see, so $v$ must be a unit vector - not just any vector.
– Hunle
Jul 28 at 19:36
Yup, suppose not, divide by the norm to make it a unit vector.
– Siong Thye Goh
Jul 29 at 1:57
I see, so $v$ must be a unit vector - not just any vector.
– Hunle
Jul 28 at 19:36
I see, so $v$ must be a unit vector - not just any vector.
– Hunle
Jul 28 at 19:36
Yup, suppose not, divide by the norm to make it a unit vector.
– Siong Thye Goh
Jul 29 at 1:57
Yup, suppose not, divide by the norm to make it a unit vector.
– Siong Thye Goh
Jul 29 at 1:57
add a comment |Â
up vote
1
down vote
We have that
$$(vv^T)b=(v^Tb)v$$
is the othogonal projection of $b$ onto $v$ then
$$b-(vv^T)b=(I-vv^T)b$$
is the projection onto the orthogonal complement.
answered Jul 28 at 6:41
gimusi
64.8k73483
only if $v$ has norm 1, it seems to me.
– Hunle
Jul 28 at 19:38
@Hunle Yes of course, we are assuming that $v$ is a direction vector with $|n|=1$.
– gimusi
Jul 28 at 19:39
add a comment |Â
up vote
1
down vote
We have that
$$(vv^T)b=(v^Tb)v$$
is the othogonal projection of $b$ onto $v$ then
$$b-(vv^T)b=(I-vv^T)b$$
is the projection onto the orthogonal complement.
answered Jul 28 at 6:41
gimusi
64.8k73483
only if $v$ has norm 1, it seems to me.
– Hunle
Jul 28 at 19:38
@Hunle Yes of course, we are assuming that $v$ is a direction vector with $|n|=1$.
– gimusi
Jul 28 at 19:39
add a comment |Â
up vote
1
down vote
up vote
1
down vote
We have that
$$(vv^T)b=(v^Tb)v$$
is the othogonal projection of $b$ onto $v$ then
$$b-(vv^T)b=(I-vv^T)b$$
is the projection onto the orthogonal complement.
answered Jul 28 at 6:41
gimusi
64.8k73483
We have that
$$(vv^T)b=(v^Tb)v$$
is the othogonal projection of $b$ onto $v$ then
$$b-(vv^T)b=(I-vv^T)b$$
is the projection onto the orthogonal complement.
answered Jul 28 at 6:41
gimusi
64.8k73483
answered Jul 28 at 6:41
gimusi
64.8k73483
answered Jul 28 at 6:41
gimusi
64.8k73483
answered Jul 28 at 6:41
gimusi
64.8k73483
64.8k73483
only if $v$ has norm 1, it seems to me.
– Hunle
Jul 28 at 19:38
@Hunle Yes of course, we are assuming that $v$ is a direction vector with $|n|=1$.
– gimusi
Jul 28 at 19:39
add a comment |Â
only if $v$ has norm 1, it seems to me.
– Hunle
Jul 28 at 19:38
@Hunle Yes of course, we are assuming that $v$ is a direction vector with $|n|=1$.
– gimusi
Jul 28 at 19:39
only if $v$ has norm 1, it seems to me.
– Hunle
Jul 28 at 19:38
only if $v$ has norm 1, it seems to me.
– Hunle
Jul 28 at 19:38
@Hunle Yes of course, we are assuming that $v$ is a direction vector with $|n|=1$.
– gimusi
Jul 28 at 19:39
@Hunle Yes of course, we are assuming that $v$ is a direction vector with $|n|=1$.
– gimusi
Jul 28 at 19:39
add a comment |Â
up vote
1
down vote
Let $v$ an unit vector and $(v,u_1,dots,u_n-1)$ an orthonormal basis of $mathbbR^n$ (Constructed by Grahm-Schmitt for example).
Then $x$ can be decomposed in that basis, i.e. : $$x=lambda v+sum_i=1^n-1 lambda_i u_i.$$
Then $v^Tx$ is the inner product and its values is $lambda$. Thus $vv^Tx= lambda v$. In other words, $vv^Tx$ is the orthogonal projection into the vector space spanned by $v$.
Then $$(I-vv^T)x=sum_i=1^n-1 lambda_i u_i.$$
What we have done is removing the $v$ component of $x$. Since our basis is orthogonal, $$v^T(I-vv^T)x=sum_i=1^n-1lambda_i v^Tu_i=0.$$
$(I-vv^T)$ fails to be a projection into $ Span(u_1,..,u_n-1)$ if $v$ is not a unit vector, as $vv^Tx=v|v|_2lambda$, hence :
$$(I-vv^T)x=(1-|v|_2)lambda v +sum_i=1^n-1 lambda_i u_i.$$
which still have a component in $v$.
answered Jul 28 at 6:42
nicomezi
3,4121819
1
Can someone explain the downvote ? Did I say something wrong ?
– nicomezi
Jul 28 at 10:40
add a comment |Â
up vote
1
down vote
Let $v$ an unit vector and $(v,u_1,dots,u_n-1)$ an orthonormal basis of $mathbbR^n$ (Constructed by Grahm-Schmitt for example).
Then $x$ can be decomposed in that basis, i.e. : $$x=lambda v+sum_i=1^n-1 lambda_i u_i.$$
Then $v^Tx$ is the inner product and its values is $lambda$. Thus $vv^Tx= lambda v$. In other words, $vv^Tx$ is the orthogonal projection into the vector space spanned by $v$.
Then $$(I-vv^T)x=sum_i=1^n-1 lambda_i u_i.$$
What we have done is removing the $v$ component of $x$. Since our basis is orthogonal, $$v^T(I-vv^T)x=sum_i=1^n-1lambda_i v^Tu_i=0.$$
$(I-vv^T)$ fails to be a projection into $ Span(u_1,..,u_n-1)$ if $v$ is not a unit vector, as $vv^Tx=v|v|_2lambda$, hence :
$$(I-vv^T)x=(1-|v|_2)lambda v +sum_i=1^n-1 lambda_i u_i.$$
which still have a component in $v$.
answered Jul 28 at 6:42
nicomezi
3,4121819
1
Can someone explain the downvote ? Did I say something wrong ?
– nicomezi
Jul 28 at 10:40
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $v$ an unit vector and $(v,u_1,dots,u_n-1)$ an orthonormal basis of $mathbbR^n$ (Constructed by Grahm-Schmitt for example).
Then $x$ can be decomposed in that basis, i.e. : $$x=lambda v+sum_i=1^n-1 lambda_i u_i.$$
Then $v^Tx$ is the inner product and its values is $lambda$. Thus $vv^Tx= lambda v$. In other words, $vv^Tx$ is the orthogonal projection into the vector space spanned by $v$.
Then $$(I-vv^T)x=sum_i=1^n-1 lambda_i u_i.$$
What we have done is removing the $v$ component of $x$. Since our basis is orthogonal, $$v^T(I-vv^T)x=sum_i=1^n-1lambda_i v^Tu_i=0.$$
$(I-vv^T)$ fails to be a projection into $ Span(u_1,..,u_n-1)$ if $v$ is not a unit vector, as $vv^Tx=v|v|_2lambda$, hence :
$$(I-vv^T)x=(1-|v|_2)lambda v +sum_i=1^n-1 lambda_i u_i.$$
which still have a component in $v$.
answered Jul 28 at 6:42
nicomezi
3,4121819
Let $v$ an unit vector and $(v,u_1,dots,u_n-1)$ an orthonormal basis of $mathbbR^n$ (Constructed by Grahm-Schmitt for example).
Then $x$ can be decomposed in that basis, i.e. : $$x=lambda v+sum_i=1^n-1 lambda_i u_i.$$
Then $v^Tx$ is the inner product and its values is $lambda$. Thus $vv^Tx= lambda v$. In other words, $vv^Tx$ is the orthogonal projection into the vector space spanned by $v$.
Then $$(I-vv^T)x=sum_i=1^n-1 lambda_i u_i.$$
What we have done is removing the $v$ component of $x$. Since our basis is orthogonal, $$v^T(I-vv^T)x=sum_i=1^n-1lambda_i v^Tu_i=0.$$
$(I-vv^T)$ fails to be a projection into $ Span(u_1,..,u_n-1)$ if $v$ is not a unit vector, as $vv^Tx=v|v|_2lambda$, hence :
$$(I-vv^T)x=(1-|v|_2)lambda v +sum_i=1^n-1 lambda_i u_i.$$
which still have a component in $v$.
answered Jul 28 at 6:42
nicomezi
3,4121819
answered Jul 28 at 6:42
nicomezi
3,4121819
answered Jul 28 at 6:42
nicomezi
3,4121819
answered Jul 28 at 6:42
nicomezi
3,4121819
3,4121819
1
Can someone explain the downvote ? Did I say something wrong ?
– nicomezi
Jul 28 at 10:40
add a comment |Â
1
Can someone explain the downvote ? Did I say something wrong ?
– nicomezi
Jul 28 at 10:40
1
1
Can someone explain the downvote ? Did I say something wrong ?
– nicomezi
Jul 28 at 10:40
Can someone explain the downvote ? Did I say something wrong ?
– nicomezi
Jul 28 at 10:40
add a comment |Â
up vote
1
down vote
Let $E$ be the orthogonal space to the line defined by $v$. The orthogonal projection $P: mathbb R^n to E$ is the unique operator such that the following conditions are satisfied:
- The image of $P$ is $E$
- $langle x - Px, E rangle = 0$
- $E$ is $P$-invariant
To see uniqueness, suppose that $P'$ was another orthogonal projection, then
$$
0 = langle Px - P'Px, E rangle = langle Px - P'x, E rangle,
$$
and since $P, P'$ both map to $E$, they are equal. (Existence is typically shown by using an orthonormal basis of $E$ that is extended to an orthonormal basis of $mathbb R^n$, using e.g. basis extension and Gram--Schmidt.) Hence, it is sufficient to show that your projection satisfies these properties.
So, let $x in mathbb R^n$.
- $I - v v^t$ maps to $E$, since $langle (I - vv^t)x, v rangle = 0$ (here we need $v$ to be a unit vector and use $langle v v^t x, v rangle = langle x, v v^t v rangle$. For reasons of dimension ($vv^t$ is a matrix of rank $1$), the image equals $E$.
- This is because $langle -v v^t x, y rangle = 0$ for $y in E$.
- This is again by orthogonality, namely $v v^t y = v langle v, yrangle = 0$ for $y in E$.
answered Jul 28 at 6:48
AlgebraicsAnonymous
66611
add a comment |Â
up vote
1
down vote
Let $E$ be the orthogonal space to the line defined by $v$. The orthogonal projection $P: mathbb R^n to E$ is the unique operator such that the following conditions are satisfied:
- The image of $P$ is $E$
- $langle x - Px, E rangle = 0$
- $E$ is $P$-invariant
To see uniqueness, suppose that $P'$ was another orthogonal projection, then
$$
0 = langle Px - P'Px, E rangle = langle Px - P'x, E rangle,
$$
and since $P, P'$ both map to $E$, they are equal. (Existence is typically shown by using an orthonormal basis of $E$ that is extended to an orthonormal basis of $mathbb R^n$, using e.g. basis extension and Gram--Schmidt.) Hence, it is sufficient to show that your projection satisfies these properties.
So, let $x in mathbb R^n$.
- $I - v v^t$ maps to $E$, since $langle (I - vv^t)x, v rangle = 0$ (here we need $v$ to be a unit vector and use $langle v v^t x, v rangle = langle x, v v^t v rangle$. For reasons of dimension ($vv^t$ is a matrix of rank $1$), the image equals $E$.
- This is because $langle -v v^t x, y rangle = 0$ for $y in E$.
- This is again by orthogonality, namely $v v^t y = v langle v, yrangle = 0$ for $y in E$.
answered Jul 28 at 6:48
AlgebraicsAnonymous
66611
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $E$ be the orthogonal space to the line defined by $v$. The orthogonal projection $P: mathbb R^n to E$ is the unique operator such that the following conditions are satisfied:
- The image of $P$ is $E$
- $langle x - Px, E rangle = 0$
- $E$ is $P$-invariant
To see uniqueness, suppose that $P'$ was another orthogonal projection, then
$$
0 = langle Px - P'Px, E rangle = langle Px - P'x, E rangle,
$$
and since $P, P'$ both map to $E$, they are equal. (Existence is typically shown by using an orthonormal basis of $E$ that is extended to an orthonormal basis of $mathbb R^n$, using e.g. basis extension and Gram--Schmidt.) Hence, it is sufficient to show that your projection satisfies these properties.
So, let $x in mathbb R^n$.
- $I - v v^t$ maps to $E$, since $langle (I - vv^t)x, v rangle = 0$ (here we need $v$ to be a unit vector and use $langle v v^t x, v rangle = langle x, v v^t v rangle$. For reasons of dimension ($vv^t$ is a matrix of rank $1$), the image equals $E$.
- This is because $langle -v v^t x, y rangle = 0$ for $y in E$.
- This is again by orthogonality, namely $v v^t y = v langle v, yrangle = 0$ for $y in E$.
answered Jul 28 at 6:48
AlgebraicsAnonymous
66611
Let $E$ be the orthogonal space to the line defined by $v$. The orthogonal projection $P: mathbb R^n to E$ is the unique operator such that the following conditions are satisfied:
- The image of $P$ is $E$
- $langle x - Px, E rangle = 0$
- $E$ is $P$-invariant
To see uniqueness, suppose that $P'$ was another orthogonal projection, then
$$
0 = langle Px - P'Px, E rangle = langle Px - P'x, E rangle,
$$
and since $P, P'$ both map to $E$, they are equal. (Existence is typically shown by using an orthonormal basis of $E$ that is extended to an orthonormal basis of $mathbb R^n$, using e.g. basis extension and Gram--Schmidt.) Hence, it is sufficient to show that your projection satisfies these properties.
So, let $x in mathbb R^n$.
- $I - v v^t$ maps to $E$, since $langle (I - vv^t)x, v rangle = 0$ (here we need $v$ to be a unit vector and use $langle v v^t x, v rangle = langle x, v v^t v rangle$. For reasons of dimension ($vv^t$ is a matrix of rank $1$), the image equals $E$.
- This is because $langle -v v^t x, y rangle = 0$ for $y in E$.
- This is again by orthogonality, namely $v v^t y = v langle v, yrangle = 0$ for $y in E$.
answered Jul 28 at 6:48
AlgebraicsAnonymous
66611
answered Jul 28 at 6:48
AlgebraicsAnonymous
66611
answered Jul 28 at 6:48
AlgebraicsAnonymous
66611
answered Jul 28 at 6:48
AlgebraicsAnonymous
66611
66611
add a comment |Â
add a comment |Â
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