The orthogonal projection of u onto v is 0?
Clash Royale CLAN TAG#URR8PPP
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u = $beginbmatrix-1\1\0\1\0endbmatrix$ and v = $beginbmatrix1\0\1\1\1endbmatrix$
Find proj_v^u.
The orthogonal projection of u onto v is equal to (u ∙ v/ v∙v)(v)
u ∙ v = 0. Can the orthogonal projection be equal to zero? How can I visualize this?
linear-algebra vector-spaces orthogonality
asked Apr 8 '16 at 18:27
H.W.
6517
add a comment |Â
up vote
2
down vote
favorite
u = $beginbmatrix-1\1\0\1\0endbmatrix$ and v = $beginbmatrix1\0\1\1\1endbmatrix$
Find proj_v^u.
The orthogonal projection of u onto v is equal to (u ∙ v/ v∙v)(v)
u ∙ v = 0. Can the orthogonal projection be equal to zero? How can I visualize this?
linear-algebra vector-spaces orthogonality
asked Apr 8 '16 at 18:27
H.W.
6517
7
Yes, it can be when $u$ is already orthogonal to $v$. To visualize, stand a pen on your table.
– Zhanxiong
Apr 8 '16 at 18:30
In the plane visualize the orthogonal projection of $(0,1)$ on $(1,0)$. Draw the picture for any orthogonal pair.
– Ethan Bolker
Apr 8 '16 at 18:31
Visualizing this directly is a bit difficult since the vectors are $5$-dimensional. Try with $[-1, 1]$ and $[1,1]$ or $[-1, 1, 1] and $[1, 0, 1]$ instead.
– Arthur
Apr 8 '16 at 18:31
Of course it can be equal to zero. If the vectors are othogonal. To better visualize, imagine an orthogonal projection of a vector $(0,1)$ on a vector $(1,0)$.
– TZakrevskiy
Apr 8 '16 at 18:31
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
u = $beginbmatrix-1\1\0\1\0endbmatrix$ and v = $beginbmatrix1\0\1\1\1endbmatrix$
Find proj_v^u.
The orthogonal projection of u onto v is equal to (u ∙ v/ v∙v)(v)
u ∙ v = 0. Can the orthogonal projection be equal to zero? How can I visualize this?
linear-algebra vector-spaces orthogonality
asked Apr 8 '16 at 18:27
H.W.
6517
u = $beginbmatrix-1\1\0\1\0endbmatrix$ and v = $beginbmatrix1\0\1\1\1endbmatrix$
Find proj_v^u.
The orthogonal projection of u onto v is equal to (u ∙ v/ v∙v)(v)
u ∙ v = 0. Can the orthogonal projection be equal to zero? How can I visualize this?
linear-algebra vector-spaces orthogonality
linear-algebra vector-spaces orthogonality
asked Apr 8 '16 at 18:27
H.W.
6517
asked Apr 8 '16 at 18:27
H.W.
6517
asked Apr 8 '16 at 18:27
H.W.
6517
asked Apr 8 '16 at 18:27
H.W.
6517
asked Apr 8 '16 at 18:27
H.W.
6517
6517
7
Yes, it can be when $u$ is already orthogonal to $v$. To visualize, stand a pen on your table.
– Zhanxiong
Apr 8 '16 at 18:30
In the plane visualize the orthogonal projection of $(0,1)$ on $(1,0)$. Draw the picture for any orthogonal pair.
– Ethan Bolker
Apr 8 '16 at 18:31
Visualizing this directly is a bit difficult since the vectors are $5$-dimensional. Try with $[-1, 1]$ and $[1,1]$ or $[-1, 1, 1] and $[1, 0, 1]$ instead.
– Arthur
Apr 8 '16 at 18:31
Of course it can be equal to zero. If the vectors are othogonal. To better visualize, imagine an orthogonal projection of a vector $(0,1)$ on a vector $(1,0)$.
– TZakrevskiy
Apr 8 '16 at 18:31
add a comment |Â
7
Yes, it can be when $u$ is already orthogonal to $v$. To visualize, stand a pen on your table.
– Zhanxiong
Apr 8 '16 at 18:30
In the plane visualize the orthogonal projection of $(0,1)$ on $(1,0)$. Draw the picture for any orthogonal pair.
– Ethan Bolker
Apr 8 '16 at 18:31
Visualizing this directly is a bit difficult since the vectors are $5$-dimensional. Try with $[-1, 1]$ and $[1,1]$ or $[-1, 1, 1] and $[1, 0, 1]$ instead.
– Arthur
Apr 8 '16 at 18:31
Of course it can be equal to zero. If the vectors are othogonal. To better visualize, imagine an orthogonal projection of a vector $(0,1)$ on a vector $(1,0)$.
– TZakrevskiy
Apr 8 '16 at 18:31
7
7
Yes, it can be when $u$ is already orthogonal to $v$. To visualize, stand a pen on your table.
– Zhanxiong
Apr 8 '16 at 18:30
Yes, it can be when $u$ is already orthogonal to $v$. To visualize, stand a pen on your table.
– Zhanxiong
Apr 8 '16 at 18:30
In the plane visualize the orthogonal projection of $(0,1)$ on $(1,0)$. Draw the picture for any orthogonal pair.
– Ethan Bolker
Apr 8 '16 at 18:31
In the plane visualize the orthogonal projection of $(0,1)$ on $(1,0)$. Draw the picture for any orthogonal pair.
– Ethan Bolker
Apr 8 '16 at 18:31
Visualizing this directly is a bit difficult since the vectors are $5$-dimensional. Try with $[-1, 1]$ and $[1,1]$ or $[-1, 1, 1] and $[1, 0, 1]$ instead.
– Arthur
Apr 8 '16 at 18:31
Visualizing this directly is a bit difficult since the vectors are $5$-dimensional. Try with $[-1, 1]$ and $[1,1]$ or $[-1, 1, 1] and $[1, 0, 1]$ instead.
– Arthur
Apr 8 '16 at 18:31
Of course it can be equal to zero. If the vectors are othogonal. To better visualize, imagine an orthogonal projection of a vector $(0,1)$ on a vector $(1,0)$.
– TZakrevskiy
Apr 8 '16 at 18:31
Of course it can be equal to zero. If the vectors are othogonal. To better visualize, imagine an orthogonal projection of a vector $(0,1)$ on a vector $(1,0)$.
– TZakrevskiy
Apr 8 '16 at 18:31
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
Yes, the projection of $u$ onto $v$ can be $0$. The projection of $u$ onto $v$ is the vector of the form $lambda v$ with smaller distance to $u$. So, asserting that the projection of $u$ onto $v$ is $0$ simply means that of all vectors of the form $lambda v$, the one which is closest to $u$ is the one for which $lambda=0$. Geometrically, this means that $u$ and $v$ are orthogonal.
answered Jul 26 '17 at 13:18
José Carlos Santos
121k16101185
add a comment |Â
up vote
0
down vote
Here is a good visualization of the dot product as a projection can be found here. The whole "essence of linear algebra" series is actually fantastic.
When you watch the video you can see that the dot product of a vector $vec v$ with a unit vector $hat u$ can be seen as the length of the projection of $vec v$ onto $hat u$. If the dot product is positive, it means the projection of $vec v$ lands in the same direction as $hat u$. If the dot product is negative, it means the projection of $vec v$ lands in the opposite direction as $hat u$. If the dot product is zero, it means the projection became the zero vector, which means $vec v$ had to be perpendicular to $hat u$.
answered Sep 15 '16 at 16:08
Chuck Larrieu
1
Your link does not work for me.
– naslundx
Jan 1 at 20:34
add a comment |Â
up vote
0
down vote
It can be zero, and in fact all linear projections will have an entire subspace of vectors which are all equal to zero under the projection. You can see this in your example by taking a scalar multiple of $u$ and noting the dot project remains zero. This subspace will always be orthogonal to the space you're projecting onto, in this case the space spanned by $v$. Visualizing orthogonal subspaces isn't too hard in $mathbbR^3$. You have a plane through the origin which is perpendicular to a line going through the origin. You can either project the plane onto the line, or the line onto the plane with the entire line or plane being mapped to zero under the projection. This is the main model I use for visualizing quotient spaces as well.
answered Nov 2 '17 at 13:43
CyclotomicField
1,6141312
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Yes, the projection of $u$ onto $v$ can be $0$. The projection of $u$ onto $v$ is the vector of the form $lambda v$ with smaller distance to $u$. So, asserting that the projection of $u$ onto $v$ is $0$ simply means that of all vectors of the form $lambda v$, the one which is closest to $u$ is the one for which $lambda=0$. Geometrically, this means that $u$ and $v$ are orthogonal.
answered Jul 26 '17 at 13:18
José Carlos Santos
121k16101185
add a comment |Â
up vote
1
down vote
Yes, the projection of $u$ onto $v$ can be $0$. The projection of $u$ onto $v$ is the vector of the form $lambda v$ with smaller distance to $u$. So, asserting that the projection of $u$ onto $v$ is $0$ simply means that of all vectors of the form $lambda v$, the one which is closest to $u$ is the one for which $lambda=0$. Geometrically, this means that $u$ and $v$ are orthogonal.
answered Jul 26 '17 at 13:18
José Carlos Santos
121k16101185
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Yes, the projection of $u$ onto $v$ can be $0$. The projection of $u$ onto $v$ is the vector of the form $lambda v$ with smaller distance to $u$. So, asserting that the projection of $u$ onto $v$ is $0$ simply means that of all vectors of the form $lambda v$, the one which is closest to $u$ is the one for which $lambda=0$. Geometrically, this means that $u$ and $v$ are orthogonal.
answered Jul 26 '17 at 13:18
José Carlos Santos
121k16101185
Yes, the projection of $u$ onto $v$ can be $0$. The projection of $u$ onto $v$ is the vector of the form $lambda v$ with smaller distance to $u$. So, asserting that the projection of $u$ onto $v$ is $0$ simply means that of all vectors of the form $lambda v$, the one which is closest to $u$ is the one for which $lambda=0$. Geometrically, this means that $u$ and $v$ are orthogonal.
answered Jul 26 '17 at 13:18
José Carlos Santos
121k16101185
answered Jul 26 '17 at 13:18
José Carlos Santos
121k16101185
answered Jul 26 '17 at 13:18
José Carlos Santos
121k16101185
answered Jul 26 '17 at 13:18
José Carlos Santos
121k16101185
121k16101185
add a comment |Â
add a comment |Â
up vote
0
down vote
Here is a good visualization of the dot product as a projection can be found here. The whole "essence of linear algebra" series is actually fantastic.
When you watch the video you can see that the dot product of a vector $vec v$ with a unit vector $hat u$ can be seen as the length of the projection of $vec v$ onto $hat u$. If the dot product is positive, it means the projection of $vec v$ lands in the same direction as $hat u$. If the dot product is negative, it means the projection of $vec v$ lands in the opposite direction as $hat u$. If the dot product is zero, it means the projection became the zero vector, which means $vec v$ had to be perpendicular to $hat u$.
answered Sep 15 '16 at 16:08
Chuck Larrieu
1
Your link does not work for me.
– naslundx
Jan 1 at 20:34
add a comment |Â
up vote
0
down vote
Here is a good visualization of the dot product as a projection can be found here. The whole "essence of linear algebra" series is actually fantastic.
When you watch the video you can see that the dot product of a vector $vec v$ with a unit vector $hat u$ can be seen as the length of the projection of $vec v$ onto $hat u$. If the dot product is positive, it means the projection of $vec v$ lands in the same direction as $hat u$. If the dot product is negative, it means the projection of $vec v$ lands in the opposite direction as $hat u$. If the dot product is zero, it means the projection became the zero vector, which means $vec v$ had to be perpendicular to $hat u$.
answered Sep 15 '16 at 16:08
Chuck Larrieu
1
Your link does not work for me.
– naslundx
Jan 1 at 20:34
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Here is a good visualization of the dot product as a projection can be found here. The whole "essence of linear algebra" series is actually fantastic.
When you watch the video you can see that the dot product of a vector $vec v$ with a unit vector $hat u$ can be seen as the length of the projection of $vec v$ onto $hat u$. If the dot product is positive, it means the projection of $vec v$ lands in the same direction as $hat u$. If the dot product is negative, it means the projection of $vec v$ lands in the opposite direction as $hat u$. If the dot product is zero, it means the projection became the zero vector, which means $vec v$ had to be perpendicular to $hat u$.
answered Sep 15 '16 at 16:08
Chuck Larrieu
1
Here is a good visualization of the dot product as a projection can be found here. The whole "essence of linear algebra" series is actually fantastic.
When you watch the video you can see that the dot product of a vector $vec v$ with a unit vector $hat u$ can be seen as the length of the projection of $vec v$ onto $hat u$. If the dot product is positive, it means the projection of $vec v$ lands in the same direction as $hat u$. If the dot product is negative, it means the projection of $vec v$ lands in the opposite direction as $hat u$. If the dot product is zero, it means the projection became the zero vector, which means $vec v$ had to be perpendicular to $hat u$.
answered Sep 15 '16 at 16:08
Chuck Larrieu
1
answered Sep 15 '16 at 16:08
Chuck Larrieu
1
answered Sep 15 '16 at 16:08
Chuck Larrieu
1
answered Sep 15 '16 at 16:08
Chuck Larrieu
1
1
Your link does not work for me.
– naslundx
Jan 1 at 20:34
add a comment |Â
Your link does not work for me.
– naslundx
Jan 1 at 20:34
Your link does not work for me.
– naslundx
Jan 1 at 20:34
Your link does not work for me.
– naslundx
Jan 1 at 20:34
add a comment |Â
up vote
0
down vote
It can be zero, and in fact all linear projections will have an entire subspace of vectors which are all equal to zero under the projection. You can see this in your example by taking a scalar multiple of $u$ and noting the dot project remains zero. This subspace will always be orthogonal to the space you're projecting onto, in this case the space spanned by $v$. Visualizing orthogonal subspaces isn't too hard in $mathbbR^3$. You have a plane through the origin which is perpendicular to a line going through the origin. You can either project the plane onto the line, or the line onto the plane with the entire line or plane being mapped to zero under the projection. This is the main model I use for visualizing quotient spaces as well.
answered Nov 2 '17 at 13:43
CyclotomicField
1,6141312
add a comment |Â
up vote
0
down vote
It can be zero, and in fact all linear projections will have an entire subspace of vectors which are all equal to zero under the projection. You can see this in your example by taking a scalar multiple of $u$ and noting the dot project remains zero. This subspace will always be orthogonal to the space you're projecting onto, in this case the space spanned by $v$. Visualizing orthogonal subspaces isn't too hard in $mathbbR^3$. You have a plane through the origin which is perpendicular to a line going through the origin. You can either project the plane onto the line, or the line onto the plane with the entire line or plane being mapped to zero under the projection. This is the main model I use for visualizing quotient spaces as well.
answered Nov 2 '17 at 13:43
CyclotomicField
1,6141312
add a comment |Â
up vote
0
down vote
up vote
0
down vote
It can be zero, and in fact all linear projections will have an entire subspace of vectors which are all equal to zero under the projection. You can see this in your example by taking a scalar multiple of $u$ and noting the dot project remains zero. This subspace will always be orthogonal to the space you're projecting onto, in this case the space spanned by $v$. Visualizing orthogonal subspaces isn't too hard in $mathbbR^3$. You have a plane through the origin which is perpendicular to a line going through the origin. You can either project the plane onto the line, or the line onto the plane with the entire line or plane being mapped to zero under the projection. This is the main model I use for visualizing quotient spaces as well.
answered Nov 2 '17 at 13:43
CyclotomicField
1,6141312
It can be zero, and in fact all linear projections will have an entire subspace of vectors which are all equal to zero under the projection. You can see this in your example by taking a scalar multiple of $u$ and noting the dot project remains zero. This subspace will always be orthogonal to the space you're projecting onto, in this case the space spanned by $v$. Visualizing orthogonal subspaces isn't too hard in $mathbbR^3$. You have a plane through the origin which is perpendicular to a line going through the origin. You can either project the plane onto the line, or the line onto the plane with the entire line or plane being mapped to zero under the projection. This is the main model I use for visualizing quotient spaces as well.
answered Nov 2 '17 at 13:43
CyclotomicField
1,6141312
answered Nov 2 '17 at 13:43
CyclotomicField
1,6141312
answered Nov 2 '17 at 13:43
CyclotomicField
1,6141312
answered Nov 2 '17 at 13:43
CyclotomicField
1,6141312
1,6141312
add a comment |Â
add a comment |Â
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7
Yes, it can be when $u$ is already orthogonal to $v$. To visualize, stand a pen on your table.
– Zhanxiong
Apr 8 '16 at 18:30
In the plane visualize the orthogonal projection of $(0,1)$ on $(1,0)$. Draw the picture for any orthogonal pair.
– Ethan Bolker
Apr 8 '16 at 18:31
Visualizing this directly is a bit difficult since the vectors are $5$-dimensional. Try with $[-1, 1]$ and $[1,1]$ or $[-1, 1, 1] and $[1, 0, 1]$ instead.
– Arthur
Apr 8 '16 at 18:31
Of course it can be equal to zero. If the vectors are othogonal. To better visualize, imagine an orthogonal projection of a vector $(0,1)$ on a vector $(1,0)$.
– TZakrevskiy
Apr 8 '16 at 18:31