Orthogonal projection over an orthogonal subspace

In summary: Orth F" is the orthogonal subspace of F; i.e. the plane.Here's another picture that explains things better.Anyway, what I asked before was why can't you draw something like this? It took me about 2 minutes to create the first drawing, and another minute to update it to the above drawing. Using pencil and paper, you could probably create a simple drawing like this in about the same amount of time.
  • #1
coltson
9
0

Homework Statement


Being F = (1,1,-1), the orthogonal projection of (2,4,1) over the orthogonal subspace of F is:

a) (1,2,3)
b) (1/3, 7/3, 8/3)
c) (1/3, 2/3, 8/3)
d) (0,0,0)
e) (1,1,1)

The correct answer is B

Homework Equations



The Attempt at a Solution



Using the orthogonal projection formula:

(1,1,-1).(2,4,-1) * (1,1,-1) = 7/3 * (1,1,-1) = (7/3,7/3,-7/3)
(1,1,-1).(1,1-1)

I am obviously misunderstanding something, I thought that maybe orthogonal subspace ain't (1,1-1), but the answer makes me believe that I should have (1,1-1) in the formula.
 
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  • #2
coltson said:
(1,1,-1).(2,4,-1) * (1,1,-1) = 7/3 * (1,1,-1) = (7/3,7/3,-7/3)
(1,1,-1).(1,1-1)

I am obviously misunderstanding something, I thought that maybe orthogonal subspace ain't (1,1-1), but the answer makes me believe that I should have (1,1-1) in the formula.

Shouldn't that be (2,4,1).
 
  • #3
coltson said:

Homework Statement


Being F = (1,1,-1), the orthogonal projection of (2,4,1) over the orthogonal subspace of F is:
The problem statement is confusing to me. I think it's asking for the projection of <2, 4, 1> onto the the orthogonal subspace of F (i.e., the plane perpendicular to F). It might be helpful to think about the geometry here.
coltson said:
a) (1,2,3)
b) (1/3, 7/3, 8/3)
c) (1/3, 2/3, 8/3)
d) (0,0,0)
e) (1,1,1)

The correct answer is B

Homework Equations



The Attempt at a Solution



Using the orthogonal projection formula:

(1,1,-1).(2,4,-1) * (1,1,-1) = 7/3 * (1,1,-1) = (7/3,7/3,-7/3)
(1,1,-1).(1,1-1)

I am obviously misunderstanding something, I thought that maybe orthogonal subspace ain't (1,1-1), but the answer makes me believe that I should have (1,1-1) in the formula.
 
  • #4
Mark44 said:
The problem statement is confusing to me. I think it's asking for the projection of <2, 4, 1> onto the the orthogonal subspace of F (i.e., the plane perpendicular to F). It might be helpful to think about the geometry here.

In that case, the dot product between F and a random vector contained in the plane would be equal to zero? That means that F is the normal of the plane?

But in the end, what he probably wants? The projection of F over a vector contained in this plane that is perpendicular to F?
 
  • #5
coltson said:
In that case, the dot product between F and a random vector contained in the plane would be equal to zero? That means that F is the normal of the plane?
Yes, that's what it means to be the orthogonal subspace to F.

coltson said:
But in the end, what he probably wants? The projection of F over a vector contained in this plane that is perpendicular to F?
It means to find the projection of <2, 4, 1> onto the plane that is orthogonal to (perpendicular to) F = <1, 1, -1>.

Again, draw a picture.
 
  • #6
Mark44 said:
Yes, that's what it means to be the orthogonal subspace to F.

It means to find the projection of <2, 4, 1> onto the plane that is orthogonal to (perpendicular to) F = <1, 1, -1>.

Again, draw a picture.

Well, I don't know how to draw it, so drawing doesn't help. I can imagine what I would like to draw if I had the skill, and all I see is that the projection is a vector contained in the plane, which means it should satisfy the plane equation.
 
  • #7
coltson said:
Well, I don't know how to draw it, so drawing doesn't help. I can imagine what I would like to draw if I had the skill, and all I see is that the projection is a vector contained in the plane, which means it should satisfy the plane equation.
You can't do something like this?
Projection.png

Not labelled, but the projection of the vector onto the plane is the one heading off to the right and down.
 

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    Projection.png
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  • #8
Mark44 said:
You can't do something like this?
View attachment 218065
Not labelled, but the projection of the vector onto the plane is the one heading off to the right and down.

I do not understand. There is three vectors and three labels, how can it not be labelled? As far as I understand, it is the one with the label "Ortho F". I also do not understand what do you mean by "heading down". Doesn't it has to be contained in the plane? How can it be heading down, since it implies that it would cross the plane?
 
  • #9
coltson said:
I do not understand. There is three vectors and three labels, how can it not be labelled? As far as I understand, it is the one with the label "Ortho F". I also do not understand what do you mean by "heading down". Doesn't it has to be contained in the plane? How can it be heading down, since it implies that it would cross the plane?
What I've labelled "Orth F" is the orthogonal subspace of F; i.e. the plane.
Here's another picture that explains things better.
Projection2.png

Anyway, what I asked before was why can't you draw something like this? It took me about 2 minutes to create the first drawing, and another minute to update it to the above drawing. Using pencil and paper, you could probably create a simple drawing like this in about the same amount of time.
What I'm calling "Proj <2, 4, 1>" is the heavy line segment that is the projection of <2, 4, 1> onto the plane. This projection is perpendicular to F (= <1, 1, -1>) and lies in the plane. Using simple trig (including the dot product), you should be able to find the vector I'm showing with the heavy line.

Clear?
 

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  • #10
Buffu said:
Shouldn't that be (2,4,1).

Here is. A simple typo. Should have realized that before.

Mark44 said:
Anyway, what I asked before was why can't you draw something like this? It took me about 2 minutes to create the first drawing, and another minute to update it to the above drawing. Using pencil and paper, you could probably create a simple drawing like this in about the same amount of time.

Because I had not thought about drawing a "rotated" square as my plane representation. And about the second picture, I also did not think about the dotted line.
 

What is an orthogonal projection over an orthogonal subspace?

An orthogonal projection over an orthogonal subspace is a mathematical process in which a vector or set of vectors is projected onto a subspace that is orthogonal, or perpendicular, to the original vector(s). This results in a new vector that is the closest possible approximation to the original vector(s) within the subspace.

How is an orthogonal projection over an orthogonal subspace calculated?

The calculation of an orthogonal projection over an orthogonal subspace involves finding the projection matrix, which is a square matrix that represents the orthogonal projection onto the subspace. This matrix can be calculated using the Gram-Schmidt process or the QR decomposition method.

What is the significance of using an orthogonal projection over an orthogonal subspace?

An orthogonal projection over an orthogonal subspace is important in many fields of science, particularly in linear algebra and geometry. It allows for the simplification of complex vector spaces and can be used to solve systems of linear equations. It is also used in various applications such as image and signal processing, data compression, and machine learning.

How does an orthogonal projection over an orthogonal subspace differ from a regular projection?

The main difference between an orthogonal projection over an orthogonal subspace and a regular projection is that the former is performed onto a subspace that is perpendicular to the original vector(s), while the latter is performed onto a general subspace without the restriction of orthogonality. This means that the orthogonal projection minimizes the distance between the original vector(s) and the new vector, while a regular projection may not.

Can an orthogonal projection over an orthogonal subspace be applied to higher dimensional spaces?

Yes, an orthogonal projection over an orthogonal subspace can be applied to any number of dimensions. The process remains the same, although the calculations may become more complex as the dimensionality increases. However, the concept and significance of the projection remain the same in higher dimensional spaces.

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