# Thread: Projections and reflections around/onto a plane

1. ## Projections and reflections around/onto a plane

Hi, I've got a model solution to this question but am not convinced by it, can anyone help?

Find a matrix for the orthogonal projection such that (1,1,1) and (-1,1,-1) map to the same point.

If these 2 vectors both map to the same point, it must be (0,0,0)

The solution says that (2,0,2) must map to this point, which I figure is because the vector joining (1,1,1) and (-1,1,-1) is a vector that spans the normal to the plane we are projecting onto.

Hence the plane is 2x+2z=0?

Next assuming this is correct I selected 2 vectors that map to themself in the plane and constructed a matrix.

My question is, is there more than one solution depending on the vectors in the plane that are selected?

Also the second part of the question:

Find a matrix for a reflection that maps (1,1,1) to (-1,-1,1)

For this the solution has used the formula 2P-I, where P is the projection matrix from the first part.

This confused me because I thought this formula could only be used if the reflection is around the plane which the matrix P is describing a projection onto.

However if the plane is the same in both parts, then as (1,1,1) is perpendicular to the plane (from the first part), then wouldnt it map to minus itself rather than (-1,-1,1)?

2. ## Re: Projections and reflections around/onto a plane

Hey darren86.

I just wanted to check whether the projection was done against a plane or against another vector.

You have suggested its a plane which means the distance of the projection is exactly the same along some standard vector.t

It means that p1 - n*<n,p1> = p2 - n*<n,p2> = p3 where n is the normal vector to the plane.

So you have two expressions for n and since ||n|| = 1 if you get 2 of the three you get the third component.

Note that p3 is the final projected point.

3. ## Re: Projections and reflections around/onto a plane

orthogonal projections are linear mappings so if:

P(1,1,1) = P(-1,1,1) then:

P(1,1,1) - P(-1,1,-1) = 0 that is:

P(2,0,2) = 0. if we are talking about an orthogonal projection on to a plane (so that dim(im(P)) = 2), this tells us that ker(P) = <(1,0,1)>, and this is a normal vector to the plane x + z = 0

(we can divide by 2 in the plane equation). let's call ker(P), U, and our plane x + z = 0, W = im(P). the fact that P is an orthogonal projection means: R3 = U⊕W = U⊕U

U has basis {(1,0,1)} and W has basis {(1,0,-1),(0,1,0)}.

since (1,0,0) = (1/2)(1,0,1) + (1/2)(1,0,-1), this tells us that P(1,0,0) = P(1/2,0,1/2) + P(1/2,0,-1/2) = (1/2,0,-1/2) <---first column of P

clearly P(0,1,0) = (0,1,0), since (0,1,0) is already in our plane W, and P fixes W.

finally, P(0,0,1) = P[(1/2)(1,0,1)] + P[(-1/2)(1,0,-1)] = (-1/2,0,1/2) <--3rd column of P, so

$\displaystyle P = \begin{bmatrix}\frac{1}{2}&0&-\frac{1}{2}\\0&1&0\\ -\frac{1}{2}&0&\frac{1}{2} \end{bmatrix}$

one can easily verify that P2 = P (this is the defining property of reflections).

now our reflection R is a "reflection about W", so for v = u+w, where u is in U, and w is in W, R(v) = R(u) + R(w) = -u + w.

we can write this in terms of P like so:

for this same pair u,w: P(v) = P(u) + P(w) = 0 + w = w.

that is: u = v - P(v), and w = P(v), so:

R(v) = -u + w = -(v - P(v)) + P(v) = -v + P(v) + P(v) = 2P(v) - I(v) = (2P- I)(v), so that R = 2P - I or:

$\displaystyle R = \begin{bmatrix}0&0&-1\\0&1&0\\-1&0&0 \end{bmatrix}$

i always like to "make sure" my P and R do what the problem says they should do.

so let's verify that P(1,1,1) = P(-1,1,-1).

P(1,1,1) = (0,1,0).

P(-1,1,-1) = (0,1,0) <---see this point is NOT (0,0,0).

P(x,y,-x) = (x,y,-x) (the points (x,y,-x) are the points in W, the plane x + z = 0, oh look! these are all eigenvectors belonging to the eigenvalue 1), P leaves W fixed.

P(x,0,x) = (0,0,0) (yep, P "wipes out" all the normal vectors to W, also: these are eigenvectors of P belonging to the eigenvalue 0).

now let's verify that R does what it should.

R(x,y,z) = (-z,y,-x). so if (x,y,z) is in U (that is (x,y,z) = (x,0,x)):

R(x,0,x) = (-x,0,-x) = -(x,0,x) (hmm....eigenvectors of R with the eigenvalue -1).

R(x,y,-x) = (x,y,-x) (hmm....eigenvectors of R with the eigenvalue 1).

R indeed fixes W, and sends every element of U to its negative (on the "opposite side of W").

one last check: what is R(1,1,1)?

well, R(1,1,1) = (-1,1,-1), which is confusing to me, since you say the problem says R should map (1,1,1) to (-1,-1,1).

4. ## Re: Projections and reflections around/onto a plane

Originally Posted by Deveno
orthogonal projections are linear mappings so if:

P(1,1,1) = P(-1,1,1) then:

P(1,1,1) - P(-1,1,-1) = 0 that is:

P(2,0,2) = 0. if we are talking about an orthogonal projection on to a plane (so that dim(im(P)) = 2), this tells us that ker(P) = <(1,0,1)>, and this is a normal vector to the plane x + z = 0

(we can divide by 2 in the plane equation). let's call ker(P), U, and our plane x + z = 0, W = im(P). the fact that P is an orthogonal projection means: R3 = U⊕W = U⊕U

U has basis {(1,0,1)} and W has basis {(1,0,-1),(0,1,0)}.

since (1,0,0) = (1/2)(1,0,1) + (1/2)(1,0,-1), this tells us that P(1,0,0) = P(1/2,0,1/2) + P(1/2,0,-1/2) = (1/2,0,-1/2) <---first column of P

clearly P(0,1,0) = (0,1,0), since (0,1,0) is already in our plane W, and P fixes W.

finally, P(0,0,1) = P[(1/2)(1,0,1)] + P[(-1/2)(1,0,-1)] = (-1/2,0,1/2) <--3rd column of P, so

$\displaystyle P = \begin{bmatrix}\frac{1}{2}&0&-\frac{1}{2}\\0&1&0\\ -\frac{1}{2}&0&\frac{1}{2} \end{bmatrix}$

one can easily verify that P2 = P (this is the defining property of reflections).

now our reflection R is a "reflection about W", so for v = u+w, where u is in U, and w is in W, R(v) = R(u) + R(w) = -u + w.

we can write this in terms of P like so:

for this same pair u,w: P(v) = P(u) + P(w) = 0 + w = w.

that is: u = v - P(v), and w = P(v), so:

R(v) = -u + w = -(v - P(v)) + P(v) = -v + P(v) + P(v) = 2P(v) - I(v) = (2P- I)(v), so that R = 2P - I or:

$\displaystyle R = \begin{bmatrix}0&0&-1\\0&1&0\\-1&0&0 \end{bmatrix}$

i always like to "make sure" my P and R do what the problem says they should do.

so let's verify that P(1,1,1) = P(-1,1,-1).

P(1,1,1) = (0,1,0).

P(-1,1,-1) = (0,1,0) <---see this point is NOT (0,0,0).

P(x,y,-x) = (x,y,-x) (the points (x,y,-x) are the points in W, the plane x + z = 0, oh look! these are all eigenvectors belonging to the eigenvalue 1), P leaves W fixed.

P(x,0,x) = (0,0,0) (yep, P "wipes out" all the normal vectors to W, also: these are eigenvectors of P belonging to the eigenvalue 0).

now let's verify that R does what it should.

R(x,y,z) = (-z,y,-x). so if (x,y,z) is in U (that is (x,y,z) = (x,0,x)):

R(x,0,x) = (-x,0,-x) = -(x,0,x) (hmm....eigenvectors of R with the eigenvalue -1).

R(x,y,-x) = (x,y,-x) (hmm....eigenvectors of R with the eigenvalue 1).

R indeed fixes W, and sends every element of U to its negative (on the "opposite side of W").

one last check: what is R(1,1,1)?

well, R(1,1,1) = (-1,1,-1), which is confusing to me, since you say the problem says R should map (1,1,1) to (-1,-1,1).
Thanks that explained things extremely well.

A few questions though: when you found the basis for W, what made you choose them? They are the same as the model solution, but any 2 linearly independent vectors in the plane could be chosen right? which would result in a different matrix?

Also for the reflection how can we presume its a reflection about the same plane as before?

As for the final answer, you got the exact same matrix for the reflection so maybe its a misprint in the question.

5. ## Re: Projections and reflections around/onto a plane

i chose the basis for W so that the standard basis vectors (1,0,0), (0,1,0) and (0,0,1) would be easy to express in terms of the basis for W.

but let's be silly, and choose a different 2. our plane is 2x + 2z = 0, let's say (that's the form you used). so lets's say we use this basis B = {(1,1,-1), (1,2,-1)} (sort of a weird basis, but whatever).

now we have:

(1,0,0) = a(1,0,1) + b(1,1,-1) + c(1,2,-1) = (a+b+c,b+2c,a-b-c)

so a+b+c = 1
b+2c = 0
a-b-c = 0

since b = -2c, re-write eq. 3:

a+2c-c = 0
a+c = 0

and re-write eq. 1:

a-2c+c = 1
a-c = 1

add these two new eqs. together:

2a = 1, a = 1/2.

now we solve for b and c:

b+c = 1/2
-2c+c = 1/2
-c = 1/2
c = -1/2, so b = 1

check:

1/2+1-1/2 = 1

1 + 2(-1/2) = 0

1/2 -1 - (-1/2) = 0. cool.

therefore:

(1,0,0) = (1/2)(1,0,1) + (1,1,-1) -(1/2)(1,2,-1), so

P(1,0,0) = (1/2)P(1,0,1) + P(1,1,-1) - (1/2)P(1,2,-1) = 0(1,0,1) + (1,1,-1) - (1/2)(1,2,-1) = (1,1,-1) + (-1/2,-1,1/2) = (1/2,0,-1/2) <--same first column see?

it doesn't really matter which basis vectors i pick to span W, what matters is that i found P in terms of the "standard" basis. if i pick different basis vectors for W, i'll get (1,0,0) as a different linear combination of them, but P will still do the same thing to (1,0,0), (0,1,0) and (0,0,1)

why do i want what P does to (1,0,0), (0,1,0) and (0,0,1)? because that's the basis our coordinates are in.