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**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: R^{3} = 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 P^{2} = 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).