Orthogonal Projection Matrices for Points on Subspaces in R^3

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The discussion focuses on finding orthogonal projection matrices for specific subspaces in R^3. For the z-axis, the transformation matrix is straightforward: [0 0 0; 0 0 0; 0 0 1]. For the line defined by x=y=2z, the projection involves calculating the projection of vectors onto the line, resulting in the first column of the matrix being [4/3, 4/3, 2/3]. To project onto the plane x+y+z=0, the normal vector is used to find the projection onto the normal line, followed by subtracting this from the original vector to obtain the projection onto the plane. The thread highlights the systematic approach to deriving these matrices through vector projections.
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Find the matrices of the transformations T which orthogonally project a point (x,y,z) on to the following subspaces of R^3.

(a) The z-axis
(b) the straight line x=y=2z
(c) the plane x+y+z=0

(a) is easy just the matrix [0 0 0;0 0 0;0 0 1]

as for (b) and (c) i have no idea how to work them out. I think (b) might have something to do with projection of a vector on to another... ((u.v)/(|u|^2))u

So maybe v = (x,y,z) and u = a(1,1,2) (a is any real number)

But I'm really stuck on what to do

Any help like usual greatly appreciated :)
 
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forty said:
Find the matrices of the transformations T which orthogonally project a point (x,y,z) on to the following subspaces of R^3.

(a) The z-axis
(b) the straight line x=y=2z
(c) the plane x+y+z=0

(a) is easy just the matrix [0 0 0;0 0 0;0 0 1]

as for (b) and (c) i have no idea how to work them out. I think (b) might have something to do with projection of a vector on to another... ((u.v)/(|u|^2))u

So maybe v = (x,y,z) and u = a(1,1,2) (a is any real number)

But I'm really stuck on what to do

Any help like usual greatly appreciated :)

The simplest way to find the matrix of a linear transformation is to apply the transformation to each basis vector in turn. That gives the columns of the matrix.
For example, projecting \vec{i}= (1, 0, 0) onto the z-axis gives (0, 0, 0). The first column of that matrix is [0 0 0].

Yes, it has everything to do with the projection of one vector onto another. One vector on the line x= y= 2z is (2, 2, 1) (take z= 1). The projection of (1, 0, 0) onto that is, using the formula you give, (2)/(3)(2,2,1)= (4/3, 4/3, 2/3). The first column of the matrix is [4/3, 4/3, 2/3]. The projection of (0, 1, 0) onto (2, 2, 1) is the same and the projection of (0, 0, 1) onto it is (1)/(3)(2, 2, 1)= (2/3, 2/3, 1/3).

To project onto a plane, project onto its normal line and then subtract. The normal vector of the plane x+ y+ z= 0 is (1, 1, 1). The projection of (1, 0, 0) onto that line is (1)/\sqrt{3}(1, 1, 1)= (\sqrt{3},\sqrt{3},\sqrt{3}) so the projection onto the plane is (1, 0, 0)- (\sqrt{3},\sqrt{3},\sqrt{3})= (1- \sqrt{3}, -\sqrt{3}, - \sqrt{3}). That is the first column of the matrix. Do the same with (0, 1, 0) and (0, 0, 1).
 
Amazing! You always make things so simple, Thanks a lot :)

by the way you seem to have a 'thing' for linear algebra...
 
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