Thread: Orthogonal projection and linear transformation

Let T : R^3 -> R^3 be orthogonal projection onto the plane x + y + z = 0.
a) Show that the closest point to (1,0,0) on the plane x + y + z = - is (2/3,-1/3,-1/3).
b) Show that the standard matrix for T is A = 1/3[2 -1 -1; -1 2 -1; -1 -1 2]

Can anyone show me how to do (a)? I'm not sure I have done it right. I first found the basis for the plane: {(-1,1,0),(-1,0,1)} (correct?) and then changed it to an orthonormal basis. Then I went on to find the orthogonal projection of (1,0,0) using the orthonormal basis vectors.

Any ideas about (b)?

The shortest distance from a point to a plane is along the line perpendicular to the plane. The normal vector to x+ y+ z= 0 at any point on the plane is <1, 1, 1>. The line through (1, 0, 0) with direction vector < 1, 1, 1> is x= t+1, y= t, z= t. Where does that line intersect the plane x+ y+ z= 0?

What (a) tells you is that (1, 0, 0) is mapped by A onto (3/2, -1/3, -1/3). Do the same thing to determine what A maps (0, 1, 0) and (0, 0, 1) to. those three vectors form the columns of A in the standard basis.

Originally Posted by mushroom

Any ideas about (b)?

Another way: according to a well known theorem the projection matrix in over the hyperplane is where .


P.S. Of course, I don't know if you have covered that theorem.