Question

Consider the linear transformation P : R3 → R3 given by orthogonal projection onto the plane 3x − y − 2z = 0, using the dot product on R3 as inner product.

Describe the eigenspaces and eigenvalues of P, giving specific reasons for your answers. (Hint: you do not need to find a matrix representing the transformation.)

Answer 1

The equation of the given plane is 3x−y−2z=0, hence the vector (3,-1,-2) is normal to this plane. The projection of the vector (x,y,z) onto the given plane is (x,y,z)- proj_(3,-1,-2) (x,y,z) = (x,y,z)- [((x,y,z).(3,-1,-2)/( (3,-1,-2). (3,-1,-2)](3,-1,-2)=(x,y,z)-[(3x-y-2z)/(9+1+4)](3,-1,-2)=(x,y,z)-((9x-3y-2z)/14,(-3x+y+2z)/14,(-6x+2y+4z)/14)=                                                   ((5x+3y+2z)/14,(3x+13y-2z)/14,(6x-2y+10z)/14).

Thus, P(x,y,z) = ((5x+3y+2z)/14,(3x+13y-2z)/14,(6x-2y+10z)/14).

Now, P(e₁)= P(1,0,0) = (5/14, 3/14,6/14) , P(e₂)= P(0,1,0) = (3/14,13/14,-2/14) and P(e₃)= P(0,0,1) =( 2/14, -2/14,10/14).

Therefore, the standard matrix of P is A =

5/14	3/14	2/14
3/14	13/14	-2/14
6/14	-2/14	10/14

The eigenvalues and the eigenvectors of P are same as those of A.

The characteristic equation of A is det(A-ʎI₃) = 0 or, ʎ³ -2ʎ² +55ʎ/49 -6/49 = 0 or, (ʎ-1)(ʎ-6/7)(ʎ-1/7) = 0. Hence the eigenvalues of P are ʎ₁ = 1, ʎ₂ = 6/7 and ʎ₃ =1/7.

The eigenvector of P corresponding to the eigenvalue 1 is solution to the equation (A-I₃)X = 0. To solve this equation, we have to reduce A-I₃ to its RREF, which is

1	-1/3	0
0	0	1
0	0	0

Now, if X = (x,y,z)^T, then the equation (A-I₃)X = 0 is equivalent to x-y/3 = 0 or, x = y/3 and z = 0 so that X = (y/3,y,0)^T = (y/3)(3,1,0)^T. Hence, the eigenvector of P corresponding to the eigenvalue 1 is v₁ = (3,1,0)^T. Similarly, the eigenvectors of P corresponding to the eigenvalues 6/7 and 1/7 are v₂ = (1/2,1/2,1)^T and v₃ = (-7/6,1/2,1)^T respectively.