Hannisch
Nov23-09, 02:32 PM
1. The problem statement, all variables and given/known data
Determine the matrix A for the linear map T: R3→R3 which is defined by that the vector u first is mapped on v×u, where v=(-9,2,9) and then reflected in the plane x=z (positively oriented ON-system). Also determine the determinant for A.
2. Relevant equations
3. The attempt at a solution
I actually started with the determinant and said that since the first mapping is a projection the determinant of that is =0 -- thus, the determinant for the whole thing is 0, since det(B*C)=det(B)*det(C) and in this case, A=S*P, where S is the determinant for the reflection (which is what we usually use in Swedish) and P is the projection.
Anyway, I started with S (for practise, if nothing else):
The plane will have the equation x-z=0 in its normal form and thus the normal to the plane is <1,0,-1>. So if I call vector w <a,b,c>, the reflection in the plane is:
<a,b,c> + t<1,0,-1> = <a+t,b,c-t>.
<a,b,c> + (t/2)<1,0,-1> needs to be on the plane and thus the coordinates for that vector need to fulfill the plane equation,
(a+t/2) - (c-t/2) = 0, t= -(a-c).
<a-(a-c),b,c+(a-c)>=<c,b,a>.
Thus, the reflection matrix is:
S = \left( \begin{array}{ccc}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0 \end{array} \right)\]
I didn't think that was too difficult, but now I'm entering the confusing part.
I can quite easily calculate v×u, if u=<x,y,z>.
(v×u = < -9y+2z, 9x+9z, -2x - 9y>.)
But (and I was thinking this from the very beginning) v×u is orthogonal to u (by definition of the cross product, because it creats a vector orthogonal to the plane containing v and u). But there can't be any projection if it's orthogonal, right? So I thought that if they all stay the same it should be the identity matrix and thus S*I=S, but that is not correct. And now I've got no idea what to do... And I didn't really want it to be the identity matrix, because that would not mean that the determinant is 0.
The conclusion is that I'm very confused.
Determine the matrix A for the linear map T: R3→R3 which is defined by that the vector u first is mapped on v×u, where v=(-9,2,9) and then reflected in the plane x=z (positively oriented ON-system). Also determine the determinant for A.
2. Relevant equations
3. The attempt at a solution
I actually started with the determinant and said that since the first mapping is a projection the determinant of that is =0 -- thus, the determinant for the whole thing is 0, since det(B*C)=det(B)*det(C) and in this case, A=S*P, where S is the determinant for the reflection (which is what we usually use in Swedish) and P is the projection.
Anyway, I started with S (for practise, if nothing else):
The plane will have the equation x-z=0 in its normal form and thus the normal to the plane is <1,0,-1>. So if I call vector w <a,b,c>, the reflection in the plane is:
<a,b,c> + t<1,0,-1> = <a+t,b,c-t>.
<a,b,c> + (t/2)<1,0,-1> needs to be on the plane and thus the coordinates for that vector need to fulfill the plane equation,
(a+t/2) - (c-t/2) = 0, t= -(a-c).
<a-(a-c),b,c+(a-c)>=<c,b,a>.
Thus, the reflection matrix is:
S = \left( \begin{array}{ccc}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0 \end{array} \right)\]
I didn't think that was too difficult, but now I'm entering the confusing part.
I can quite easily calculate v×u, if u=<x,y,z>.
(v×u = < -9y+2z, 9x+9z, -2x - 9y>.)
But (and I was thinking this from the very beginning) v×u is orthogonal to u (by definition of the cross product, because it creats a vector orthogonal to the plane containing v and u). But there can't be any projection if it's orthogonal, right? So I thought that if they all stay the same it should be the identity matrix and thus S*I=S, but that is not correct. And now I've got no idea what to do... And I didn't really want it to be the identity matrix, because that would not mean that the determinant is 0.
The conclusion is that I'm very confused.