Matrix A for Linear Map T: R3→R3

[Hannisch]

Homework Statement

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.

Homework Equations

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.

 

[mighty2000]

Thank you for your post. I understand your confusion and would be happy to provide some guidance.

Firstly, let's address your concern about the projection. You are correct that the cross product of v and u will be orthogonal to both v and u. However, this does not mean that there is no projection happening. In fact, the cross product itself is a form of projection. It takes a vector and projects it onto a plane perpendicular to both v and u.

Now, to find the matrix A for this linear map, we need to consider the two parts of the transformation separately: the first mapping of u onto v×u, and then the reflection in the plane x=z.

For the first part, we can use the fact that the cross product can be represented by a matrix operation. This matrix is given by:

[v×u] = \left( \begin{array}{ccc}
0 & -v_z & v_y \\
v_z & 0 & -v_x \\
-v_y & v_x & 0 \end{array} \right)\]

where v_x, v_y, and v_z are the components of v.

So, the first part of the transformation can be represented by the matrix [v×u]. Now, for the reflection in the plane x=z, we can use a similar approach to what you did in your attempt. The normal to this plane is <1,0,-1>, so the reflection matrix would be:

S = \left( \begin{array}{ccc}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0 \end{array} \right)\]

Now, to find the matrix A for the entire transformation, we need to multiply these two matrices together. Remember that matrix multiplication is not commutative, so the order matters.

A = S[v×u] = \left( \begin{array}{ccc}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0 \end{array} \right)\] \left( \begin{array}{ccc}
0 & -v_z & v_y \\
v_z & 0 & -v_x \\
-v_y & v_x & 0 \end{array} \right)\]

Multiplying these two matrices together will give you the matrix A for the entire transformation.

I hope