# matrice

• Dec 2nd 2012, 02:01 PM
Petrus
matrice
Determine the matrix of the linear mapping https://webwork.math.su.se/webwork2_...e043270961.png defined by the vector https://webwork.math.su.se/webwork2_...fd8fbc7ea1.png first mapped to https://webwork.math.su.se/webwork2_...c3c12e01a1.png where https://webwork.math.su.se/webwork2_...c42a1ee611.png and then mirrored in the plane https://webwork.math.su.se/webwork2_...5317b0abb1.png. (positively oriented ON-systems).

i got no progress, dont know how to solve it
• Dec 2nd 2012, 02:47 PM
Plato
Re: matrice
Quote:

Originally Posted by Petrus

Can you explain what "then mirrored in the plane $\displaystyle x=z$ (positively oriented ON-systems)" means.
• Dec 2nd 2012, 02:51 PM
jakncoke
Re: matrice
See where the mapping take the basis vectors $\displaystyle \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} , \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$. The adjoin the 3 transformed basis vectors to get your transformation.
• Dec 2nd 2012, 02:56 PM
jakncoke
Re: matrice
Quote:

Originally Posted by Plato
Can you explain what "then mirrored in the plane $\displaystyle x=z$ (positively oriented ON-systems)" means.

I think he means to project the vector on the plane x = z.
• Dec 2nd 2012, 03:31 PM
Plato
Re: matrice
Quote:

Originally Posted by jakncoke
I think he means to project the vector on the plane x = z.

Yes I got that much. But what does that mean?
In particular, "positively oriented ON-systems)"?
• Dec 2nd 2012, 09:52 PM
Deveno
Re: matrice
presumably "positively oriented ON-systems" means we are using the "right-hand rule" to determine an orientation (order) of the basis {e1,e2,e3} (the xyz-axes).

(i am guessing ON stands for orthonormal).

reflection across the plane x = z is definitely NOT the same is "projection on the line x = z" the reflection is an invertible linear map, the projection is NOT.

what i believe he means by the plane "x = z" is the 2-dimensional subspace of R3 spanned by {(1,0,1),(0,1,0)}.

let's work out what this reflection does to the vectors (1,0,0), (0,1,0), and (0,0,1). clearly it does nothing to (0,1,0) since that is already in the plane.

suppose v is in the orthogonal complement of P = span({(1,0,1),(0,1,0)}). then our reflection takes v to -v. let's determine P.

if (x,y,z).(1,0,1) = 0, then x = -z.
if (x,y,z).(0,1,0) = 0, then y = 0. so the orthogonal complement is spanned by (1,0,-1).

so let's write (1,0,0) in the basis {(1,0,1),(0,1,0),(1,0,-1)}. this gives:

(1,0,0) = a(1,0,1) + b(0,1,0) + c(1,0,-1) = (a+c,b,a-c) giving:

a+c = 1
b = 0
a-c = 0

thus a = 1/2, b = 0, c = 1/2.

since (1,0,0) = (1/2)(1,0,1) + (1/2)(1,0,-1), applying our reflection to (1,0,0) we see it fixes (1,0,1) and sends (1,0,-1) to (-1,0,1).

thus (1,0,0) gets mapped to (1/2)(1,0,1) - (1/2)(1,0,-1) = (0,0,1).

similarly, if:

(0,0,1) = a(1,0,1) + b(0,1,0) + c(1,0,-1) = (a+c,b,a-c)

a+c = 0
b = 0
a-c = 1

so a = 1/2, b = 0, c = -1/2.

thus (0,0,1) = (1/2)(1,0,1) - (1/2)(1,0,-1) gets mapped to (1/2)(1,0,1) + (1/2)(1,0,-1) = (1,0,0).

so the matrix for our reflection is:

$\displaystyle \begin{bmatrix}0&0&1\\0&1&0\\1&0&0 \end{bmatrix}$.

now we need to determine the matrix for the linear transformation L(u) = (-1,2,1) x u. so we compute L((1,0,0)), L((0,1,0)) and L((0,0,1)).

L((1,0,0)) = (-1,2,1) x (1,0,0) = (2*0-1*0, 1*1-(-1)*0, (-1)*0-2*1) = (0,1,-2)

L((0,1,0)) = (-1,2,1) x (0,1,0) = (2*0-1*1, 1*0-(-1)*0, (-1)*1-2*0) = (-1,0,-1)

L((0,0,1)) = (-1,2,1) x (0,0,1) = (2*1-1*0, 1*0-(-1)*1, (-1)*0-2*0) = (2,1,0)

so the matrix for L is:

$\displaystyle \begin{bmatrix}0&-1&2\\1&0&1\\-2&1&0 \end{bmatrix}$

(note this matrix is anti-symmetric, as we would expect since the cross product is anti-commutative).

if we call the matrix for our reflection M, and the matrix for L, N, then the matrix for T is given by MN.