Thread: matrice

Determine the matrix of the linear mapping defined by the vector first mapped to where and then mirrored in the plane . (positively oriented ON-systems).

i got no progress, dont know how to solve it

Originally Posted by Petrus

Determine the matrix of the linear mapping defined by the vector first mapped to where and then mirrored in the plane . (positively oriented ON-systems).

Can you explain what "then mirrored in the plane (positively oriented ON-systems)" means.

See where the mapping take the basis vectors . The adjoin the 3 transformed basis vectors to get your transformation.

Originally Posted by Plato

Can you explain what "then mirrored in the plane (positively oriented ON-systems)" means.

I think he means to project the vector on the plane x = z.

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)"?

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:

.

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:



(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.