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    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
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    Re: matrice

    Quote Originally Posted by Petrus View Post
    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 x=z (positively oriented ON-systems)" means.
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    Re: matrice

    See where the mapping take the basis vectors  \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.
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    Re: matrice

    Quote Originally Posted by Plato View Post
    Can you explain what "then mirrored in the plane x=z (positively oriented ON-systems)" means.
    I think he means to project the vector on the plane x = z.
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    Re: matrice

    Quote Originally Posted by jakncoke View Post
    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)"?
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  6. #6
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    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:

    \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:

    \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.
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