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Math Help - how can this evaluate to a matrix?

  1. #1
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    how can this evaluate to a matrix?

    Hello I'm trying to understand the formula for getting a rotation matrix for a rotation about an axis. I just need to understand it enough so I can implement it in some code. Any extra gems of knowledge you have are welcome of course : ).

    The specific bit I don't get is the uu^T bit in the second line. Does that mean the dot product of u with the transpose of u? Why do they transpose it? Is taking the dot product of a vector with the transpose of itself a normal and reasonable thing to do?

    This is from page 754 of the OpenGL programming guide:

    v is the axis of rotation and a is the angle of rotation.

    Let  v=(x,y,z)^T<br />
and <br />
u=\frac{v}{ \parallel v \parallel } = (x', y', z')^T<br />
    Also let
    <br />
S=<br />
\begin{bmatrix}<br />
0 & -z' & y' \\<br />
z' & 0 & -x' \\<br />
-y' & x' & 0 \end{bmatrix}<br />
and <br />
M = uu^T + (\cos a)(I-uu^T) + (\sin a) S<br />

    Also does the "I" represent the identity matrix so that (I-uu^T) is simple scalar subtraction from the identity matrix (assuming uu^T is the dot product producing a scalar value)?

    Thanks alot for having a look.
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  2. #2
    MHF Contributor

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    Quote Originally Posted by tleave2000 View Post
    Hello I'm trying to understand the formula for getting a rotation matrix for a rotation about an axis. I just need to understand it enough so I can implement it in some code. Any extra gems of knowledge you have are welcome of course : ).

    The specific bit I don't get is the uu^T bit in the second line. Does that mean the dot product of u with the transpose of u? Why do they transpose it? Is taking the dot product of a vector with the transpose of itself a normal and reasonable thing to do?
    No, it's not "the dot product of a vector with the transpose of itself". If u is written as a "column matrix", say \begin{bmatrix}u_x \\ u_y \\ u_z\end{bmatrix} then its transpose is the "row matrix", \begin{bmatrix} u_x & u_y & u_z\end{bmatrix} and the product you indicate is the matrix product u u^T= \begin{bmatrix}u_x \\ u_y \\ u_z\end{bmatrix}\begin{bmatrix} u_x & u_y & u_z\end{bmatrix}= \begin{bmatrix}u_x^2 & u_xu_y & u_xu_z \\ u_yu_x & u_y^2 & u_yu_z \\ u_zu_x & u_zu_y & u_z^2\end{bmatrix}.

    In fact, a common way of denoting the dot product of a vector with itself is the other way around: u^T u= \begin{bmatrix} u_x & u_y & u_z\end{bmatrix}\begin{bmatrix}u_x \\ u_y \\ u_z\end{bmatrix}= \begin{bmatrix}u_x^2+ u_y^2+ u_z^2\end{bmatrix}.

    This is from page 754 of the OpenGL programming guide:

    v is the axis of rotation and a is the angle of rotation.

    Let  v=(x,y,z)^T<br />
and <br />
u=\frac{v}{ \parallel v \parallel } = (x', y', z')^T<br />
    Also let
    <br />
S=<br />
\begin{bmatrix}<br />
0 & -z' & y' \\<br />
z' & 0 & -x' \\<br />
-y' & x' & 0 \end{bmatrix}<br />
and <br />
M = uu^T + (\cos a)(I-uu^T) + (\sin a) S<br />

    Also does the "I" represent the identity matrix so that (I-uu^T) is simple scalar subtraction from the identity matrix (assuming uu^T is the dot product producing a scalar value)?

    Thanks alot for having a look.
    Yes, I is the identity matrix but there is no such thing as "scalar subtraction"- you can't subtract a scalar from a matrix. However, as I said above, uu^T is not a dot product and so not a scalar value. It is a matrix which can be subtracted from the identity matrix. With u as above, I- uu^T= \begin{bmatrix}1 -u_x^2 & -u_xu_y & -u_xu_z \\ -u_yu_x & 1 - u_y^2 & -u_yu_z \\ -u_zu_x & -u_zu_y & 1 -u_z^2\end{bmatrix}
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  3. #3
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    Apr 2009
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    Thanks that helped alot.

    Ok so it's the outer product (once I knew how it expanded I had an idea what to look out for on wiki) which can be looked at as a special case of the kronecker product. Is that why it gets around the problem with matrix multiplication (at least the kind I knew about) not normally being defined between matrices where the number of columns in the first matrix is different to the number of rows in the second? I think that's what threw me off, I'd never really seen the outer product before.

    Is all that more or less right?
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