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    #1
    Hey!!

    Which is the geometric interpretation of the following maps?

    $$v\mapsto \begin{pmatrix} 0&-1&0\\ 1&0&0\\ 0&0&-1\end{pmatrix}v$$ and $$v\mapsto \begin{pmatrix} 1& 0&0\\0&\frac{1}{2} &-\frac{\sqrt{3}}{2}\\ 0&\frac{\sqrt{3}}{2}&\frac{1}{2}\end{pmatrix}v$$

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    #2
    Quote Originally Posted by mathmari View Post
    Hey!!

    Which is the geometric interpretation of the following maps?

    $$v\mapsto \begin{pmatrix} 0&-1&0\\ 1&0&0\\ 0&0&-1\end{pmatrix}v$$ and $$v\mapsto \begin{pmatrix} 1& 0&0\\0&\frac{1}{2} &-\frac{\sqrt{3}}{2}\\ 0&\frac{\sqrt{3}}{2}&\frac{1}{2}\end{pmatrix}v$$
    Hey mathmari!!

    Can we think of some geometric properties that we can verify?

    For instance, if a map is a projection, we expect that we have fixpoint vectors in a plane and a non-zero kernel, don't we?
    For a rotation we would expect a fixpoint vector along the axis, and that the lengths and angles of vectors are preserved.
    For a reflection we would expect fixpoint vectors in a plane, and a normal vector that is mapped to its opposite.

    Do those maps have fixpoint vectors?
    What is their kernel?
    Is there a vector that is mapped to its opposite?

    We might also characterize the matrices by the eigenvalues and eigenvectors.
    For instance a projection in 3D will have eigenvalues 1, 1, and 0.
    And a reflection will have eigenvalues 1,1, and -1.
    Can we find those?

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    #3 Thread Author
    For the kernel we have: \begin{equation*}\begin{pmatrix}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix} \ \overset{Z_1\leftrightarrow Z_2}{\longrightarrow } \ \begin{pmatrix}1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1\end{pmatrix}\end{equation*}

    So the kernel contains only the zero vector. Therefore we don't have a projection.


    For the set of fixed points we have: \begin{align*}\begin{pmatrix}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}v_1 \\ v_2\\ v_3\end{pmatrix}=\begin{pmatrix}v_1 \\ v_2\\ v_3\end{pmatrix} &\longrightarrow \begin{pmatrix}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}v_1 \\ v_2\\ v_3\end{pmatrix}-\begin{pmatrix}v_1 \\ v_2\\ v_3\end{pmatrix}=\begin{pmatrix}0 \\ 0\\ 0\end{pmatrix} \\ & \longrightarrow \begin{pmatrix}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}v_1 \\ v_2\\ v_3\end{pmatrix}-\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}v_1 \\ v_2\\ v_3\end{pmatrix}=\begin{pmatrix}0 \\ 0\\ 0\end{pmatrix} \\ & \longrightarrow \left (\begin{pmatrix}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix}-\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}\right )\begin{pmatrix}v_1 \\ v_2\\ v_3\end{pmatrix}=\begin{pmatrix}0 \\ 0\\ 0\end{pmatrix}
    \\ & \longrightarrow \begin{pmatrix}-1 & -1 & 0 \\ 1 & -1 & 0 \\ 0 & 0 & 0\end{pmatrix}\begin{pmatrix}v_1 \\ v_2\\ v_3\end{pmatrix}=\begin{pmatrix}0 \\ 0\\ 0\end{pmatrix}\end{align*}
    The echelon form of the matric ist:
    \begin{equation*}\begin{pmatrix}-1 & -1 & 0 \\ 1 & -1 & 0 \\ 0 & 0 & 0\end{pmatrix} \ \longrightarrow \begin{pmatrix}-1 & -1 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 0\end{pmatrix}\end{equation*}
    We get the equations
    \begin{align*}-v_1-v_2=&0 \\ -2v_2=&0\end{align*}
    The set of fixed vectors is therefore \begin{equation*}\text{Fix}=\left \{v_3\begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix}\mid v_3\in \mathbb{R}\right \}\end{equation*}
    Is this a fixpoint vector along the axis, and that the lengths and angles of vectors are preserved and so we have a rotation?

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    #4
    Quote Originally Posted by mathmari View Post
    Hey!!

    Which is the geometric interpretation of the following maps?

    $$v\mapsto \begin{pmatrix} 0&-1&0\\ 1&0&0\\ 0&0&-1\end{pmatrix}v$$ and
    (1, 0, 0) is mapped to (0, 1, 0), (0, 1, 0) is mapped to (-1, 0, 0), and (0, 0, -1). Geometrically, a three dimensional figure is rotated 90 degrees in the xy-plane and the z-direction is reversed.

    Quote Quote:
    $$v\mapsto \begin{pmatrix} 1& 0&0\\0&\frac{1}{2} &-\frac{\sqrt{3}}{2}\\ 0&\frac{\sqrt{3}}{2}&\frac{1}{2}\end{pmatrix}v$$
    $cos(-60)= \frac{1}{2}$ and $sin(-60)= -\frac{\sqrt{3}}{2}$ so this is a rotation by -60 degrees (60 degrees counter-clockwise) in the yz-plane.

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    #5
    Quote Originally Posted by mathmari View Post
    For the kernel we have: \begin{equation*}\begin{pmatrix}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix} \ \overset{Z_1\leftrightarrow Z_2}{\longrightarrow } \ \begin{pmatrix}1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1\end{pmatrix}\end{equation*}
    Shouldn't the matrix be:
    \begin{pmatrix}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & {\color{red}-\enclose{circle}{1}}\end{pmatrix}

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