1

I am trying to solve the following exercise:

Let $M=\begin{bmatrix} -\frac{1}{2} & -\frac{1}{2} & \frac{\sqrt{2}}{2}\\ \frac{1}{2} & \frac{1}{2} & \frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} & 0 \end{bmatrix}$.

(a) Show that it is orthogonal.
(b) Classify it: does it represent a rotation or a reflection?
(c) If it represents a rotation, find the axis of rotation and the angle; if it represents a reflection, find the plane of reflection and, if present, the rotation around the normal to the plane".

What I have done:

(a) $M^TM=I$ so the matrix is orthogonal.

(b) $\det(M)=-1$ so the matrix represents a reflection.

EDIT:

c) If $\vec{v_{\perp}}$ is a vector perpendicular to the plane of reflection the relation $M\vec{v_{\perp}}=-\vec{v_{\perp}}$ must hold i.e. $\vec{v_{\perp}}$ is the eigenvector corresponding to the eigenvalue $\lambda =-1$. So we have $M-(-1)I=M+I=\begin{bmatrix} \frac{1}{2} & -\frac{1}{2} & \frac{\sqrt{2}}{2}\\ \frac{1}{2} & \frac{3}{2} & \frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} & 1 \end{bmatrix}\vec{v_{\perp}}=\begin{bmatrix}0\\ 0\\ 0 \end{bmatrix}\Rightarrow\vec{v_{\perp}}=\begin{bmatrix}\sqrt{2}\\ 0\\ -1\end{bmatrix}$ thus M reflects with respect to the plane $\sqrt{2}x-z=0$. We can build the rotation matrix around the normal $\vec{v_{\perp}}$ by finding two vectors perpendicular to it, which is easy to see are $\vec{v_2}=\begin{bmatrix}0\\ 1\\ 0\end{bmatrix}$ and $\vec{v_3}=\begin{bmatrix}1\\ 0\\ \sqrt{2}\end{bmatrix}$ thus $C=\begin{bmatrix} 0 & 1/\sqrt{3} & \sqrt{2/3}\\ 1 & 0 & 0 \\ 0 & \sqrt{2/3} & -1/\sqrt{3} \end{bmatrix}$ is the rotation matrix and $C^{-1} MC=\begin{bmatrix} 1/2 & \sqrt{3}/2 & 0\\ -\sqrt{3}/2 & 1/2 & 0 \\ 0 & 0 & -1\end{bmatrix}$ which represents (in the new coordinate system with the z-axis coinciding with $\vec{v_{\perp}}$) a reflection with respect to the new $x,y$ plane and a rotation around the $z$ axis with an angle of rotation of $5\pi/6$ (since $\cos(\theta)=1/2$ and $\sin(\theta)=-\sqrt{3}/2$).

So, I would be very grateful if someone could check what I have done and give me some feedback.

Thank you very much for your time and help.

lorenzo
  • 4,032
  • 1
    I think answers it: https://math.stackexchange.com/questions/68119/why-does-ata-i-det-a-1-mean-a-is-a-rotation-matrix – stoic-santiago May 13 '21 at 15:22
  • @epsilon-emperor thank you for your interest in my question; it does answer it but only in part (it explains for example why $\det(M)=1$ and $M^T M=1$ imply that $M$ is a rotation ) but it doesn't contain the actual method to compute for example the axis of rotation or the plane of reflection and so on. – lorenzo May 13 '21 at 15:30
  • You can see that if $x = (x_1,x_2,x_3)^T$, then range $Mx = (a, -a, b)^T$. This means $M:R^3\to V$ where $V = span((1,-1,0),(0,0,1))$ – Snowball May 13 '21 at 15:31
  • @ThomasAndrews You are right: it was a typo, now fixed. Thank you! – lorenzo May 13 '21 at 15:33
  • 3
    @lorenzo Note that the axis of reflection/rotation of an orthogonal matrix will be an eigenvector – Ben Grossmann May 13 '21 at 16:02
  • @BenGrossmann thank you for your interest in my question; so, I have found the eigenvector $\vec{v}=(\sqrt{2},0,-1)$ associated to the unique eigenvalue $\lambda=-2$ thus the line spanned by $\vec{v}$ is the line of reflection, am I right? But, if it is a reflection, shouldn't I have found two eigenvalues whose associated eigenvectors span a plane? Is there something I am missing? – lorenzo May 13 '21 at 18:12
  • @lorenzo First of all, you seem to have made an error with that eigenvector. Second, not necessarily: a "reflection" (according to the definition used in this exercise) is allowed to consist of a straight reflection followed by a rotation about the normal to the plane. When a rotation in the plane is present (by an angle besides $0$ or $180^\circ$), there will be only one real eigenvalue. – Ben Grossmann May 13 '21 at 18:55
  • @BenGrossmann I made a mistake with the eigenvalue, which should have been $\lambda = -1$ but the eigenvector was correct (see https://www.wolframalpha.com/input/?i=eigenvectors+%7B%7B-1%2F2%2C-1%2F2%2Csqrt%282%29%2F2%7D%2C+%7B1%2F2%2C1%2F2%2Csqrt%282%29%2F2%7D%2C+%7Bsqrt%282%29%2F2%2C+-sqrt%282%29%2F2%2C+0%7D%7D , you put in an extra factor of $1/2$). I have edited my question and tried to do part (c) and I would like to know your opinion about it, if you would be so kind to take a look at it, thanks. – lorenzo May 13 '21 at 22:19
  • 1
    If $M$ is a reflection then $M^{2} = I$. – Elmex80s May 14 '21 at 00:22
  • @Elmex80s That is not true for the definition of "reflection" used in this exercise – Ben Grossmann May 14 '21 at 15:03
  • 1
    @lorenzo Your answer for part (c) looks perfect – Ben Grossmann May 14 '21 at 15:05

0 Answers0