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$A=\frac{1}{9} \begin{pmatrix} -7 & 4 & 4\\ 4 & -1 & 8\\ 4 & 8 & -1 \end{pmatrix}$

How do I prove that A is a rotation ? How do I find the rotation axis and the rotation angle ?

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    No. Lots of matrices have $\det A = 1$ without being rotations. See the answers below. Basically you need to check $A^T A = I$.2012-12-18

6 Answers 6

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You have $A^T A = I$. Hence $A$ is a rotation. Since $\det A = 1$, it is proper.

By inspection, $A \begin{bmatrix} 1 \\ 2 \\ 2\end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\2\end{bmatrix}$, which gives the axis of rotation.

Inspection also shows that $\begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 2 \\ 4 \\ -5 \end{bmatrix}$ are orthogonal eigenvectors corresponding to the (repeated) eigenvalue $-1$. Hence we see that the rotation angle is $\pi$.

Explicitly, if we let $R = \begin{bmatrix} 1 & 2 & 2 \\ 2 & -1 & 4 \\ 2 & 0 & -5 \end{bmatrix}$, then $R^{-1} = \frac{1}{405} \begin{bmatrix} 45 & 90 & 90 \\ 162 & -81 & 0 \\ 18 & 36 & -45\end{bmatrix}$, and $R^{-1} A R = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$, from which we see that the rotation angle is $\pi$.

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If a linear transformation $T:\Bbb R^3\to\Bbb R^3$ is a non-trivial rotation, then the set $\{x\in\Bbb R^3:T(x)=x\}$ will be the axis of rotation, since non-trivial rotation about an axis moves every point except the points on the axis. Also, if the determinant of $T$ isn't $1,$ then it isn't a rotation (why?), though you've already seen that $\det(A)=1$ in this case.

Here, we're working with the transformation $T(x)=Ax$, so the set $\{x\in\Bbb R^3:T(x)=x\}$ is just the eigenspace of $A$ corresponding to the eigenvalue $1$. If $A$ didn't have $1$ as an eigenvalue, we'd know it wasn't a rotation at all (in this case, it does have $1$ as an eigenvalue). If the eigenspace's dimension were greater than $1$, then either it'd be a reflection matrix (if dimension $2$), or the identity matrix (if dimension $3$). The latter is clearly not the case, so it's either a rotation matrix or a reflection matrix. However, if it were a reflection matrix, its determinant would be $-1,$ instead (why?), and so it is a rotation matrix.


Side Note: Given any two non-zero vectors $x,y$ in $\Bbb R^3$ with the angle from $x$ to $y$ being $\theta$, we have the following formulas (where $\cdot$ is the dot product and $\times$ is the cross product): $x\cdot y=\lVert x\rVert\lVert y\rVert\cos\theta\tag{1}$ $\lVert x\times y\rVert=\lVert x\rVert\lVert y\rVert\sin\theta\tag{2}$

To see where $(1)$ and $(2)$ come from, see here and here.


In general, let's suppose we've been given some matrix $A$ corresponding to a rotation in $\Bbb R^3$, and that we want to find its angle of rotation. First, find a basis $\{w\}$ for the axis of rotation (found as above), let $x$ be any non-zero unit vector orthogonal (perpendicular) to $w$, let $y=Ax$. Then both $x$ and $y=Ax$ will be unit vectors. (Do you see why $y$ is a unit vector?), so the formulas $(1)$ and $(2)$ yield the following alternate formulas for our particular $x,y$: $x\cdot y=\cos\theta\tag{$1'$}$ $\lVert x\times y\rVert=\sin\theta\tag{$2'$}$ Here, $\theta$ is the angle of rotation of $A$. (Do you see why?) From there, we can determine $\theta$. (Do you see how?)

Alternatively, start with $w$ (as above), normalize it to $\hat w$, and then determine an orthonormal basis $B=\{\hat w,v_2,v_3\}$ for $\Bbb R^3$ with the Gram-Schmidt process. Then $(\hat w\: v_2\: v_3)^TA(\hat w\: v_2\: v_3)=\left(\begin{array}{ccc}1 & 0 & 0\\0 & \cos\theta & -\sin\theta\\0 & \sin\theta & \cos\theta\end{array}\right),$ which gives us another way to find $\theta$.

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The simplest way is to take the trace of the matrix, the sum of the diagonal elements, by the answer above this equals 1+ 2 Cos(theta) where theta is the angle of rotation. theta can then be determined up to sign which will depend on the orientation of the axis of rotation chosen.

The axis of rotation can be obtained from the skew-symmetric part of the rotation matrix, S = .5*(R-Transpose(R));

Then if S=(aij), rotation axis is (a21,a20,a01);

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    Does this method extend to n by n matrices?2018-12-18
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Wolfram Alpha tells me there is an eigenspace with eigenvalue $-1$ generated by $(-2, 0, 1)$ and $(-2, 1, 0)$ and an eigenspace with eigenvalue $1$ generated by $(1, 2, 2)$. (You could do this by hand). The eigenspaces are orthogonal, so this is a rotation by 180 degrees about the axis $(1, 2, 2)$.

The problem is a little atypical, since if you rotate by any angle that's not an integer multiple of $\pi$ you will have complex eigenvalues.

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    I believe the rotation is by a multiple of $\pi$?2012-12-18
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A rotation matrix has unit determinant. Such a matrix that has all non-zero entries may be decomposed into three rotation matrices, each representing a rotation about an orthogonal coordinate axis.

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The document: orthogonal matrices is an excellence reference for this problem.

The steps are as follows.

  1. Show that the determinant is 1. Matrices with determinant -1 are reflections no rotations.
  2. Find the eigenvalues. The three eigenvalues of the matrix are $1, \text{e}^{-i \theta}$, where $\theta$ is the angle of rotation.

  3. find the eigenvector for the eigenvalue 1. This is the axis of rotation.

Solution: That the determinant is 1 can be directly checked.

The eigenvalues are $\lambda=1, \pm 1$. So the angle of rotation is $0$.

The eigenvector for the eigenvalue $1$ is $(1/2,1,1)$. This is the axis of rotation.

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    You've miscalculated the eigenvalues, I'm afraid. If $1$ were a repeated eigenvalue, then the determinant would be $-1,$ instead.2017-01-23