Three-dimensional orthogonal maps

Orthogonal and symmetric maps: Classification of orthogonal maps

Three-dimensional orthogonal maps

For the study of three-dimensional orthogonal maps we will use the $(2\times2)$ -matrices

$D_{\varphi} = \matrix{\cos(\varphi)&-\sin(\varphi)\\ \sin(\varphi)& \cos(\varphi)}$ for certain $\varphi$ . These are the rotation matrices of Two-dimensional orthogonal maps.

Orthogonal maps in three dimensions

Let $V$ be a $3$ -dimensional inner product space and $L:V\to V$ an orthogonal map.

If $L$ is special orthogonal, then it is a rotation, that is to say, there is a line passing through the origin, such that each vector along that line is fixed by $L$ , and the restriction of $L$ to the plane through the origin perpendicular to that line is a rotation. There is an orthonormal basis $\alpha$ of $V$ and an angle $\varphi$ such that $L_\alpha = \matrix{1&0\\ 0 & D_{\varphi}}$
If $L$ is non-special orthogonal, then it is an improper rotation, that is to say, there is a line passing through the origin which is the eigenspace of $L$ corresponding to eigenvalue $-1$ , and the restriction of $L$ to the plane through the origin perpendicular to that line is a rotation. There is an orthonormal basis $\alpha$ of $V$ and an angle $\varphi$ such that $L_\alpha = \matrix{-1&0\\ 0 & D_{\varphi}}$

In both cases, the line is called an axis of rotation and the plane a plane of rotation or rotation plane.

If $L$ does not equal $I_V$ or $-I_V$ , then its axis and rotation plane are unique.

Each orthogonal map $V\to V$ is either a rotation or an improper rotation. An improper rotation is the composition of an orthogonal reflection and a rotation in the reflection plane (mirror plane).

Let $V$ be a $3$ -dimensional inner product space and $L:V\to V$ an orthogonal map. Suppose that $\lambda$ is a non-real eigenvalue of $L$ . Then also the complex conjugate $\overline\lambda$ is a non-real eigenvalue. Because the total number of eigenvalues is $3$ , the map $L$ must have at least one real eigenvalue. We know from property 4 of orthogonal maps that every real eigenvalue of $L$ equals $1$ or $-1$ . If $L$ is special orthogonal, this real eigenvalue of $L$ equals $1=\det(L)$ ; otherwise (that is, if $L$ is non-special orthogonal), this real eigenvalue of $L$ equals $-1=\det(L)$ . Let $\vec{a}_1$ be a normalized eigenvector of $L$ corresponding to this real eigenvalue, $\varepsilon = \det(L)$ . Then, the restriction of $L$ to $\vec{a}_1^{\perp}$ is a special orthogonal map (after all, the determinant of this restriction is $1$ ). Thus, there is an orthonormal basis $\basis{\vec{a}_2,\vec{a}_3}$ of $\vec{a}_1^{\perp}$ such that the matrix of the restriction of $L$ with respect to this basis is $D_{\varphi}$ for a suitable angle $\varphi$ . Now $\alpha = \basis{\vec{a}_1,\vec{a}_2,\vec{a}_3}$ is an orthonormal basis for $V$ so

$L_\alpha = \matrix{\varepsilon&0\\ 0 & D_{\varphi}}$

If $\varepsilon = 1$ , then the map $L$ is a rotation whose axis is the span of the first basis vector of $\alpha$ . In all cases except the identity, the axis is the $1$ -dimensional eigenspace of $L$ with eigenvalue $1$ , so it is unique.

If $\varepsilon = -1$ , then the determinant of $L$ equals $-1$ . The map is an improper rotation whose axis is the span of the first basis vector of $\alpha$ . If $L$ has only real eigenvalues, then $L_\alpha$ is either equal to the diagonal matrix with diagonal $-1,1,1$ or to $-I_3$ . The cases of a rotation about $0$ , respectively, $\pi$ radians, correspond to the above two matrices with real eigenvalues. In all cases except scalar multiplication by $-1$ , the axis is the $1$ -dimensional eigenspace of $L$ with eigenvalue $-1$ , so it is unique.

If the trace of $L$ is known, for example from a matrix of $L$ with respect to a different basis, this information can be used for a swift calculation of the rotation angle $\varphi$ by setting the trace equal to $2\cos(\varphi)+1$ if $L$ is a rotation, and to $2\cos(\varphi)-1$ if $L$ is an improper rotation. The cases of a rotation over $0$ , respectively, $\pi$ radians correspond to the two matrices with real eigenvalues when the determinant of $L$ is $1$ .

In order to distinguish between rotations and improper rotations, we sometimes also refer to $3$ -dimensional rotations as proper rotations.

Determine whether or not the orthogonal map $L_A:\mathbb{R}^3\to\mathbb{R}^3$ determined by the $(3\times3)$ -matrix

$A={{1}\over{23}}\,\matrix{-13 & -6 & -18 \\ -18 & -3 & 14 \\ -6 & 22 & -3 \\ }$ is special orthogonal or non-special orthogonal.

special orthogonal

An orthogonal map $L$ is special orthogonal if $\det (L_A)=1$ and non-special orthogonal if $\det (L_A)=-1$ . To find the answer, we determine the determinant of $L_A$ , which is equal to $\det(A)$ . For this purpose we reduce $A$ to reduced echelon form, as shown below. Here we record in the second column, by which number the newly obtained matrix has to be multiplied in order to get the determinant of the previous matrix. In the third column we record the product of these factors.

$\begin{array}{c|r|r} \text{matrix}&\text{step}&\text{total}\\ \hline \matrix{-{{18}\over{23}} & -{{3}\over{23}} & {{14}\over{23}} \\ -{{13}\over{23}} & -{{6}\over{23}} & -{{18}\over{23}} \\ -{{6}\over{23}} & {{22}\over{23}} & -{{3}\over{23}} \\ } & -1 & -1 \\ \matrix{1 & {{1}\over{6}} & -{{7}\over{9}} \\ -{{13}\over{23}} & -{{6}\over{23}} & -{{18}\over{23}} \\ -{{6}\over{23}} & {{22}\over{23}} & -{{3}\over{23}} \\ } & -{{23}\over{18}} & {{23}\over{18}} \\ \matrix{1 & {{1}\over{6}} & -{{7}\over{9}} \\ 0 & -{{1}\over{6}} & -{{11}\over{9}} \\ -{{6}\over{23}} & {{22}\over{23}} & -{{3}\over{23}} \\ } & 1 & {{23}\over{18}} \\ \matrix{1 & {{1}\over{6}} & -{{7}\over{9}} \\ 0 & -{{1}\over{6}} & -{{11}\over{9}} \\ 0 & 1 & -{{1}\over{3}} \\ } & 1 & {{23}\over{18}} \\ \matrix{1 & {{1}\over{6}} & -{{7}\over{9}} \\ 0 & 1 & -{{1}\over{3}} \\ 0 & -{{1}\over{6}} & -{{11}\over{9}} \\ } & -1 & -{{23}\over{18}} \\ \matrix{1 & 0 & -{{13}\over{18}} \\ 0 & 1 & -{{1}\over{3}} \\ 0 & -{{1}\over{6}} & -{{11}\over{9}} \\ } & 1 & -{{23}\over{18}} \\ \matrix{1 & 0 & -{{13}\over{18}} \\ 0 & 1 & -{{1}\over{3}} \\ 0 & 0 & -{{23}\over{18}} \\ } & 1 & -{{23}\over{18}} \\ \matrix{1 & 0 & -{{13}\over{18}} \\ 0 & 1 & -{{1}\over{3}} \\ 0 & 0 & 1 \\ } & -{{18}\over{23}} & 1 \\ \matrix{1 & 0 & 0 \\ 0 & 1 & -{{1}\over{3}} \\ 0 & 0 & 1 \\ } & 1 & 1 \\ \matrix{1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ } & 1 & 1 \end{array}$ We conclude that the determinant of $A$ equals $1$ so the answer is: special orthogonal.

New example