Lowdimensional orthogonal maps

Orthogonal and symmetric maps: Classification of orthogonal maps

Lowdimensional orthogonal maps

We saw examples of orthogonal maps like rotations, reflections, and combinations thereof. We will see that in case the dimension is $2$ , the orthogonal maps can be divided into reflections and rotations.

For each finite-dimensional inner product space $V$ and orthogonal map $L:V\to V$ we have $\det(L)=\pm 1$

We write $n$ for the dimension of $V$ and choose an orthonormal basis $\alpha$ for the vector space $V$ . Let $A=L_{\alpha}$ By Orthogonality criteria for matrices, the matrix $A$ is orthogonal, that is, $A^{\top}A=I_n$ . For the determinant of $A$ we have the following equalities: $\begin{array}{rcl}\det(A)^2 &=&\det(A^\top)\cdot \det(A)\\ &&\phantom{xxx}\color{blue}{\det(A)=\det(A^\top)}\\ &=& \det(A^\top\,A)\\&&\phantom{xxx}\color{blue}{\det(X\, Y)=\det(X)\cdot\det(Y)}\\&=& \det(I_n)\\&&\phantom{xxx}\color{blue}{A^\top\, A = I_n}\\&=&1\end{array}$ From the above we conclude that $\det(L)=\det(A)=\pm 1$ .

Let $V$ and $L$ be as in the theorem and write $n=\dim{V}$ . We know that the determinant $\det(L)$ is equal to the product of the complex eigenvalues $\lambda_1,\ldots,\lambda_n$ of $L$ . Because the length is invariant under the map $L$ , each eigenvalue lies on the unit circle. If $\lambda$ is a non-real eigenvalue, it is accompanied by the eigenvalue $\overline{\lambda}$ , while the product of these two eigenvalues equals $1$ . If $\det(L) = -1$ , at least one eigenvalue must be equal to $-1$ .

We will be using this theorem to divide orthogonal maps on finite-dimensional vector spaces into two groups.

Let $V$ be a finite-dimensional vector space and $L$ an orthogonal map $L:V\to V$ . We define the following division in the orthogonal maps:

If $\det(L)=1$ , then $L$ is called a special orthogonal map.
If $\det(L)=-1$ , then $L$ is called a non-special orthogonal map.

Previously we saw that any orthogonal map $L:V\rightarrow V$ with $\dim {V}=1$ is either the identity map or multiplication by $-1$ . In the first case, $L$ is special orthogonal; in the second case, non-special orthogonal.

Let $V$ and $L$ be as in the theorem and suppose $\dim{V}=2$ . If $L$ is non-special orthogonal, then $L$ has an eigenvalue $-1$ (see above), and so also an eigenvalue $1$ . In this case, the eigenspaces for the various eigenvalues are mutually perpendicular by property 5 of orthogonal maps, so $L$ is an orthogonal reflection about the eigenspace corresponding to the eigenvalue $1$ . If $L$ is special orthogonal and distinct from $I_2$ and from $-I_2$ , then $L$ has two different, complex conjugate, eigenvalues $\lambda$ and $\overline\lambda$ . We shall see below that then $L$ is a rotation with angle $\varphi=\arccos\left(\frac{\lambda+\overline{\lambda}}2\right)$

The following theorem shows concretely that these two groups correspond to reflections and rotations in the two-dimensional case.

Let $V$ be an inner product space of dimension $2$ , and $L:V\to V$ an orthogonal map.

If $L$ is special orthogonal, then there exists an orthonormal basis $\alpha$ for $V$ such that $L_\alpha=\matrix{\cos(\varphi )& -\sin(\varphi)\\ \sin(\varphi )& \cos(\varphi)}$ where the angle $\varphi$ belongs to $\ivcc{0}{\pi}$ . In this case we call $L$ a rotation with oriented angle $\varphi$ .
If $L$ is non-special orthogonal, then there exists an orthonormal basis $\beta$ for $V$ such that $L_{\beta}=\matrix{1&0\\0&-1}$ Then $L$ is a reflection.

Choose an orthonormal basis $\alpha=\basis{\vec{a}_1,\vec{a}_2}$ of $V$ . We note that a vector in $\mathbb{R}^2$ of length $1$ can be written as $\rv{\cos(\varphi),\sin(\varphi)}$ . After all, $\norm{\rv{\cos(\varphi),\sin(\varphi)}}=\sqrt{\cos(\varphi)^2+\sin(\varphi)^2}=1$ Therefore, we can write the matrix of $L$ with respect to $\basis{\vec{a}_1,\vec{a}_2}$ as $L_\alpha=\left(\,\begin{array}{ll} \cos(\varphi )& \cos(\psi)\\ \sin(\varphi) & \sin(\psi) \end{array}\,\right)$ where $\varphi$ and $\psi$ belong to the interval $\ivcc{-\pi}{\pi}$ . Indeed, the columns of $L_\alpha$ both have length $1$ . The other requirement is that the inner product of the columns be zero. This gives $\cos(\varphi)\cdot \cos(\psi)+\sin(\varphi)\cdot\sin(\psi) =\cos (\psi-\varphi )=0$ This implies $\psi -\varphi =\pm \frac{\pi}{2}+k\cdot 2\pi$ for some $k\in\mathbb{Z}$ , so
$\psi=\varphi\pm\frac{\pi}{2}+k\cdot 2\pi$ We first consider the case where $\psi=\varphi+\frac{\pi}{2}+k\cdot 2\pi$ . Then
$L_\alpha=\matrix{ \cos(\varphi) & \cos(\varphi+\frac{\pi}{2})\\ \sin(\varphi) & \sin(\varphi+\frac{\pi}{2})} =\ \matrix{ \cos(\varphi )& -\sin(\varphi)\\ \sin(\varphi )& \cos(\varphi) }$ The determinant is $1$ ; the rotation is special orthogonal. It only remains for us to show that $\varphi$ can always be chosen in $\ivcc{0}{\pi}$ .

Suppose $\varphi\in\ivco{-\pi}{0}$ . Then we have $-\varphi\in\ivoc{0}{\pi}$ and interchanging the two vectors of the orthonormal basis again gives an orthonormal basis with respect to which the matrix of $L$ equals $\begin{array}{rcl}\matrix{0&1\\ 1&0}\,\matrix{ \cos(\varphi )& -\sin(\varphi)\\ \sin(\varphi )& \cos(\varphi)}\, \matrix{0&1\\ 1&0}^{-1}&=&\matrix{ \cos(\varphi )& \sin(\varphi)\\ -\sin(\varphi )& \cos(\varphi)}\\&=&\matrix{ \cos(-\varphi )& -\sin(-\varphi)\\ \sin(-\varphi )& \cos(-\varphi)}\end{array}$ This shows that we can always choose $\varphi\in\ivcc{0}{\pi}$ .

Now consider the case where $\psi =\varphi-\frac{\pi}{2}+k\cdot 2\pi$ . This leads to
$L_\alpha= \matrix{ \cos(\varphi) & \sin(\varphi)\\ \sin(\varphi) & -\cos(\varphi) }$ This matrix has determinant $-1$ , which implies that $L$ has an eigenvalue $\lambda_1=-1$ . Since $\det\left(L\right)=\lambda_1\cdot\lambda_2=-1$ , the other eigenvalue must be $\lambda_2=1$ . Consequently, there is a basis $\beta =\basis{\vec{b}_1,\vec{b}_2}$ of eigenvectors with eigenvalues $-1$ and $1$ . The matrix of $L$ with respect to this basis is $L_\beta=\left(\,\begin{array}{rr} -1 & 0\\ 0 & 1 \end{array}\,\right)$ The eigenvectors $\vec{b}_1$ and $\vec{b}_2$ are perpendicular to each other, for $\dotprod{\vec{b}_1}{\vec{b}_2}=\dotprod{L(\vec{b}_1)}{L(\vec{b}_2)}=\dotprod{(-\,\vec{b}_1)}{\vec{b}_2}=-\,(\dotprod{\vec{b}_1}{\vec{b}_2})$ It follows that $\dotprod{\vec{b}_1}{\vec{b}_2}=0$ . With respect to the basis $\beta=\basis{\vec{b}_2,\vec{b}_1}$ the map $L$ has matrix $L_\beta=\matrix{1&0\\0&-1}$

GeometryWe will explain further how these matrices correspond to rotations and reflections.

Rotation $L_\alpha=\matrix{\cos(\varphi )& -\sin(\varphi)\\ \sin(\varphi )& \cos(\varphi)}$ : We consider the image of the vector $\vec{x}=\lambda\, \vec{a}_1 +\mu\, \vec{a}_2$ where $\alpha=\basis{\vec{a}_1,\vec{a}_2}$ is an orthonormal basis of $V$ . We now have $\begin{array}{rcl}\dotprod{\vec{x}}{L(\vec{x})}&=&\dotprod{\vec{x}}{L\left(\lambda\,\vec{a}_1+\mu\,\vec{a}_2\right)}\\&&\phantom{xx}\color{blue}{\vec{x}\text{ substituted in the argument of }L}\\&=&\dotprod{\vec{x}}{\left(\lambda\,L(\vec{a}_1)+\mu\,L(\vec{a}_2)\right)}\\&&\phantom{xx}\color{blue}{\text{linearity of }L}\\&=&\dotprod{\vec{x}}{\left(\lambda\left(\cos(\varphi)\vec{a}_1+\sin(\varphi)\vec{a}_2\right)+\mu\left(-\sin(\varphi)\vec{a}_1+\cos(\varphi)\vec{a}_2\right)\right)}\\&&\phantom{xx}\color{blue}{\text{first column of }L_\alpha\text{ is the image of }\vec{a}_1\text{ with respect to }\alpha}\\&&\phantom{xx}\color{blue}{\text{second column of }L_\alpha\text{ is the image of }\vec{a}_2\text{ with respect to }\alpha}\\&=&\dotprod{\left(\lambda\vec{a}_1 +\mu \vec{a}_2\right)}{\left((\lambda \cos (\varphi) -\mu \sin(\varphi))\vec{a}_1 + (\lambda \sin(\varphi) +\mu \cos(\varphi))\vec{a}_2\right)}\\&&\phantom{xx}\color{blue}{\vec{x}\text{ substituted on the left and reordered terms on the right}}\\&=&\lambda\cdot (\lambda \cos (\varphi) -\mu \sin(\varphi))+\mu \cdot (\lambda \sin(\varphi) +\mu \cos(\varphi))\\&&\phantom{xx}\color{blue}{\text{bilinearity of the inner product and orthornomality of } \alpha}\\&=&(\lambda^2+\mu^2)\cdot \cos (\varphi) \\&&\phantom{xx}\color{blue}{\text{expression simplified}}\\&=&\norm{\vec{x}}^2\cdot \cos(\varphi)\\&&\phantom{xx}\color{blue}{\text{definition of norm}}\end{array}$ Therefore, the cosine of the angle between the two vectors $\vec{x}$ and $L(\vec{x})$ is given by $\dfrac{\dotprod{\vec{x}}{L(\vec{x})}}{\norm{\vec{x}}\cdot\norm{L(\vec{x})}}=\dfrac{\norm{\vec{x}}^2\cdot \cos(\varphi)}{\norm{\vec{x}}\cdot \norm{L(\vec{x})}}=\cos(\varphi)$ Here we have made use of the fact that $\norm{\vec{x}}=\norm{L(\vec{x})}$ . We see that $L$ indeed describes a rotation over the angle $\varphi$ .

Reflection $L_{\beta}=\matrix{1&0\\0&-1}$ : This is the reflection in the line spanned by $\vec{b}_1$ , where $\beta = \basis{\vec{b}_1,\vec{b}_2}$ is an orthonormal basis: a vector $\vec{x}=x_1 \vec{b}_1 +x_2 \vec{b}_2$ with components $x_1\vec{b}_1$ along the line spanned by $\vec{b}_1$ and $x_2\vec{b}_2$ perpendicular to that line is mapped onto $x_1\vec{b}_1+x_2\vec{b}_2$ .

Consider the matrix $A = \matrix{ \cos(\varphi) & -\sin(\varphi)\\ \sin(\varphi) & \cos(\varphi) }$ of a rotation about the origin in the plane over an oriented angle $\varphi\in\ivoo{\pi}{2\pi}$ . We can conjugate $A$ by one of the two following matrices $\matrix{0&1\\ 1&0},\phantom {xxx} \matrix{1&0\\ 0 & -1}$ to the matrix $\matrix{ \cos(-\varphi) & -\sin(-\varphi)\\ \sin(-\varphi) & \cos(-\varphi) }$ of a rotation with oriented angle $2\pi-\varphi\in\ivoo{0}{\pi}$ .

The first conjugator corresponds to interchanging the two basis vectors, the second to replacing the second basis vector by its negative.

We can read off from the matrix that the characteristic polynomial of a $2$ -dimensional orthogonal map equals $x^2-2\cos(\varphi)\cdot x +1$ in case of a rotation and $x^2-1$ in case of a reflection.

These polynomials factor into linear terms as follows: $\begin{array}{rcl} x^2-2\cos(\varphi)\cdot x +1&=&(x-\ee^{\varphi\ii})\cdot(x-\ee^{-\varphi\ii})\\x^2-1&=&(x-1)\cdot(x+1)\\ \end{array}$ The two factors only coincide if we are the first case, with $\sin(\varphi) = 0$ . In this case, the matrix $L_\alpha$ is diagonal. If $\sin(\varphi)\ne0$ , then the two factors are distinct, and so $L$ is complex diagonalizable. Due to Recognizing diagonalizability using the minimal polynomial, it turns out that $L$ is always complex diagonalizable.

In the second case, the matrix is already diagonal. In the first case, the matrix $T=\matrix{1&\ii \\ 1&-\ii}$ conjugates $L_{\alpha}$ to the diagonal matrix with diagonal entries $\ee^{\varphi\ii},\ee^{-\varphi\ii}$ : $\begin{array}{rcl}TL_\alpha T^{-1}&=&\dfrac{1}{2}\matrix{1&\ii \\ 1&-\ii}\matrix{\cos(\varphi )& -\sin(\varphi)\\ \sin(\varphi)& \cos(\varphi)}\matrix{1&1\\-\ii&\ii}\\&=&\dfrac{1}{2}\matrix{1&\ii\\1&-\ii}\matrix{\cos(\varphi)+\ii\sin(\varphi)&\cos(\varphi)-\ii\sin(\varphi)\\\sin(\varphi)-\ii\cos(\varphi)&\sin(\varphi)+\ii\cos(\varphi)}\\&=&\matrix{\cos(\varphi)+\ii\sin(\varphi)&0\\0&\cos(\varphi)-\ii\sin(\varphi)}\\&=&\matrix{\ee^{\varphi\ii}&0\\0&\ee^{-\varphi\ii}}\end{array}$

If $L$ is special orthogonal, then $L$ is a rotation, as we have seen. The above expression for $L_\alpha$ immediately shows that the trace of this matrix equals $\text{tr}(L_\alpha)=2\cos(\varphi)$ . The trace is invariant under basis transformations, so we may write $\text{tr}(L)=2\cos(\varphi)$ . This result is consistent with the fact that there exists a basis $\gamma$ with respect to which the matrix of $L$ is diagonal with diagonal elements $\lambda=\ee^{\varphi\ii}$ and $\overline{\lambda}=\ee^{-\varphi\ii}$ , as we saw in the previous tab Diagonalizability: $\text{tr}(L)=\text{tr}(L_\gamma)=\lambda+\overline{\lambda}=\ee^{\varphi\ii}+\ee^{-\varphi\ii}=\cos(\varphi)+\ii\sin(\varphi)+\cos(\varphi)-\ii\sin(\varphi)=2\cos(\varphi)$ Vice versa, we can express the angle of rotation $\varphi$ in terms of the trace of the map $L$ : $\varphi=\arccos\left(\frac{\text{tr}(L)}{2}\right)$

We will denote the $(2\times 2)$ -matrix of rotation by $D_{\varphi}=\matrix{ \cos(\varphi) & -\sin(\varphi)\\ \sin(\varphi) & \cos(\varphi) }$ This notation will be used later in the discussion of the orthogonal Jordan normal form.

Consider the matrix: $A = \dfrac{1}{25}\, \matrix{7 & 24 \\ -24 & 7 \\ }$ This matrix is orthogonal and its determinant is $1$ , so $A$ is a rotation.

Determine the angle $\varphi$ of this rotation in degrees. Round to the nearest integer.

$\varphi \approx$ $74$ ${}^\circ$

The vector $\rv{1,0}$ is mapped by $A$ onto $\left[ {{7}\over{25}} , -{{24}\over{25}} \right]$ . Because these vectors have length $1$ , the angle $\varphi$ between these two vectors is given by $\cos(\varphi) = \dotprod{\rv{1,0}}{\left[ {{7}\over{25}} , -{{24}\over{25}} \right] }={{7}\over{25}}$ Therefore, $\varphi = \arccos\left({{7}\over{25}}\right)\approx 1.287 = \left(\frac{1.287\cdot 180}{\pi}\right) ^\circ\approx \left({{232.2}\over{\pi}} \right)^\circ\approx 74^\circ$

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