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Special Types of Linear Maps: Orthogonal Tensors

Orthogonal Tensors

Definition

Let Q:\mathbb{R}^n\rightarrow\mathbb{R}^n. Q is called an orthogonal tensor if Q^TQ=I.

Properties

Using the above definition, the following five main properties of Orthogonal Tensors can be directly deduced:

Property 1: Orthogonal tensors preserve the norm (length) of vectors and the dot product between vectors:

\forall a,b\in\mathbb{R}^n we have:

    \[ a\cdot b=a\cdot Q^TQb=Qa\cdot Qb \]

Property 2: Orthogonal tensors are invertible and \forall orthogonal tensor Q: \det(Q)=\pm 1:

The easiest way to see this is to assume that Q is not invertible, which implies \exists a\in \mathbb{R}^n and a \neq 0 while Qa=0. This implies that Q^T (Qa)=0 which contradicts that Q^T(Qa)=Q^TQa=a\neq 0.

Another way to show that Q is invertible is to rely on the determinant function. Since \det(Q^T Q)=\det(Q^T)\det(Q)=\det(Q)\det(Q)=(\det(Q))^2=1, therefore, \det(Q)=\pm 1\neq0\Rightarrow Q is invertible.
If \det(Q)= 1 then Q is called a proper orthogonal tensor, and If \det(Q)= -1 then Q is called an improper orthogonal tensor.

Since Q is invertible we can denote its inverse by Q^{-1} which leads to Q^TQ=I\Rightarrow Q^TQ Q^{-1}=Q^{-1}. Therefore:

    \[Q^T=Q^{-1}\]

Property 3: The rows of the matrix representation of Q are orthonormal:

This is a direct consequence of the fact that QQ^T=I

Property 4: The columns of the matrix representation of Q are orthonormal:

This is a direct consequence of the fact that Q^TQ=I

Property 5: The product of two orthogonal tensors is again orthogonal:

Indeed, let Q_1 and Q_2 be two orthogonal tensors, therefore:

    \[ Q_1Q_2\left(Q_1Q_2\right)^T=Q_1Q_2Q_2^TQ_1^T=Q_1\left(Q_2Q_2^T\right)Q_1^T=Q_1\left(I\right)Q_1^T=I \]

Therefore, the product Q_1Q_2 is orthogonal.

Orthogonal Tensors in \mathbb{R}^2

Assume that the matrix representation of an orthogonal tensor has the following representation:

    \[ Q=\left(\begin{array}{cc} a_1 & a_2 \\b_1&b_2\end{array}\right)\]

Then, using the properties above, we reach the following relations between the components:

    \[ a_1^2+a_2^2=1\]

    \[ b_1^2+b_2^2=1\]

    \[ a_1b_1+a_2b_2=0\]

    \[ a_1b_2-a_2b_1=\pm 1 \]

These relationships assure the existence of an angle 0\leq\theta\leq 2\pi such that Q admits one of the the following two representations:

(1)   \begin{equation*} \mbox{If } \det(Q)=1 \Rightarrow Q=\left(\begin{array}{cc} \cos(\theta) & \sin(\theta)\\-\sin(\theta)&\cos(\theta)\end{array}\right) \end{equation*}

(2)   \begin{equation*} \mbox{If } \det(Q)=-1 \Rightarrow Q=\left(\begin{array}{cc} \cos(\theta) & \sin(\theta)\\\sin(\theta)&-\cos(\theta)\end{array}\right) \end{equation*}

Proof:

Since a_1^2+a_2^2=1\Rightarrow\exists 0\leq\theta\leq 2\pi such that: a_1=\cos(\theta) and a_2=\sin(\theta).
Since a_1b_1+a_2b_2=0 \mbox{ and } b_1^2+b_2^2=1 \Rightarrow b_1=\mp \sin(\theta) and b_2=\pm \cos(\theta).

\blacksquare

Orthogonal tensors in \mathbb{R}^2 are either rotations or reflections. If \det(Q)=1, then Q is called a rotation and as shown below, represents a geometric rotation of elements of \mathbb{R}^2. If \det(Q)=-1, then Q is called a reflection and as shown below, represents a geometric reflection of elements of \mathbb{R}^2. It is important to note that this is only true for elements of \mathbb{R}^2 and as will be shown later, if \det(Q)=-1 for higher dimensions, Q does not necessarily represent a reflection but rather an improper rotation or “rotoinversion”.

Examples

The following are rotation matrices

    \[ I=\left(\begin{array}{cc} 1 & 0\\0&1\end{array}\right) \mbox{ and } Q_1=\left(\begin{array}{cc} 0.6 & 0.8 \\ -0.8 &0.6\end{array}\right) \]

The following are reflection matrices

    \[ Q_2=\left(\begin{array}{cc} 1 & 0\\0&-1\end{array}\right) \mbox{ and } Q_3=\left(\begin{array}{cc} 0.6 & 0.8\\ 0.8 & -0.6\end{array}\right) \]

Representation of Rotation Tensors in \mathbb{R}^2

Let \theta be the angle of rotation associated with a rotation matrix Q:\mathbb{R}^2\rightarrow\mathbb{R}^2. Then, given any two orthonormal vectors p, q\in\mathbb{R}^2, Q admits the following representation:

(3)   \begin{equation*} Q=(p\otimes p + q\otimes q) \cos(\theta) + (p\otimes q - q\otimes p) \sin(\theta) \end{equation*}

Proof:

Since p and q are orthonormal and QQ^T=Q^TQ=I, then the following relationships hold:

    \[ p\cdot q = Qp \cdot Qq = 0 \]

    \[ p\cdot p = q\cdot q = Qp \cdot Qp = Qq\cdot Qq=1 \]

Therefore, \exists 0\leq\theta\leq 2\pi such that

    \[ Qp = \cos(\theta) p - \sin (\theta)q \]

    \[ Qq = \sin(\theta) p + \cos(\theta)q \]

Therefore, \forall a\in\mathbb{R}^2, a=(a\cdot p)p+(a\cdot q)q:

    \[ \begin{split} Qa&=Q(a\cdot p)p + Q(a\cdot q)q\\ &=\cos(\theta) (a\cdot p) p - \sin (\theta)(a\cdot p) q+\sin(\theta) (a\cdot q)p + \cos(\theta)(a\cdot q)q\\ &=\left((p\otimes p + q\otimes q) \cos(\theta) + (p\otimes q - q\otimes p) \sin(\theta)\right)a \end{split} \]

Therefore:

    \[ Q=(p\otimes p + q\otimes q) \cos(\theta) + (p\otimes q - q\otimes p) \sin(\theta) \]

Notice that the above relationship represents a clockwise rotation of an angle \theta. By replacing \theta with \phi=-\theta, the counterclockwise rotation of an angle \phi can be represented by the form:

    \[ Q=(p\otimes p + q\otimes q) \cos(\phi) - (p\otimes q - q\otimes p) \sin(\phi) \]

\blacksquare

Geometric Representation of Rotation Tensors in \mathbb{R}^2

Using the matrix representation in (1) for Q the following example applies a rotation of \theta to the blue triangle to produce a red triangle. The angle \theta is illustrated by the black arc. Notice that the matrix shown in (1) rotates the object clockwise!

Reflection Tensors in \mathbb{R}^2

Reflection tensors represent the operation of reflecting elements in \mathbb{R}^2 across a line of reflection.

Assertion 1:Eigenvalues of \mathbb{R}^2 Reflection Tensors:

Reflection tensors in \mathbb{R}^2 have the two eigenvalues 1 and -1 and the associated eigenvectors are orthogonal.

Proof:

It suffices to show that if Q is a reflection tensor in \mathbb{R}^2 then \det(Q-I)=\det(Q+I)=0.
Indeed:

    \[ \begin{split}\det(Q-I)&=\det(Q-Q^TQ)=\det((I-Q^T)Q)\\ &=\det(Q^T-I)\det(-Q)\\ &=\det(Q^T-I)\det(Q)\\ &=\det(Q^T-I)(-1)\\ &=-\det(Q-I) \end{split}\]

Therefore:

    \[\det(Q-I)=0\]

Similarly,

    \[\det(Q+I)=0\]

Let a,b\in\mathbb{R}^2 be the eigenvectors associated with -1,1 respectively. Therefore,

    \[a\cdot b=Qa \cdot Qb = -a \cdot b\Rightarrow a\cdot b = 0\]

\blacksquare

Assertion 2: Reflection tensors in \mathbb{R}^2 are symmetric.

Proof:

This follows directly from having two orthogonal eigenvectors (See symmetric matrices).

\blacksquare

Representation of Reflection Tensors in \mathbb{R}^2

From assertion 2 above, a reflection tensor Q in \mathbb{R}^2 has two eigenvectors a and b associated with the eigenvalues -1 and 1 respectively. Therefore, \forall c\in \mathbb{R}^2:c=(c\cdot a)a+(c\cdot b)b and

    \[Qc=Q(c\cdot a)a+Q(c\cdot b)b=-(c\cdot a)a+(c\cdot b)b=(-a\otimes a + b\otimes b)c\]

Therefore,

(4)   \begin{equation*} Q=-a\otimes a + b\otimes b \end{equation*}

Matrix Representations of Reflection Tensors in \mathbb{R}^2

In addition to the representations (2) and (4), a reflection matrix has various other representations. Let Q be a reflection tensor in \mathbb{R}^2. In a coordinate system whose basis vectors are the eigenvectors a and b of Q associated with the eigenvalues -1 and 1, respectively, we denote the matrix representation of Q by Q' which from (4) admits the form :

    \[Q'=\left(\begin{array}{cc} -1 & 0\\0&1\end{array}\right)\]

In a general coordinate system whose basis vectors are e_1 and e_2, we first apply a coordinate transformation P into the coordinate system of the eigenvectors of Q:

    \[P=\left(\begin{array}{cc} a_1 & a_2 \\b_1&b_2\end{array}\right)\]

Then, Q admits the following form:

    \[Q=P^TQ'P=\left(\begin{array}{cc} -a_1^2+b_1^2 & -a_1a_2+b_1b_2 \\-a_1a_2+b_1b_2&-a_2^2+b_2^2\end{array}\right)\]

where, a_1, a_2, b_1 and b_2 are the components of the vectors a and b.
Notice that we can view a as the perpendicular to the line of reflection since Q reflects a (Qa=-a). Also, b\perp a and keeps its direction (Qb=b), so it lies on the line of reflection. The components of b can be chosen such that b_1=a_2 while b_2=-a_1. In this case Q admits the representation:

    \[Q=\left(\begin{array}{cc} -a_1^2+a_2^2 & -2a_1a_2 \\-2a_1a_2&a_1^2-a_2^2\end{array}\right)\]

Since a and b form an orthonormal basis in \mathbb{R}^2 then P admits the representation:

    \[P=\left(\begin{array}{cc} \cos(\theta_a) & \sin(\theta_a) \\-\sin(\theta_a) &\cos(\theta_a)\end{array}\right)\]

Where \theta_a represents the geometric angle (with positive being the counter-clockwise direction) between the eigenvector a and the basis vector e_1. Therefore, Q admits the representation:

(5)   \begin{equation*} Q=\left(\begin{array}{cc} -\cos(2\theta_a) & -\sin(2\theta_a)\\-\sin(2\theta_a)&\cos(2\theta_a)\end{array}\right) \end{equation*}

Comparing (5) with (2) shows that \theta_a={{\theta - \pi}\over 2}. This indicates, that the angle \theta appearing in the reflection matrix representation (2) is equal to double the angle of inclination of the line of reflection.

In the following illustrative example the effect of varying the angle of inclination of the vector a, namely \theta_a on the reflection of the blue triangle is shown. The vector a is illustrated by the thick black arrow, while the line of reflection is represented in green. \theta_a, \theta/2 and \theta are represented in black, green and red, respectively. The two equivalent matrix representations (5) and (2) are shown underneath the image.
Can you use the example below to find out the approximate inclination of the line of symmetry of the shown triangle?

Orthogonal Tensors in \mathbb{R}^3

Assertion 1: Eigenvalues of Orthogonal Tensors in \mathbb{R}^3:

Proper and improper orthogonal tensors in \mathbb{R}^3 have at least one eigenvalue that is equal to 1 or to -1 respectively.

Proof:

Let Q be a proper orthogonal tensor in \mathbb{R}^3, then \det(Q)=1 and \det(I-Q^T)=-\det(Q^T-I). Therefore:

    \[\begin{split} \det(Q-I)&=\det(Q-QQ^T)=\det(Q)\det(I-Q^T)=\det(-(Q^T-I))\\ &=-\det(Q^T-I)=-\det(Q-I)\\ &\Rightarrow\det(Q-I)=0 \end{split}\]

Therefore, 1 is an eigenvalue associated with every proper orthogonal tensor.

Similarly, if Q is an improper orthogonal tensor then:

    \[\begin{split} \det(Q+I)&=\det(Q+QQ^T)=\det(Q)\det(I+Q^T)=-\det(Q^T+I)=-\det(Q+I)\\ &\Rightarrow\det(Q+I)=0 \end{split}\]

Therefore, -1 is an eigenvalue associated with every improper orthogonal tensor.

\blacksquare

Representation of Orthogonal Tensors in \mathbb{R}^3

From the assertion above, if Q is an orthogonal tensor, then \exists p\in\mathbb{R} such that Qp=\pm p where the positive and negative signs correspond to a proper or an improper orthogonal tensor respectively.
Let q,r\in\mathbb{R}^3 form with p a right hand oriented orthonormal basis set for \mathbb{R}^3. Then, the following relationships hold with the positive and negative sign corresponding to proper and improper orthogonal tensors respectively:

    \[ Qp=\pm p = Q^Tp \]

    \[ p\cdot Qp=\pm p\cdot p=\pm 1 \]

    \[ p\cdot q =p\cdot r = Q^Tp\cdot q=p\cdot Qq=Q^Tp\cdot r=p\cdot Qr=0 \]

    \[ q\cdot r = Qq\cdot Qr = 0 \]

    \[ p\cdot p = Qp \cdot Qp = q\cdot q = Qq \cdot Qq= r\cdot r = Qr \cdot Qr=1 \]

Therefore, \exists 0\leq\theta\leq 2 \pi such that

    \[ Qq = \cos(\theta) q - \sin (\theta)r \]

    \[ Qr = \sin(\theta) q + \cos(\theta)r \]

Therefore, \forall a\in\mathbb{R}^3, a =(a\cdot p)p + (a\cdot q) q + (a\cdot r) r:

    \[ Qa=Q(a\cdot p)p + Q(a\cdot q)q+Q(a\cdot r)r=\left(\pm p\otimes p + (q\otimes q + r\otimes r) \cos(\theta) + (q\otimes r - r\otimes q) \sin(\theta)\right)a \]

Therefore, Q admits the following representation:

(6)   \begin{equation*} Q=\pm p\otimes p + (q\otimes q + r\otimes r) \cos(\theta) + (q\otimes r - r\otimes q) \sin(\theta) \end{equation*}

Where p is the eigenvector associated with 1 and -1 for proper and improper orthogonal tensors respectively, q and r are two vectors that form with p a right handed orthonormal basis set.

The following example shows a proper orthogonal (rotation) tensor in \mathbb{R}^3. You can vary the coordinates of the vector p and the angle of rotation \theta. The code then normalizes p (shown as a blue arrow) and finds two vectors q and r (shown as red arrows) that are perpendicular to p. Then, the proper orthogonal tensor Q is formed using the tensor representation in (6). The rotation is then applied to a sphere. Notice that the above form of the tensor representation rotates the sphere in a clockwise direction around p.

Unlike orthogonal tensors in \mathbb{R}^2, an orthogonal tensor with a determinant equal to -1 in \mathbb{R}^3 is not necessarily associated with a reflection, but rather it represents a “rotoinversion” or an improper rotation.
The following example illustrates the action of an improper orthogonal tensor on a stack of boxes. When the angle \theta in (6) is chosen to be zero, Q represents a reflection across the plane perpendicular to p (The plane formed by the two red arrows). The angle \theta represents a rotation around p and thus, the action of Q constitutes a rotation and an inversion and hence the term “rotoinversion”. You can change the components of the vector p and the angle \theta to see the effect on the resulting transformation.

Matrix Representation of Orthogonal Tensors in \mathbb{R}^3

The tensor representation in (6) can be viewed in matrix form as follows. Given a normal vector p=\{p_1,p_2,p_3\} such that p_1^2+p_2^2+p_3^2=1, two normalized vectors q and r perpendicular to p can be chosen. Assuming that p, q and r form a right handed orthonormal set, then, the matrix form of a proper orthogonal tensor Q is given by:

(7)   \begin{equation*} Q=\left(\begin{array}{ccc} p_1^2(1-\cos(\theta)) + \cos(\theta) & p_1p_2(1-\cos(\theta))+p_3 \sin(\theta) & p_1p_3 (1-\cos(\theta))-p_2 \sin(\theta)\\ p_1p_2(1-\cos(\theta))-p_3 \sin(\theta)&p_2^2(1-\cos(\theta)) + \cos(\theta) &p_2p_3(1-\cos(\theta))+p_1 \sin(\theta)\\ p_1p_3 (1-\cos(\theta))+p_2 \sin(\theta)&p_2p_3 (1-\cos(\theta))-p_1 \sin(\theta)&p_3^2(1-\cos(\theta))+\cos(\theta)\end{array}\right) \end{equation*}

The trace of a proper orthogonal matrix in \mathbb{R}^3 is equal to 1+2\cos(\theta).
The matrix form of an improper orthogonal tensor Q is given by:

(8)   \begin{equation*} Q=\left(\begin{array}{ccc} -p_1^2(1+\cos(\theta)) + \cos(\theta) & -p_1p_2(1+\cos(\theta))+p_3 \sin(\theta) & -p_1p_3 (1+\cos(\theta))-p_2 \sin(\theta)\\ -p_1p_2(1+\cos(\theta))-p_3 \sin(\theta)&-p_2^2(1+\cos(\theta)) + \cos(\theta) &-p_2p_3(1+\cos(\theta)+p_1 \sin(\theta)\\ -p_1p_3 (1+\cos(\theta))+p_2 \sin(\theta)&-p_2p_3 (1+\cos(\theta))-p_1 \sin(\theta)&-p_3^2(1+\cos(\theta))+\cos(\theta)\end{array}\right) \end{equation*}

The trace of an improper orthogonal matrix in \mathbb{R}^3 is equal to -1+2\cos(\theta).
When the angle \theta in (8) is 0 Degrees, the matrix represents a geometric reflection across the plane perpendicular to the vector p. In this case, the matrix representation is given by:

(9)   \begin{equation*} Q=\left(\begin{array}{ccc} -2p_1^2 +1 & -2p_1p_2 & -2p_1p_3 \\ -2p_1p_2&-2p_2^2+ 1&-2p_2p_3\\ -2p_1p_3 &-2p_2p_3 &-2p_3^2+1\end{array}\right) \end{equation*}

The tensor representation (6) asserts that any rotation matrix can be viewed as a rotation around an axis p. Any rotation can also be viewed using Euler’s angles as consecutive rotations around each of the basis vectors of the coordinate system. Clockwise rotations with an angles \theta_a, \theta_b, \theta_c around the basis vectors e_1, e_2 and e_3 are given by the following matrices Q_a, Q_b and Q_c, respectively:

    \[ Q_a=\left(\begin{array}{ccc} 1& 0& 0\\ 0& \cos(\theta_a) &\sin(\theta_a)\\ 0&- \sin(\theta_a)&\cos(\theta_a)\end{array}\right) \]

    \[ Q_b=\left(\begin{array}{ccc} \cos(\theta_b)& 0& -\sin(\theta_b)\\ 0& 1&0\\ \sin(\theta_b) & 0 &\cos(\theta_b)\end{array}\right)\]

    \[Q_c=\left(\begin{array}{ccc} \cos(\theta_c)& \sin(\theta_c)&0\\ -\sin(\theta_c) &\cos(\theta_c)&0\\0&0&1\end{array}\right) \]

It is important to notice that the order of rotation changes the final position of the rotated object. See the example in the rigid body rotation section.

Problems

What values for the angle \theta would make the matrices in (7) and (8) symmetric?.
Find the axis and angle of rotation of the rotation matrix Q_1.
Find the plane of inversion and the angle of rotation of the improper orthogonal matrices Q_2 and Q_3.
Find the corresponding \theta_1, \theta_2 and \theta_3 if the rotation matrix Q_1 is viewed as a rotation around e_1 followed by e_2 then e_3.
Find the corresponding \theta_1, \theta_2 and \theta_3 if the rotation matrix Q_1 is viewed as a rotation around e_3 followed by e_2 then e_1.

    \[ Q_1=\left(\begin{array}{ccc} {1\over3}(1+\sqrt2)&{1\over6}(2-\sqrt2-\sqrt6)&{1\over6}(2-\sqrt2+\sqrt6)\\ {1\over6}(2-\sqrt2+\sqrt6)&{1\over3}(1+\sqrt2)&{1\over6}(2-\sqrt2-\sqrt6)\\ {1\over6}(2-\sqrt2-\sqrt6)&{1\over6}(2-\sqrt2+\sqrt6)&{1\over3}(1+\sqrt2) \end{array}\right) \]

    \[ Q_2=\left(\begin{array}{ccc} {1\over3}(-1+\sqrt2)&{1\over6}(-2-\sqrt2+\sqrt6)&{1\over6}(-2-\sqrt2-\sqrt6)\\ {1\over6}(-2-\sqrt2-\sqrt6)&{1\over3}(-1+\sqrt2)&{1\over6}(-2-\sqrt2+\sqrt6)\\ {1\over6}(-2-\sqrt2+\sqrt6)&{1\over6}(-2-\sqrt2-\sqrt6)&{1\over3}(-1+\sqrt2) \end{array}\right) \]

    \[ Q_3=\left(\begin{array}{ccc} {1\over3}&{-2\over3}&{-2\over3}\\ {-2\over3}&{1\over3}&{-2\over3}\\ {-2\over3}&{-2\over3}&{1\over3} \end{array}\right) \]

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