# $3\times 3$ Orthogonal Matrices with an Analysis of Eigenvalues

It is said that one can prove that all 3x3 orthogonal matrices correspond to linear operators on $R^3$ of the following types:

1. Rotations about lines through he origin
2. Reflections about planes through the origin
3. A rotation about a line through the origin followed by a reflection about the plane through the origin that is perpendicular to the line

Obviously, we can tell whether a 2x2 matrix $A$ represents a rotation or reflection by looking at its determinant: that is, a rotation if $det(A)=1$ and a reflection if $det(A)=-1$. But for a 3x3 matrix, this is different. My textbook says that if $det(A)=1$, it is a rotation. If $det(A)=-1$, it is either of type 2 or 3.

My question is: How to tell whether a 3x3 orthogonal matrix with determinant $-1$ represents a type 2 or a type 3 ? I heard that this has to do with an analysis of eigenvectors and eigenvalus, but could anyone shed some light on this please?

For example, how can one tell whether the following operator is of type 2 or type 3? (Actually I was quite sure, as I "invented" it, if no mistake, that the following linear operator is neither a rotation, nor a reflection$-$it is of type 3.) \begin{bmatrix} \frac{1}{3} & 0 & \frac{4\sqrt{2}}{6} \\ \frac{2}{3} & \frac{1}{\sqrt{2}} & \frac{-\sqrt{2}}{6} \\ \frac{2}{3} & \frac{-1}{\sqrt{2}} & \frac{-\sqrt{2}}{6} \\ \end{bmatrix}

Eigenvalues (or strictly speaking, eigendecomposition) can be used to establish the fact that the three listed cases are the only possible ones. They are not necessary in differentiating type 2 and type 3 matrices.

If $A$ is a reflection about a plane through the origin, then by applying $A$ again, the reflected image will be reflected back to its original. Therefore $A^2=I$.

If $A$ is a rotation followed by a reflection through the rotation plane, pick any vector $u$ that has a nonzero component on the plane of rotation (i.e. pick any $u$ whose orthogonal projection on the rotation plane is nonzero). Decompose $u$ into the sum $v+w$, where $v$ lies on the plane of rotation and $w$ is parallel to the rotation axis (and hence orthogonal to $v$. Then $A^2u=A^2v+w$. When the angle of rotation is not an odd multiple of $\pi$, we have $A^2v\ne v$ and hence $A^2u\ne u$ and hence $A^2\ne I$. When the angle of rotation is an odd multiple of $\pi$, we get $A=-I$.

Therefore, when $A$ is real orthogonal and $\det A=-1$, it is of type 2 if and only if $A^2=I$ and $A\ne-I$, and it is of type 3 if and only if $A^2\ne I$ or $A=-I$.

• I have read your answer and amd's answer already. As you pointed out, we can determine the type of the matrix operator by checking if $A^2 = I$ and $A \ne I$ or not; if this is the case, it is of type2, if this is not the case, it is of another type. But I was just wondering how can I relate this to an analysis of eigenvectors and eigenvalues as amd's pointed out in his post. – IgNite Apr 16 '16 at 8:02
• Or more precisely, why the matrix operator is of type 3 if one of its eigenvalues is $1$ and the other two aren't $\pm1$? – IgNite Apr 16 '16 at 8:03
• @Ignite What you say cannot happen. As $\det A=-1$ and one of the eigenvalues must be -1 (because a reflection occurs), if one of the eigenvalues is 1, the other two must be -1 and 1. – user1551 Apr 16 '16 at 8:40
• @Ignite Anyway, take a unit vector $u$ parallel to the axis of rotation and take two mutually orthogonal unit vectors $v$ and $w$ from the plane normal to $u$. Then $\{u,v,w\}$ is an orthonormal basis of $\mathbb R^3$. With respect to this basis, the matrix becomes $\pmatrix{-1\\ &R}$, where $R$ is a 2x2 rotation matrix. $R=I$ when it is of type 2, and $R\ne I$ when it is of type 3. The eigenvalues of a 2x2 rotation matrix are $\cos\theta\pm i\sin\theta$, where $\theta$ is the angle of rotation. If $R^2=I$ but $R\ne I$, $\theta$ must be an odd multiple of $\pi$, i.e. $R=-I$. – user1551 Apr 16 '16 at 8:44
• Oh I'm very sorry, there's a typo. It should be why the matrix operator is of type 3 if one of its eigenvalues is $-1$ and the other two aren't $\pm1$? – IgNite Apr 16 '16 at 8:55

If there’s a rotation involved, two of the eigenvalues will be a complex conjugate pair. If it’s a pure reflection, all of the eigenvalues will be real. There’s one exception to this—a rotation through an angle of $\pi$—but I’ll cover that below.

In more detail, a pure 3-D rotation matrix will have eigenvalues $1$ and $\cos\theta\pm i\sin\theta$, where $\theta$ is the rotation angle. If $\theta=n\pi$, then this complex conjugate pair will show up as $\pm1$. If positive, you just have the identity transformation; if negative, you have a 180-degree rotation (which is equivalent to a reflection relative the axis of rotation). The eigenspace of $1$ in the non-identity case is one-dimensional and represents the rotation axis. If you follow this rotation with a reflection in the plane normal to this axis, the eigenvalue $1$ will become $-1$.

On the other hand, the eigenvalues of a reflection will all be $\pm1$. The multiplicity of $-1$ tells you what sort of a reflection this is: a single $-1$ is a reflection in a plane, with the corresponding eigenspace normal to this plane; a multiplicity of 2 is a reflection in a line, which is equivalent to a rotation through an angle of $\pi$; a multiplicity of 3 represents releflection through the origin, which we see can be decomposed into a 180-degree rotation followed by a reflection relative to a plane.

To take your example, the matrix has $-1$ as an eigenvalue with multiplicity one, while the other eigenvalues aren’t $\pm1$, so it’s type 3. The axis of rotation is found by computing the kernel of $A+I$, giving $(2+\sqrt2,-2,-2-2\sqrt2)^T$. The angle of rotation can be found by finding the other eigenvalues and converting to polar form. (I did some quick calculations in a CAS program and it looks like the angle might be $\frac\pi4$.)

• Working on this. I'm figuring out, in the third paragraph, why the number of the multiplicity of eigenvalues $-1$ can tell what sort of a reflection is... – IgNite Apr 16 '16 at 7:31
• Could you explain a little bit more why the matrix operator is of type 3 if one of its eigenvalues is $-1$ and the other two aren't $\pm1$? – IgNite Apr 16 '16 at 8:56
• @Ignite Looks like user1551 covered it in the comments to his answer. Another way to understand this is to recall the motivation for the concept of eigenvectors: what lines are mapped to themselves? The corresponding eigenvalues are the “scale factors” for those fixed lines. For an orthogonal matrix, each $-1$ corresponds to an axis that is flipped by the transformation and each $1$ leaves the corresponding axis unchanged. If there’s some other eigenvalue, then something else—a rotation in this case—is going on. – amd Apr 16 '16 at 16:51