Rotation Matrix for 180-Degree Turn About an Axis
A linear operator $A$ rotates every vector in $\mathbb{R}^3$ by 180 degrees around an axis defined by the unit vector $\hat{u} = [a,\, b,\, c]^T$ with $a^2 + b^2 + c^2 = 1$.
- Find the matrix representation of $A$.
2. What are the properties of this matrix? Specifically: - Is $A$ symmetric? - Does $A^2 = I$? - Is $A$ orthogonal? - What is $\det(A)$? - What are the eigenvalues and eigenspaces of $A$?
Hints
- Decompose any vector into its component along the rotation axis and its component perpendicular to the axis. What does a 180-degree rotation do to each part?
- The projection of $x$ onto $\hat{u}$ is $\hat{u}(\hat{u}^T x)$. The parallel part is preserved and the perpendicular part is negated -- write this as a single matrix expression.
- You should arrive at $A = 2\hat{u}\hat{u}^T - I$. To verify properties, use $\hat{u}^T\hat{u} = 1$ to compute $A^2$ and note that symmetry plus $A^2 = I$ immediately gives orthogonality.
Worked Solution
How to Think About It: A 180-degree rotation about an axis $\hat{u}$ does two things: it leaves $\hat{u}$ alone and flips every vector in the plane perpendicular to $\hat{u}$. So you need a matrix whose eigenvalue is $+1$ along $\hat{u}$ and $-1$ in the two-dimensional subspace orthogonal to $\hat{u}$. If you remember that the projection onto $\hat{u}$ is $\hat{u}\hat{u}^T$, you can build the answer from that projection. The component along $\hat{u}$ stays, the component perpendicular gets negated -- that is the whole matrix.
Quick Estimate: Before writing anything, sanity-check the formula you expect. The matrix should satisfy: $A\hat{u} = \hat{u}$, and for any $v \perp \hat{u}$, $Av = -v$. It should be orthogonal with determinant $+1$ (proper rotation), and since rotating 180 degrees twice returns you to the start, $A^2 = I$. These checks will confirm whatever formula you derive.
Approach: Decompose any vector into its component along $\hat{u}$ and perpendicular to $\hat{u}$, then apply the rotation to each part.
Formal Solution:
Any vector $x \in \mathbb{R}^3$ splits as:
$x = (\hat{u}^T x)\hat{u} + (x - (\hat{u}^T x)\hat{u})$
The first term is the projection onto $\hat{u}$, which stays fixed. The second is the perpendicular component, which gets negated by a 180-degree rotation:
$Ax = (\hat{u}^T x)\hat{u} - (x - (\hat{u}^T x)\hat{u}) = 2(\hat{u}^T x)\hat{u} - x$
In matrix form:
$A = 2\hat{u}\hat{u}^T - I$
Writing this out entry by entry with $\hat{u} = [a,\, b,\, c]^T$:
$A = \begin{pmatrix} 2a^2 - 1 & 2ab & 2ac \\ 2ab & 2b^2 - 1 & 2bc \\ 2ac & 2bc & 2c^2 - 1 \end{pmatrix}$
Verification (Rodrigues' formula): The general rotation matrix for angle $\theta$ about axis $\hat{u}$ is $R = I + (\sin\theta)\, K + (1 - \cos\theta)\, K^2$, where $K$ is the skew-symmetric cross-product matrix of $\hat{u}$. At $\theta = \pi$: $\sin\pi = 0$ and
Properties:
- Symmetric: $A^T = (2\hat{u}\hat{u}^T - I)^T = 2\hat{u}\hat{u}^T - I = A$. Yes, $A$ is symmetric.
- Involution ($A^2 = I$):
$A^2 = (2\hat{u}\hat{u}^T - I)^2 = 4\hat{u}\hat{u}^T\hat{u}\hat{u}^T - 4\hat{u}\hat{u}^T + I = 4\hat{u}\hat{u}^T - 4\hat{u}\hat{u}^T + I = I$
using $\hat{u}^T\hat{u} = 1$. So yes, $A^2 = I$.
- Orthogonal: Since $A$ is symmetric and $A^2 = I$, we get $A^T A = A \cdot A = I$. So $A$ is orthogonal, and $A^{-1} = A^T = A$.
- Determinant: $\det(A) = (+1)(-1)(-1) = +1$. This confirms $A$ is a proper rotation (belongs to $SO(3)$).
5. Eigenvalues and eigenspaces: - Eigenvalue $+1$ with eigenvector $\hat{u}$ (the rotation axis is fixed). - Eigenvalue $-1$ with multiplicity 2 (eigenspace is the plane perpendicular to $\hat{u}$, since every vector in that plane gets flipped).
Answer: The rotation matrix is $A = 2\hat{u}\hat{u}^T - I$, which is symmetric, orthogonal, self-inverse ($A^2 = I$, $A^{-1} = A$), has determinant $+1$, eigenvalue $+1$ along the axis $\hat{u}$, and eigenvalue $-1$ with multiplicity 2 in the plane perpendicular to $\hat{u}$.
Intuition
The key insight is that any rotation in 3D can be understood through projection. A 180-degree rotation about an axis keeps the axial component and negates everything perpendicular. The projection matrix $\hat{u}\hat{u}^T$ is the building block: it extracts the axial part, and $I - \hat{u}\hat{u}^T$ extracts the perpendicular part. The rotation is then just "keep the first, negate the second," which gives $\hat{u}\hat{u}^T - (I - \hat{u}\hat{u}^T) = 2\hat{u}\hat{u}^T - I$. This formula is closely related to the Householder reflection $H = I - 2\hat{u}\hat{u}^T$, which reflects through the plane perpendicular to $\hat{u}$. In fact $A = -H$: a 180-degree rotation about an axis is the negative of a reflection through the plane perpendicular to that axis. The difference is that the rotation has $\det = +1$ (proper rotation in $SO(3)$) while the reflection has $\det = -1$. In practice, this decomposition-by-projection trick shows up constantly -- in PCA (projecting onto principal components), in Gram-Schmidt, and whenever you need to split a signal into components along different directions.