2D Decompositions

Example 1: Using Mathematica, we verify the commutation laws for each of these basic three transformations, (1) -- (4). We start with rotation matrices:

Multiplying two rotation matrices, we get So we see that rotation matrices commute . Now we define a scaling matrix Since S is a diagonal matrix, we conclude that scaling matrices also commute .     Now we check products of rotation and scaling matrices: and From the above two matrix multiplications, we see that rotations and rescalings do not commute because products S R and R S are not equal.

We consider two basic reflection matrices

Their products are the same, as Mathematicaconfirms Since refX.refY = refY.refX = −I, these two basic reflection matrices commute . What about arbitrary reflections? To answer this question, we choose two noncolinear unit vectors n = (n₁, n₂) and m = (m₁, m₂) that are orthogonal to two straight lines. Assuming that these unit vectors are written in column form (n, m ∈ ℝ^2×1), the corresponding reflection matrices with respect to these two lines can be written in succinct form \[ \mathbf{R}_n = \mathbf{I} - 2\,\mathbf{n}\cdot \mathbf{n}^{\mathrm T} \qquad \mbox{and} \qquad \mathbf{R}_m = \mathbf{I} - 2\,\mathbf{m}\cdot \mathbf{m}^{\mathrm T} . \] Their product is \[ \mathbf{R}_n \mathbf{R}_m = \mathbf{I} - 2\,\mathbf{n}\cdot \mathbf{n}^{\mathrm T} - 2\,\mathbf{m}\cdot \mathbf{m}^{\mathrm T} + 4\, \mathbf{n}\cdot \mathbf{n}^{\mathrm T} \,\mathbf{m}\cdot \mathbf{m}^{\mathrm T} . \] From this formula, it follows that reflection matrices R_n and R_m commute if and only if \[ \left( \mathbf{n}\cdot \mathbf{n}^{\mathrm T} \right) \left( \mathbf{m}\cdot \mathbf{m}^{\mathrm T} \right) = \left( \mathbf{m}\cdot \mathbf{m}^{\mathrm T} \right) \left( \mathbf{n}\cdot \mathbf{n}^{\mathrm T} \right) . \tag{1.1} \] We rewrite Eq.(1.1) in coordinate form: \[ \begin{bmatrix} n_1^2 & n_1 n_2 \\ n_2 n_1 & n_2^2 \end{bmatrix} \cdot \begin{bmatrix} m_1^2 & m_1 m_2 \\ n_2 m_1 & m_2^2 \end{bmatrix} = \begin{bmatrix} m_1^2 & m_1 m_2 \\ n_2 m_1 & m_2^2 \end{bmatrix} \cdot \begin{bmatrix} n_1^2 & n_1n_2 \\ n_2 n_1 & n_2^2 \end{bmatrix} . \tag{1.2} \] Generally speaking, Eq.(1.2) does not hold and we conclude that reflection matrices do not commute unless corresponding lines are orthogonal. For instance, let us choose two lines y = x√3 and y = −x√3. Then these lines have the corresponding normal vectors: \[ \mathbf{n} = \frac{1}{2} \begin{pmatrix} -1 \\ \sqrt{3} \end{pmatrix} , \qquad \mathbf{m} = \frac{1}{2} \begin{pmatrix} 1 \\ \sqrt{3} \end{pmatrix} \] Then (upon dropping multiple ¼) their Householder matrices become \[ \mathbf{A} = \mathbf{n} \,\mathbf{n}^{\mathrm T} = \frac{1}{4} \begin{pmatrix} 1 & -\sqrt{3} \\ -\sqrt{3} & 3 \end{pmatrix} , \] and \[ \mathbf{B} = \mathbf{m} \,\mathbf{m}^{\mathrm T} = \frac{1}{4} \begin{pmatrix} 1 & \sqrt{3} \\ \sqrt{3} & 3 \end{pmatrix} . \] Mathematica evaluates their commutator [A, B] = A B − B A to be Since for our matrices \[ \begin{pmatrix} 1 & -\sqrt{3} \\ -\sqrt{3} & 3 \end{pmatrix} \begin{pmatrix} 1 & \sqrt{3} \\ \sqrt{3} & 3 \end{pmatrix} - \begin{pmatrix} 1 & \sqrt{3} \\ \sqrt{3} & 3 \end{pmatrix} \begin{pmatrix} 1 & -\sqrt{3} \\ -\sqrt{3} & 3 \end{pmatrix} = \begin{pmatrix} 0 & - 4\sqrt{3} \\ 4\sqrt{3} & 0 \end{pmatrix} , \] we conclude that reflection matrices with respect to lines having normal vectors n and m do not commute .

Now we multiply basic reflection matrices by scaling matrices

and So scaling and reflection with respect to x axes commute . Similar conclusion is valid for reflection with respect to another axes

On the other hand, products of reflection matrices and the rotation one show different output:

and Similarly, and Therefore, we conclude that rotation and reflection matrices do not commute .    ■

Example 2: Let us perform this decomposition with \[ {\bf A} = \begin{bmatrix} 5& 2 \\ 3& 1 \end{bmatrix} . \] We extract its column vectors Their (Euclidean) norms are Angles are approximately and We check whether our calculations are correct. Now we define auxiliary angles and We check our calculations Now we see directly that our decomposition of a linear map into a product of rotations and general rescalings works. In particular, we can write A as A = A₁A₂A₃A₄ with \begin{align*} \mathbf{A}_1 &= \begin{bmatrix} \cos\psi & - \sin\psi \\ \sin\psi & \cos\psi \end{bmatrix} \approx \begin{bmatrix} 0.876606 & -0.481209 \\ 0.481209 & 0.876606 \end{bmatrix} , \\ \mathbf{A}_2 &= \begin{bmatrix} \sqrt{2}\cos \chi & 0 \\ 0 & \sqrt{2}\sin\chi \end{bmatrix} \approx \begin{bmatrix} 1.41317 & 0 \\ 0 & 0.0542726 \end{bmatrix} , \\ \mathbf{A}_3 &= \begin{bmatrix} \cos \left( \frac{\pi}{4} \right) & -\sin \left( \frac{\pi}{4} \right) \\ \cos \left( \frac{\pi}{4} \right) & \sin \left( \frac{\pi}{4} \right) \end{bmatrix} \approx \begin{bmatrix} 0.707107 & - 0.707107 \\ 0.707107 & 0.707107 \end{bmatrix} , \\ \mathbf{A}_4 &= \begin{bmatrix} \sqrt{34} & 0 \\ 0 & -\sqrt{5} \end{bmatrix} \approx \begin{bmatrix} 5.83095 & 0 \\ 0 & 2.23607 \end{bmatrix} . \end{align*} This decomposition cannot be reordered, as you are asled to verify in Ecercise 2. We check with Mathematica decomposition of the given matrix A into product of four matrices.    ■

1. Express each of the folowing matrices in the product given in Eq.\eqref{Eq2D.5}.
   \[ {\bf (a) \ \ } \ \begin{bmatrix} -\sqrt{5} & 0 \\ 2 & 5 \end{bmatrix} , \qquad \quad {\bf (b) \ \ } \ \begin{bmatrix} - \frac{1}{\sqrt{2}} & -\frac{3}{\sqrt{2}} \\ \frac{3}{\sqrt{2}} & - \frac{3}{\sqrt{2}} \end{bmatrix} , \]
   \[ {\bf (c) \ \ } \ \begin{bmatrix} 5\sqrt{2} & - \frac{13}{\sqrt{2}} \\ 5\sqrt{2} & \frac{13}{\sqrt{2}} \end{bmatrix} , \qquad {\bf (d) \ \ } \ \begin{bmatrix} 0 & -\frac{\sqrt{3}}{6} \\ \frac{1}{2} & - \frac{1}{6} \end{bmatrix} , \]
2. In Example 2, it was shown that the given matrix A can be decomposed into prooduct of four matrices AA₁A₂A₃A₄. Determine whether any of the following reorderings agree with the decomposition of matrix A.
   \[ {\bf (a) \ \ } \ \mathbf{A}_4 \mathbf{A}_3 \mathbf{A}_2 \mathbf{A}_1 , \qquad {\bf (b) \ \ } \ \mathbf{A}_2 \mathbf{A}_1 \mathbf{A}_2 \mathbf{A}_1 , \qquad {\bf (c) \ \ } \ \mathbf{A}_3 \mathbf{A}_2 \mathbf{A}_1 \mathbf{A}_4 . \]
3. If A ∈ ℝ^2×2 is a singular matrix (so detA = 0), can it still be decomposed into product \eqref{Eq2D.5} ?
4. Decompose the 2 × 2 off-diagonal matrix \( \displaystyle \quad \begin{bmatrix} 0&b \\ c&0 \end{bmatrix} . \)
5. Consider a linear transformation that maps a point (x, y) into (2x − y, 0). Express the corresponding matrix as the product \eqref{Eq2D.5}.
6. Let
   \[ \mathbf{A} = \begin{bmatrix} -5 + 8\mathbf{j} & 2 - \mathbf{j} \\ 3 - 2 \mathbf{j} & -1 + 4 \mathbf{j} \end{bmatrix} = \begin{bmatrix} -5 & 2 \\ 3 & -1\end{bmatrix} + {\bf j} \begin{bmatrix} 8 & -1 \\ 3 & 4 \end{bmatrix} , \]
   where j is the imaginary unit vector on complex plane ℂ, so j ² = −1. Decompose the real part and the imaginary part of matrix A according to formula \eqref{Eq2D.5} .

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