How do I find 2x2 orthonormal diagonalizing matrices using only trigonometry?

Question

I have a matrix $A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ (where all values are known), and I eventually want to diagonalize it into: $$ A=UDV^T $$ for orthonormal U and V. If I represent U and V as: $$ U=\begin{bmatrix} cos(\theta) & sin(\theta) \\ -sin(\theta) & cos(\theta) \end{bmatrix} , V=\begin{bmatrix} cos(\phi) & sin(\phi) \\ -sin(\phi) & cos(\phi) \end{bmatrix} $$ Is there a geometric way to quickly find $\theta$ and $\phi$ using $a$, $b$, $c$, and $d$?

I realize that I could find the eigenvalues and work from there algebraically, but I'm wondering if there's a clean visual way of "seeing" $\theta$ and $\phi$, and then going from there.

I was thinking they would be related to the angles of triangles made up of the values of $A$, but I can't seem to figure out which angles they are.

Thanks!

EDIT: I realized I miswrote $U$ and $V$. $U$ and $V$ need to be rotation matrices for this to work. Any ideas?

Answer 1

Problem statement and restrictions

The problem you propose is this $$ \begin{align} \mathbf{A} &= \mathbf{U} \, \mathbf{S} \, \mathbf{V}^{*} \\ % A \left[ \begin{array}{cc} a & b \\ c & d \\ \end{array} \right] % &= % % U \left[ \begin{array}{cr} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{array} \right] % S \left[ \begin{array}{cc} \sigma_{1} & 0 \\ 0 & \sigma_{2}\\ \end{array} \right] % V* \left[ \begin{array}{rr} \cos \phi & \sin \phi \\ -\sin \phi & \cos \phi \\ \end{array} \right] \\[3pt] % &= % svd \left[ \begin{array}{cc} \sigma_{1}\cos \theta \cos \phi+\sigma_{2}\sin \theta \sin \phi & \sigma_{1}\cos \theta \sin \phi-\sigma_{2}\cos \phi \sin \theta \\ \sigma_{1}\cos \phi \sin \theta-\sigma_{2} \cos \theta \sin \phi & \sigma_{2}\cos \theta \cos \phi+\sigma_{1}\sin \theta \sin \phi \\ \end{array} \right] % \end{align} $$ The task is to connect the Roman alphabetics to the Greeks. Note this restricts us to class of problems where the domain matrices are rotation matrices. This excludes reflection matrices and matrices which are compositions with reflections.

To proceed, require

1. rank $\left( \mathbf{A} \right)$ = 2 $ \quad \Rightarrow \quad$ $\det \mathbf{A} \ne 0$.

2. The variables $a$, $b$, $c$, and $d$ are real.

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Linear regression example

Consider a closely related example which has been solved before. The problem is from linear regression $$ y(x) = a_{0} + a_{1} x $$ Given a sequence of $m$ measurements of the form $\left\{ x_{k}, y_{k} \right\}_{k=1}^{m}$, one can assemble to systems matrix $\mathbf{A}\in\mathbb{C}^{m\times 2}_{2}$ and the data vector $y\in\mathbb{C}^{m}$ which is not in the null space: $y\notin\mathcal{N}\left( \mathbf{A}^{*} \right)$. The linear system is $$ \begin{align} \mathbf{A} a & = y\\ % A \left[ \begin{array}{cc} 1 & x_{1} \\ \vdots & \vdots \\ 1 & x_{m} \\ \end{array} \right] % a \left[ \begin{array}{c} a_{0} \\ a_{1} \end{array} \right] % & = % \left[ \begin{array}{c} y_{1} \\ \vdots \\ y_{m} \end{array} \right] % \end{align} $$ The dimensions of this matrix $\mathbf{A}$ here do not match those of your example, there are too many rows. But this may not be a problem, because once we compute the matrices $\mathbf{S}$ and $\mathbf{V}$, we can construct the matrix $\mathbf{U}$ via $$ \mathbf{S}^{-1} \mathbf{A} \, \mathbf{V} = \mathbf{U}. $$

Eigensystem

The first step in decomposition is to construct the product matrix $\mathbf{A}^{*}\, \mathbf{A}$ and resolve the eigensystem. $$ \mathbf{A}^{*}\, \mathbf{A} = \left[ \begin{array}{cc} \mathbf{1} \cdot \mathbf{1} & \mathbf{1} \cdot x \\ x \cdot \mathbf{1} & x \cdot x \end{array} \right] $$ The characteristic polynomial is $$ \begin{align} p(\lambda) &= \lambda^{2} - \lambda \, \text{tr} \left( \mathbf{A}^{*} \mathbf{A}\right) + \det \left( \mathbf{A}^{*} \mathbf{A}\right) \\ \end{align} $$ The roots $p\left( \lambda_{\pm} \right)$ are the eigenvalues. The square root of the eigenvalues produces the singular values $$ \sigma = \sqrt{\frac{1}{2} \left( \mathbf{1}\cdot\mathbf{1} + x \cdot x \pm \sqrt{4 \left( \mathbf{1}\cdot x \right)^{2} - \left( \mathbf{1}\cdot\mathbf{1} - x \cdot x \right)^{2}} \right)} $$

Introduce rotation matrix

Onto computing the domain matrix, and this is where your insight will help. Use the relationship $$ \begin{align} \mathbf{A}^{*} \mathbf{A} &= \mathbf{V} \, \mathbf{S}^{2} \, \mathbf{V}^{*} \\ % &= % V \left[ \begin{array}{cr} \cos \phi & \sin -\phi \\ \sin \phi & \cos \phi \\ \end{array} \right] % \mathbf{S}^{2} % V* \left[ \begin{array}{rc} \cos \phi & \sin \phi \\ -\sin \phi & \cos \phi \\ \end{array} \right] \\[3pt] % \left[ \begin{array}{cc} \mathbf{1} \cdot \mathbf{1} & \mathbf{1} \cdot x \\ x \cdot \mathbf{1} & x \cdot x \end{array} \right] % &= % \left[ \begin{array}{cc} \sigma_{1}^{2} \cos^{2} \phi + \sigma_{2}^{2} \sin^{2} \phi & \left(\sigma_{1}^{2} - \sigma_{2}^{2} \right) \cos \phi \sin \phi \\ \left(\sigma_{1}^{2} - \sigma_{2}^{2} \right) \cos \phi \sin \phi & \sigma_{2}^{2} \cos^{2} \phi + \sigma_{1}^{2} \sin^{2} \phi \end{array} \right] % \end{align} $$

Solution

We find solutions like $$ \cos \phi = \sqrt{\frac{ \left(\mathbf{1}\cdot \mathbf{1}\right) \sigma_{1}^{2} - \left( x \cdot x \right) \sigma_{2}^{2}} {\sigma_{1}^{4} - \sigma_{2}^{4}}}, \qquad % \sin \phi = \sqrt{\frac{ \left(\mathbf{1}\cdot \mathbf{1}\right) \sigma_{2}^{2} - \left( x \cdot x \right) \sigma_{1}^{2}} {\sigma_{1}^{4} - \sigma_{2}^{4}}} $$ This constructs the angles in the rotation matrix for the domain interms of the data.

Answer 2

Without loss of generality, consider $A$ with $\det A = 1$. Let $D = \begin{bmatrix} d_1 & 0 \\ 0 & d_2 \end{bmatrix}$. If $d_1 = d_2$ the problem is simple. We will also continue the argument with $d_1 > d_2$ (the argument for $d_2 > d_1$ is the same with small modifications).

Here is my intuition.

Let $X$ be the set of points of the unit circle. The matrix $V^{T}$ will just rotate the members of $X$, so $$A X = U D V^T X = U D X$$ $D$ will elongate the circle into an ellipse with the ellipse's major axis (the long one) pointing in the $x$-axis's direction (because $d_1 > d_2$). Finally $U$ will rotate the ellipse so that the major axis points in the direction of $\theta$. What was the point of $X$ that became the tip of the major axis? Since it was $(1, 0)$ in $V^T X$ it must have been the unit vector pointing in the $\phi$ direction originally.

So for a given $A$, $\theta$ and $\phi$ are tied into geometrically the vector whose length is modified the most, (That is maximizes $\frac{|Ax|}{|x|}$.) which by linearity we can assume to be unit length.