Computing angle of rotation and scaling factor directly from given transformation matrix

Question

I have a simple 2D transformation matrix performing rotation combined with scaling \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix} Where the scalar $r$ is multiplied with \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} Is there a simple method that can be performed on the input matrix to compute $r$ and $\theta$ directly?

Answer 1

An alternative to Henrik's version, perhaps more easily understandable is the use of

{Q, R} = QRDecomposition[{{1, 1}, {-1, 1}}]
(* {{{1/Sqrt[2], -(1/Sqrt[2])}, {1/Sqrt[2], 1/Sqrt[2]}}, {{Sqrt[2],0}, {0, Sqrt[2]}}} *)

Q is a rotation matrix and R contains the scaling

Solve[RotationMatrix[\[CurlyPhi] ] == Q, \[CurlyPhi]][[1]] /.C[1] -> 0 
(*{\[CurlyPhi] -> \[Pi]/4} *)

Answer 2

m = {{1, 1}, {-1, 1}};
FullSimplify @ Solve[Transpose[RotationMatrix[θ]].ScalingMatrix[{s1, s2}] == m, 
  {s1, s2, θ}, Reals] /. C[1] -> 0 // TeXForm

$\left\{\left\{\text{s1}\to -\sqrt{2},\text{s2}\to -\sqrt{2},\theta \to -\frac{3 \pi }{4}\right\},\left\{\text{s1}\to \sqrt{2},\text{s2}\to \sqrt{2},\theta \to \frac{\pi }{4}\right\}\right\}$

Answer 3

Solution one

This is your mat:

MatrixForm[mat = {{1, 1}, {-1, 1}}]

$\small \left( \begin{array}{cc} 1 & 1 \\ -1 & 1 \\ \end{array} \right)$

And this is the angle of rotation(Positive is counterclockwise, negative is counterclockwise):

Arg[mat.{1, 0}.{1, I}]

$-\frac{\pi }{4}$

This is the scale on each axis:

Eigenvalues[mat.Inverse[mat/Sqrt[Det[mat]]]]

$\left\{\sqrt{2},\sqrt{2}\right\}$

---

Solution two

As this comment, this is rotation matrix is:

R = Inverse[Transpose[mat]].MatrixPower[Transpose[mat].mat, 1/2]

$\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \end{pmatrix}$

We can get two scaling matrix generally:

scaleLeft = MatrixPower[mat.Transpose[mat], 1/2]

$\begin{pmatrix} \sqrt{2} & 0 \\ 0 & \sqrt{2} \\ \end{pmatrix}$

This mean $\rm mat=scaleLeft.R$

scaleRight = MatrixPower[Transpose[mat].mat, 1/2]

$\begin{pmatrix} \sqrt{2} & 0 \\ 0 & \sqrt{2} \\ \end{pmatrix}$

This mean $\rm mat=R.scaleRight $. But in this case, the scaleRight equal to scaleLeft.

Answer 4

You may try to apply this function:

{#[[2, 1, 1]], -ArcTan @@ (#[[1]].#Transpose[[[3]]])[[1]]} &[SingularValueDecomposition[#]] &

A bit simpler, using complex numbers:

f = AbsArg[({1, I}.#)[[1]]] &
f[{{1, 1}, {-1, 1}}]

$\left\{\sqrt{2},-\frac{\pi }{4}\right\}$

General case:

A random scale-rotation:

scale = If[Variance[Diagonal[S]] < $MachinePrecision^2,
  Mean[Diagonal[S]],
  $Failed
  ]

First we compute the polar decomposition of M

{U, Σ, V} = SingularValueDecomposition[M];
R = U.Transpose[V];
S = V.Σ.Transpose[V];

Checking its validity:

Norm[R.S - M, "Frobenius"]
Norm[Transpose[R].R - IdentityMatrix[dim], "Frobenius"]
Norm[S - Transpose[S], "Frobenius"]

2.01949*10^-15

6.36666*10^-16

7.11972*10^-16

Everything seems to be good. R is the rotation matrix we are looking for. If S is a mupltiple of the identy matrix then this multiple is the scaling factor. Otherwise, we have non-uniform scalings.

scale = If[
  Norm[Mean[Diagonal[S]] IdentityMatrix[dim] - S, 
    "Frobenius"] < $MachinePrecision^2,
  Mean[Diagonal[S]],
  $Failed
  ]