Question about decomposition of matrices in SL(2,R)

[mnb96]

Hello,

we are given a 2×2 matrix S such that det(S)=1.
I would like to find a 2x2 invertible matrix A such that: A S A^{-1} = R, where R is an orthogonal matrix.

Note that the problem can be alternatively reformulated as: Is it possible to decompose a matrix S∈SL(2,ℝ) in the following way: S=A^{-1}R Awhere R is orthogonal and A is invertible?

Is this a well-known problem? To be honest, I don't have many ideas on how to tackle this problem, so even a suggestion that could get me on the right track would be very welcome.

 

[micromass]

Every orthogonal matrix can be diagonalized. Not every matrix in $SL(2,\mathbb{R})$ can be diagonalized. So what you ask is impossible.

 

[fresh_42]

You may count the numbers of degrees of freedom you have in either matrix.

 

[mnb96]

Why did you mention the diagonalizability of a matrix?
I am not asking if S is similar to a diagonal matrix. I am rather asking if S is similar to an orthogonal matrix.

 

[fresh_42]

$R$ orthogonal ⇒ $R$ diagonalizable ⇒ (if $R$ similar to $S$) $S$ diagonalizable
$S = \begin{pmatrix} 1 && 1 \\ 0 && 1 \end{pmatrix} \in SL(2,ℝ)$ not diagonalizable, contradiction
⇒ $S$ not similar to $R$

 

[mnb96]

$R$ orthogonal ⇒ $R$ diagonalizable

I thought orthogonal matrices were not diagonalizable over ℝ in general.Btw, I actually found a counterexample of your (micromass' and fresh's) above statements: consider the matrix <br /> S = \begin{pmatrix}<br /> 1 &amp; 2\\<br /> -1 &amp; -1<br /> \end{pmatrix}<br />.
We can easily see that det(S)=1, and we can still decompose that matrix into: <br /> S = \underbrace{\begin{pmatrix}<br /> -1 &amp; -1\\<br /> 0 &amp; 1<br /> \end{pmatrix} }_{A^{-1}}\;<br /> <br /> \underbrace{\begin{pmatrix}<br /> 0 &amp; -1\\<br /> 1 &amp; 0<br /> \end{pmatrix}}_R\;<br /> <br /> \underbrace{\begin{pmatrix}<br /> -1 &amp; -1\\<br /> 0 &amp; 1<br /> \end{pmatrix}}_A<br />

where A is invertible and R is orthogonal, as required. This proves that there exist matrices in SL(2,ℝ) that are similar to orthogonal matrices. Note also that S is not diagonalizable over ℝ.

 

[micromass]

I thought orthogonal matrices were not diagonalizable over ℝ in general.

They are over $\mathbb{C}$.

Btw, I actually found a counterexample of your (micromass' and fresh's) above statements: consider the matrix <br /> S = \begin{pmatrix}<br /> 1 &amp; 2\\<br /> -1 &amp; -1<br /> \end{pmatrix}<br />.
We can easily see that det(S)=1, and we can still decompose that matrix into: <br /> S = \underbrace{\begin{pmatrix}<br /> -1 &amp; -1\\<br /> 0 &amp; 1<br /> \end{pmatrix} }_{A^{-1}}\;<br /> <br /> \underbrace{\begin{pmatrix}<br /> 0 &amp; -1\\<br /> 1 &amp; 0<br /> \end{pmatrix}}_R\;<br /> <br /> \underbrace{\begin{pmatrix}<br /> -1 &amp; -1\\<br /> 0 &amp; 1<br /> \end{pmatrix}}_A<br />

where A is invertible and R is orthogonal, as required. This proves that there exist matrices in SL(2,ℝ) that are similar to orthogonal matrices. Note also that S is not diagonalizable over ℝ.

$S$ is diagnalizable over $\mathbb{C}$. And of course there exists matrices in $SL(2,\mathbb{R})$ similar to orthogonal matrices. The identity matrix would be an example of this. The point is that not all matrices in $SL(2,\mathbb{R})$ are similar to orthogonal matrices.

 

[micromass]

Furthermore, what you ask is possible exactly for those matrices in $SL(2,\mathbb{R})$ whose eigenvalues have $|\lambda|=1$ (and could be complex).

 

[mnb96]

Thanks micromass,

I think I understand your remarks, but I still have a doubt.

Given a matrix \mathrm{S} \in SL(2,\mathbb{R}), and assuming there exists a 2×2 real matrix matrix A such that A^{-1}SA = R, we can deduce that S must be diagonalizable (over ℂ), since every rotation matrix R is. That means that we can write: (AC)^{-1}\,S\,(AC) = \Lambda where we used the diagonalization R=C\,\Lambda\,C^{-1} (note that both C and Λ are in general complex).Now, starting from the sole knowledge of S we could perform an eigendecomposition of S and we would obtain S=Q^{-1}\Lambda Q, where Q=AC. But then, how do we extract the real matrix A from the complex matrix Q?

For instance, in the example I gave above where <br /> S = \begin{pmatrix}<br /> 1 &amp; 2\\<br /> -1 &amp; -1<br /> \end{pmatrix}<br /> the eigendecomposition of S would give S=Q \Lambda Q^{-1} where: Q=<br /> \frac{\sqrt{6}}{6}\begin{pmatrix}<br /> 2 &amp; 2\\<br /> -1+i &amp; -1-i<br /> \end{pmatrix}<br />
\Lambda=<br /> \begin{pmatrix}<br /> i &amp; 0\\<br /> 0 &amp; -i<br /> \end{pmatrix}<br />

How do we extract A=<br /> \begin{pmatrix}<br /> -1 &amp; -1\\<br /> 0 &amp; 1<br /> \end{pmatrix}<br /> from the complex matrix Q ?

 

[mnb96]

Ok, I think this question has been almost answered.

Summarizing, given a matrix S\in SL(2,\mathbb{R}) we ask if there exist one invertible matrix A\in GL(2,\mathbb{R}) and a rotation matrix R\in SO(2,\mathbb{R}) such that A^{-1}SA=R.

Since R is diagonalizable over ℂ (i.e. R=C\Lambda C^{-1}), we have that:

S=(AC)\Lambda (C^{-1}A^{-1}) = Q\Lambda Q^{-1} \qquad\qquad\qquad \mathrm{(1)}
From the above equation we can deduce that S must be diagonalizable and must have eigenvalues having modulo 1. These are necessary conditions. They would be also sufficient conditions if we add the requirement that the eigenvectors of S must have the same real part.

Since Q=AC we can find a matrix A=QC^{-1}, where the columns of C contain the eigenvectors of R, for instance C=\begin{pmatrix}<br /> 1 &amp; 1\\<br /> i &amp; -i<br /> \end{pmatrix}<br />.
It can be verified that when the eigenvectors of S have the same real part, then the matrix A is real.

However, the matrix A is not unique. For example, we can easily see that by multiplying A with an invertible matrix M that commutes with S (i.e. MS=SM) Equation (1) would still hold.