Any linear transformation is a sum of a rotation and streching. So Is it possible to decompose any $A$ real matrix into such $R$ and $S$ components?

Question

The Question

Let $A \in \mathbb{R}^{n \times n}$ be the matrix of a linear transformation.

I have learned that any linear transformation is either a rotation, a streching, or a mirroring (with the latter one being a special type of stretching with a multiplier of $-1$).

So is it possible to algebraically decompose $A$ into a purely rotational part $R$ and a purely stretching part $S$?

Definitions

A square matrix $R$ is a rotation matrix exactly if it is orthogonal $(R^T = R^{-1})$ and $\det(R)=1$.

An $n$ by $n$ square matrix $S$ is a stretch matrix exactly if it has $n$ real eigenvalues (counted with algebraic multiplicity). As such, $S$ has a real eigendecomposition $S = U^T \Lambda U$, where $U$ is orthogonal and $\Lambda = \text{diag}(\lambda_1,\dots,\lambda_n)$ with $\lambda_1,\dots,\lambda_n \in \mathbb{R}$

Algebraic phrasing of the Question

Given $A \in \mathbb{R}^{n \times n}$, does there exist $R,S \in \mathbb{R}^{n \times n}$ such that

1. $A = RS$
2. $R^T = R^{-1}$
3. $\det(R)=1$
4. $S = U^T \Lambda U$
5. $U^T = U^{-1}$
6. $\Lambda$ is a diagonal matrix with real entries

Furthermore, is this decomposition unique?

Answer 1

The closest decomposition I see to your issue is the so-called $QR$ decomposition.

It must be said at first that we have to adopt a multiplicative convention for the composition of two operations : we have to write is $RS$ instead of $R+S$ .

The $QR$ decomposition of $A$ is the fact to write $A= QR$ with unique (up to a simultaneous change of sign in $Q$ and $R$) factors $Q$ (your $R$) and $R$ (your $S$) where:

• $Q$ is an orthogonal matrix (not necessarily a rotation matrix),

• $R$ is an upper triangular matrix (which is what you call a "stretch matrix").

Answer 2

$$A = \begin{pmatrix} 0 & -1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 2 & 0\\ 0 & 0 & 0 & 2 \end{pmatrix}$$ is a linear transformation that is neither a rotation, nor a stretching (according to your definition which is strange), nor either a mirroring.