Separating Out Parts of a Matrix (Translation, Rotation, Scaling)

Question

Is there any way to separate out parts of matrices?

given a Matrix = Translation * Rotation * Scaling

How can I find out only one part of the matrix?

For instance if I only wanted the rotation element of the matrix?

Answer 1

The Iwasawa decomposition asserts that an invertible matrix $M$ can be written in the form $M = KAN$, where $K$ is an orthogonal matrix, $A$ is a diagonal matrix, and $N$ is an upper-triangular matrix with $1$'s along the diagonal. Carrying this out in practice is nothing more than Gram-Schmidt orthogonalization, which produces an orthogonal basis from an arbitrary basis.

Let $F$ be a field (think of $F = \mathbb{R}$ if that's more familiar). Given $M \in GL_n(F)$ (i.e. an invertible $n\times n$ matrix), take the columns as a basis for $F^n$; write them as $v_1, \dots, v_n$. Gram-Schmidt will produce a new, orthonormal basis $u_1, \dots, u_n$, with the property that $$u_k = \sum_{i=1}^k a_{ik} v_i$$ for $a_{ik} \in F$. If we think about change-of-basis matrices, we recognize that the expression above tells us that the change-of-basis matrix taking $v_1,\dots,v_n$ to $u_1,\dots,u_n$ is upper-triangular, because $u_k$ is formed only from $v_1, \dots, v_k$. Any upper-triangular matrix is the product of a diagonal matrix and an upper-triangular matrix with $1$'s on the diagonal (this is called a unipotent matrix); call this product $AN$.

We can think of the original $M$ as taking us from the standard basis to the $v$ basis. Gram-Schmidt then takes us from the $v$-basis to another orthonormal basis $u$, which is represented by an orthogonal matrix $K$. What this means in practice is that we have the identity $$ K^{-1}ANM = I; $$ equivalently $$ M^{-1} = K^{-1}AN. $$ Of course, the inverse of a diagonal matrix $A$ or a unipotent matrix $N$ will be of the same form, so this gives you your decomposition for $M^{-1}$ ; to get this for $M$, just apply Gram-Schmidt to $M^{-1}$.