## Polar Decomposition Example

Apr 01 2016

This is a series of notes taken during my review of linear algebra, using Axler's excellent textbook Linear Algera Done Right, which will be heavily referenced.

Every complex number can be written as a nonnegative real number $$r$$ times a number $$e^{i \theta}$$ on the unit circle:

$$z = (\frac{z}{\bar{z}}) |z| = (\frac{z}{\bar{z}}) \sqrt{\bar{z}z} = e^{i \theta} r$$

Similarly, every operator $$T$$ can be written as an isometry $$S$$ times a positive semidefinite (positive definite when $$T$$ is invertible) operator:

$$T = S \sqrt{T^*T}$$

This theorem has a nice physical manifestation: in continuum mechanics, any deformation can be separated into a pure rotation effect and pure compression effect (as an imprecise analogy, a Rasengan should to hurled at the opponent with the maximum rotation and velocity in order to deliver the greatest damage).

Example, exercise 7.D.7: Define $$T \in \mathcal{L}(\mathbb{F}^3)$$ by $$T(z_1, z_2, z_3) = (z_3,2z_1,3z_2)$$. Find (explicitly) an isometry $$S \in \mathcal{L}(\mathbb{F}^3)$$ such that $$T = S \sqrt{T^*T}$$.

Solution: with respect to the standard basis,

$$\mathcal{M}(T) = \begin{bmatrix} 0 & 0 & 1\\ 2 & 0 & 0 \\ 0 & 3 & 0 \end{bmatrix}, \mathcal{M}(T^*) = \begin{bmatrix} 0 & 2 & 0\\ 0 & 0 & 3 \\ 1 & 0 & 0 \end{bmatrix}, \mathcal{M}(T^*T) = \begin{bmatrix} 4 & 0 & 0\\ 0 & 9 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \mathcal{M}(\sqrt{T^*T}) = \begin{bmatrix} 2 & 0 & 0\\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

By inspection, $$\mathcal{M}(\sqrt{T^*T})$$ differs from $$\mathcal{M}(T)$$ only in the order of the entries, so if we define $$S$$ by $$S(z_1,z_2,z_3) = (z_3,z_1,z_2)$$, then $$S$$ is certainly an isometry, and we arrive at the beautiful decomposition:

$$\mathcal{M}(S) \mathcal{M}(\sqrt{T^*T})= \mathcal{M}(T) = \begin{bmatrix} 0 & 0 & 1\\ 2 & 0 & 0 \\ 0 & 3 & 0 \end{bmatrix}= \begin{bmatrix} 0 & 0 & 1\\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 2 & 0 & 0\\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

We have decomposed the matrix description of $$T$$ into a pure rotation matrix $$\mathcal{M}(S)$$ and a (positive) "stretching" matrix $$\mathcal{M}(\sqrt{T^*T})$$