Chapter 6 extra note 7

polar decomposition

Example:

Given

$A\in {\mathbb{M}}_{3\times 3}(\mathbb{R})$ as

$$\begin{array}{}\text{(1)}& A=\left[\begin{array}{ccc}0& \sqrt{3}& 0\\ 0& 0& \sqrt{15}\\ 0& 0& 0\end{array}\right],\end{array}$$
we want to calculate the polar decomposition of

$A$, namely, to find an unitary matrix

$U$ such that

$A=U|A|$.

At first, let us find

$|A|$. We calculate

$A^{\ast }A$ and determine

$|A|$. We have

$$\begin{array}{}\text{(2)}& A^{\ast }A=\left[\begin{array}{ccc}0& 0& 0\\ 0& 3& 0\\ 0& 0& 15\end{array}\right],|A|=\left[\begin{array}{ccc}0& 0& 0\\ 0& \sqrt{3}& 0\\ 0& 0& \sqrt{15}\end{array}\right].\end{array}$$
It should be clear that the eigenvector of

$A^{\ast }A$ and

$|A|$ are

${\mathbf{v}}_1=[0,0,1{]}^T$,

${\mathbf{v}}_2=[0,1,0{]}^T$.

Next, we require

$A{\mathbf{v}}_i=U|A|{\mathbf{v}}_i$ for

$i=1,2$.

• $i=1$:
  
  $$\begin{array}{}\text{(3)}& A\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]=\left[\begin{array}{ccc}0& \sqrt{3}& 0\\ 0& 0& \sqrt{15}\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]=U\left[\begin{array}{ccc}0& 0& 0\\ 0& \sqrt{3}& 0\\ 0& 0& \sqrt{15}\end{array}\right]\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right],\end{array}$$
  that gives the equation
  
  $$\begin{array}{}\text{(4)}& \left[\begin{array}{c}\sqrt{3}\\ 0\\ 0\end{array}\right]=U\left[\begin{array}{c}0\\ \sqrt{3}\\ 0\end{array}\right].\end{array}$$
• $i=2$
  
  $$\begin{array}{}\text{(5)}& A\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]=\left[\begin{array}{ccc}0& \sqrt{3}& 0\\ 0& 0& \sqrt{15}\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]=U\left[\begin{array}{ccc}0& 0& 0\\ 0& \sqrt{3}& 0\\ 0& 0& \sqrt{15}\end{array}\right]\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right],\end{array}$$
  that gives the equation
  
  $$\begin{array}{}\text{(6)}& \left[\begin{array}{c}0\\ \sqrt{15}\\ 0\end{array}\right]=U\left[\begin{array}{c}0\\ 0\\ \sqrt{15}\end{array}\right].\end{array}$$

Combining equations (4) and (6) we see that

$U$ should be in the following form:

$$\begin{array}{}\text{(7)}& U=\left[\begin{array}{ccc}?& 1& 0\\ ?& 0& 1\\ ?& 0& 0\end{array}\right].\end{array}$$

One might feels that there exists the third equation by using

${\mathbf{v}}_0=[1,0,0{]}^T$, but, as

${\mathbf{v}}_0\in \text{Ker}(A)$, it has no contribution:

$$\begin{array}{}\text{(8)}& A\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]=\left[\begin{array}{ccc}0& \sqrt{3}& 0\\ 0& 0& \sqrt{15}\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]=U\left[\begin{array}{ccc}0& 0& 0\\ 0& \sqrt{3}& 0\\ 0& 0& \sqrt{15}\end{array}\right]\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right].\end{array}$$
Then we get

$\mathbf{0}=U\mathbf{0}$ which is always true.

Finally, given that

$U$ is unitary, we have two solutions for

$U$ as

$$\begin{array}{}\text{(9)}& U_1=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right],U_1=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -1& 0& 0\end{array}\right].\end{array}$$

So the polar decomposition is not unique, we have

$$\begin{array}{}\text{(10)}& A=U_1|A|=U_2|A|.\end{array}$$