---
title: Ch6-7
tags: Linear algebra
GA: G-77TT93X4N1
---

# Chapter 6 extra note 7

> polar decomposition

**Example:**

Given $A\in\mathbb{M}_{3\times 3}(\mathbb{R})$ as
$$
\tag{1}
A = \begin{bmatrix}
0 & \sqrt{3} & 0\\
0 & 0 & \sqrt{15} \\
0 & 0 & 0
\end{bmatrix},
$$
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 $A^*A$ and determine $|A|$. We have
$$
\tag{2}
A^*A = \begin{bmatrix}
0 & 0 & 0\\
0 & 3 & 0 \\
0 & 0 & 15
\end{bmatrix}, \quad
|A| = \begin{bmatrix}
0 & 0 & 0\\
0 & \sqrt{3} & 0 \\
0 & 0 & \sqrt{15}
\end{bmatrix}.
$$
It should be clear that the eigenvector of $A^*A$ and $|A|$ are ${\bf v}_1 = [0, 0, 1]^T$, ${\bf v}_2=[0, 1, 0]^T$.

Next, we require $A{\bf v}_i=U|A|{\bf v}_i$ for $i=1, 2$.
* $i=1$:
$$
\tag{3}
A \begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix}= \begin{bmatrix}
0 & \sqrt{3} & 0\\
0 & 0 & \sqrt{15} \\
0 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix} = U\begin{bmatrix}
0 & 0 & 0\\
0 & \sqrt{3} & 0 \\
0 & 0 & \sqrt{15}
\end{bmatrix}
\begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix},
$$
that gives the equation
$$
\tag{4}
\begin{bmatrix}
\sqrt{3} \\
0 \\
0
\end{bmatrix}
= U\begin{bmatrix}
0\\
\sqrt{3} \\
0
\end{bmatrix}.
$$
* $i=2$
$$
\tag{5}
A \begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}= \begin{bmatrix}
0 & \sqrt{3} & 0\\
0 & 0 & \sqrt{15} \\
0 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix} = U\begin{bmatrix}
0 & 0 & 0\\
0 & \sqrt{3} & 0 \\
0 & 0 & \sqrt{15}
\end{bmatrix}
\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix},
$$
that gives the equation
$$
\tag{6}
\begin{bmatrix}
0 \\
\sqrt{15} \\
0
\end{bmatrix}
= U\begin{bmatrix}
0\\
0\\
\sqrt{15}
\end{bmatrix}.
$$

Combining equations (4) and (6) we see that $U$ should be in the following form:
$$
\tag{7}
U=\begin{bmatrix}
? & 1 & 0\\
? & 0 & 1 \\
? & 0 & 0
\end{bmatrix}.
$$

> One might feels that there exists the third equation by using ${\bf v}_0=[1,0,0]^T$, but, as ${\bf v}_0\in\text{Ker}(A)$, it has no contribution:
> $$
> \tag{8}
> A \begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}= \begin{bmatrix}
0 & \sqrt{3} & 0\\
0 & 0 & \sqrt{15} \\
0 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix} = U\begin{bmatrix}
0 & 0 & 0\\
0 & \sqrt{3} & 0 \\
0 & 0 & \sqrt{15}
\end{bmatrix}
\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}.
> $$
> Then we get ${\bf 0}=U{\bf 0}$ which is always true.

Finally, given that $U$ is unitary, we have two solutions for $U$ as
$$
\tag{9}
U_1 = \begin{bmatrix}
0 & 1 & 0\\
0 & 0 & 1 \\
1 & 0 & 0
\end{bmatrix}, \quad
U_1 = \begin{bmatrix}
0 & 1 & 0\\
0 & 0 & 1 \\
-1 & 0 & 0
\end{bmatrix}.
$$

So the polar decomposition is not unique, we have
$$
\tag{10}
A = U_1|A| = U_2|A|.
$$