Polar Decomposition

Proof. $\Leftarrow$: If $M=B^{\ast }B$, then $x^{\ast }\mathit{Mx}=\left(x^{\ast }{B}^{\ast }\right)\left(\mathit{Bx}\right)=\left|\right|\mathit{Bx}|{|}^2\ge 0$ so $M$ is positive definite.
$\Rightarrow$: Suppose $M$ is positive semidefinite, then $M$ must be a Hermitian matrix and can be expressed in the form $U^{-1}\mathit{DU}$ where $U$ is unitary and $D$ is diagonal (it is a theorem but we omit it for now). Moreover, $M$ is positive semidefinite, the eigenvalues are nonnegative, we can therefore construct $D^{\frac{1}{2}}$ by square rooting each elements of $D$, so $D^{\frac{1}{2}}$ is also a diagonal matrix and ${\left(D^{\frac{1}{2}}\right)}^2=D$. Then

$$M=U^{-1}\mathit{DU}=U^{\ast }{D}^{\frac{1}{2}}{D}^{\frac{1}{2}}U=\left(U^{\ast }{D}^{\frac{1}{2}}\right)\left(D^{\frac{1}{2}}U\right)=B^{\ast }B$$

Proof. $\Rightarrow$: Let $A=[a_1,a_2,\cdots ,a_n]$ be an $n\times n$ unitary matrix, by definition of unitary matrix,

$$I_n=A^{\ast }A=\left(\begin{array}{c}\hfill a_1^{\ast }\hfill \\ \hfill a_2^{\ast }\hfill \\ \hfill ⋮<!--\; FUNCTION\; APPLICATION\; -->\hfill \\ \hfill a_n^{\ast }\hfill \end{array}\right)\left(\begin{array}{cccc}\hfill a_1\hfill & \hfill a_2\hfill & \hfill \cdots \hfill & \hfill a_n\hfill \end{array}\right)$$

then $⟨a_i,a_j⟩={\delta }_{\mathit{ij}}$, the Kronecker delta, for all $i,j$. Hence the column vectors of $A$ are orthogonal.
$\Leftarrow$: Since the column vectors of $A$ are orthogonal, then

$${\delta }_{i,j}=⟨a_i,a_j⟩\text{and}\left(\begin{array}{c}\hfill a_1^{\ast }\hfill \\ \hfill a_2^{\ast }\hfill \\ \hfill ⋮<!--\; FUNCTION\; APPLICATION\; -->\hfill \\ \hfill a_n^{\ast }\hfill \end{array}\right)\left(\begin{array}{cccc}\hfill a_1\hfill & \hfill a_2\hfill & \hfill \cdots \hfill & \hfill a_n\hfill \end{array}\right)=A^{\ast }A\Rightarrow I_n=A^{\ast }A$$

Proof. Case 1: If $A^{\ast }A=\mathit{diag}[{\lambda }_1,{\lambda }_2,\cdots ,{\lambda }_n]$, where

$$\mathit{diag}[{\lambda }_1,{\lambda }_2,\cdots ,{\lambda }_n]=\left(\begin{array}{cccc}\hfill {\lambda }_1\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill {\lambda }_2\hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \ddots \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill {\lambda }_n\hfill \end{array}\right),\text{a diagonal matrix.}$$

Note that $A^{\ast }A$ is a Hermitian matrix since ${\left(A^{\ast }A\right)}^{\ast }=A^{\ast }{\left(A^{\ast }\right)}^{\ast }=A^{\ast }A$.
Since $A^{\ast }A$ is Hermitian matrix, $A^{\ast }A$ is positive semidefinite by lemma 2.5, then ${\lambda }_j\ge 0$ for all $1\le j\le n$. Suppose that ${\lambda }_1\ge {\lambda }_2\cdots \ge {\lambda }_k>0$, $k\le n$, and ${\lambda }_j=0$ for all $k+1\le j\le n$.
Without loss of generality, let

$$\begin{array}{llll}\hfill & A=\left(\begin{array}{cccc}\hfill a_1\hfill & \hfill a_2\hfill & \hfill \cdots \hfill & \hfill a_n\hfill \end{array}\right)\Rightarrow A^{\ast }=\left(\begin{array}{c}\hfill {\overline{a_1}}^T\hfill \\ \hfill {\overline{a_2}}^T\hfill \\ \hfill ⋮<!--\; FUNCTION\; APPLICATION\; -->\hfill \\ \hfill {\overline{a_n}}^T\hfill \end{array}\right)& \hfill & \\ \hfill \Rightarrow & \left(\begin{array}{ccccccc}\hfill {\lambda }_1\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill {\lambda }_2\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \ddots \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill {\lambda }_k\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill 0\hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \ddots \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill 0\hfill \end{array}\right)=A^{\ast }A=\left(\begin{array}{ccc}\hfill ⟨a_1,\overline{a_1}⟩\hfill & \hfill \cdots \hfill & \hfill ⟨a_n,\overline{a_1}⟩\hfill \\ \hfill ⋮<!--\; FUNCTION\; APPLICATION\; -->\hfill & \hfill \ddots \hfill & \hfill ⋮<!--\; FUNCTION\; APPLICATION\; -->\hfill \\ \hfill ⟨a_1,\overline{a_n}⟩\hfill & \hfill \cdots \hfill & \hfill ⟨a_n,\overline{a_n}⟩\hfill \end{array}\right)& \hfill & \\ \hfill \Rightarrow & \{\begin{array}{ccc}\hfill & \left|\right|a_{i,j}\left|\right|=\sqrt{{\lambda }_j},\text{for all}1\le j\le k\hfill & \hfill \\ \hfill & a_j=0,\text{for all}k+1\le j\le n\hfill \\ \hfill & ⟨a_i,a_j⟩=0\text{for all}i\ne j\hfill \end{array}\Rightarrow \{\frac{a_1}{\sqrt{{\lambda }_1}},\frac{a_2}{\sqrt{{\lambda }_2}},\cdots ,\frac{a_k}{\sqrt{{\lambda }_k}}\}\text{is orthonormal.}& \hfill & \end{array}$$

Extend it to orthonormal basis in ${ℂ}^n$, that is, $\{\frac{a_1}{\sqrt{{\lambda }_1}},\frac{a_2}{\sqrt{{\lambda }_2}},\cdots ,\frac{a_k}{\sqrt{{\lambda }_k}},u_{k+1},\cdots ,u_n\}$.
Let $U=[\frac{a_1}{\sqrt{{\lambda }_1}},\frac{a_2}{\sqrt{{\lambda }_2}},\cdots ,\frac{a_k}{\sqrt{{\lambda }_k}},u_{k+1},\cdots ,u_n]$, by lemma 2.6, $U$ is unitary and

$$\begin{array}{llll}\hfill U\left|A\right|& =\left(\begin{array}{ccccccc}\hfill \frac{a_1}{\sqrt{{\lambda }_1}}\hfill & \hfill \frac{a_2}{\sqrt{{\lambda }_2}}\hfill & \hfill \cdots \hfill & \hfill \frac{a_k}{\sqrt{{\lambda }_k}}\hfill & \hfill u_{k+1}\hfill & \hfill \cdots \hfill & \hfill u_n\hfill \end{array}\right)\left(\begin{array}{ccccccc}\hfill \sqrt{{\lambda }_1}\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \sqrt{{\lambda }_2}\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \ddots \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \sqrt{{\lambda }_k}\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill 0\hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \ddots \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill 0\hfill \end{array}\right)& \hfill & \\ \hfill & =\left(\begin{array}{ccccccc}\hfill a_1\hfill & \hfill a_2\hfill & \hfill \cdots \hfill & \hfill a_k\hfill & \hfill 0\hfill & \hfill \cdots \hfill & \hfill 0\hfill \end{array}\right)=A& \hfill & \end{array}$$

Case 2: For any complex matrix $A$, $A^{\ast }A$ is a Hermitian matrix, then

$$A^{\ast }A=V^{\ast }\left(\begin{array}{cccc}\hfill {\lambda }_1\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill {\lambda }_2\hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \ddots \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill {\lambda }_n\hfill \end{array}\right)V\text{where }V\text{ is unitary and}{\lambda }_j\ge 0\text{for all}1\le j\le n$$

So

$$\begin{array}{lll}\hfill \left(\begin{array}{cccc}\hfill {\lambda }_1\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill {\lambda }_2\hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \ddots \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill {\lambda }_n\hfill \end{array}\right)=V{A}^{\ast }A{V}^{\ast }=\left(V{A}^{\ast }{V}^{\ast }\right)\left(VA{V}^{\ast }\right)={\left(A^{\prime }\right)}^{\ast }\left(A^{\prime }\right),\text{where}{A}^{\prime }=VA{V}^{\ast }& & \hfill \end{array}$$

Since

$$\left|A\right|=V^{\ast }\left(\begin{array}{cccc}\hfill \sqrt{{\lambda }_1}\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \sqrt{{\lambda }_2}\hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \ddots \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \sqrt{{\lambda }_n}\hfill \end{array}\right)V\Rightarrow \left|A^{\prime }\right|=\left(\begin{array}{cccc}\hfill \sqrt{{\lambda }_1}\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \sqrt{{\lambda }_2}\hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \ddots \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \sqrt{{\lambda }_n}\hfill \end{array}\right)=V\left|A\right|V^{\ast }$$

By Case 1, we have $A^{\prime }=U\left|A^{\prime }\right|$, where $U$ is unitary, hence

$$VA{V}^{\ast }=\mathit{UV}\left|A\right|V^{\ast }\Rightarrow A=V^{\ast }\mathit{UV}\left|A\right|$$

Note that if $A,B$ are unitary, then $\mathit{AB}$ is unitary since ${\left(\mathit{AB}\right)}^{\ast }\mathit{AB}=B^{\ast }{A}^{\ast }\mathit{AB}=I$.