Find an Isometry $ \ S \ $ such that $ \ T=S \sqrt{T^* T} \ $ Let $  \ T : \  \mathbb{C}^2 \to \mathbb{C}^2 \ $ be given by $ \ T(z_1,z_2)=\begin{bmatrix} 2 & -1 \\ 4 & -2 \end{bmatrix} \begin{bmatrix} z_1 \\ z_2\end{bmatrix} $
Find an Isometry $ \ S \ $ such that $ \ T=S \sqrt{T^* T} \ $ , where $ \ T^* \ $ is the conjugate transpose of $ \ T \ $
Answer:
Here, 
$ T= \begin{bmatrix} 2 & -1 \\ 4 & -2 \end{bmatrix} \\ T^*=\begin{bmatrix} 2 & 4 \\ -1 & -2 \end{bmatrix} $
Thus, 
$ T^* T=\begin{bmatrix} 2 & 4 \\ -1 & -2 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 4 & -2 \end{bmatrix} \ = \begin{bmatrix} 20 & -10 \\ -10 & 5 \end{bmatrix} $ 
$ T^*T= \begin{bmatrix} 20 & -10 \\ -10 & 5 \end{bmatrix} \ $
Now we have to find $ \ P \ $ such that $ \ P=\sqrt {T^* T} \ \ \  \Rightarrow P^2=T^*T \ \ \  $
For this let us first find the Eigen values and Eigen vectors of $ \ T^*T \ $ as follows : 
The Eigen values of $ T^* T \ $ are $ \ 0,\ 25 \ $ 
The corresponding Eigen vectors are $ \ \vec a_1=\begin{bmatrix} 1 \\ 2 \end{bmatrix} \ $  and $ \vec a_2=\begin{bmatrix} 2 \\ -1 \end{bmatrix} \ $ 
This gives the basis of $ \mathbb{C}^2 \ $ . 
So let $ \ P: \mathbb{C}^2 \to \mathbb{C}^2 \ $ be given by  $ \ P(a_1)=\sqrt 0 \vec a_1 , \\ P(a_2)=\sqrt{25} \vec a_2 \ $ , 
Then $ \ P \ $ will be  the unique positive root of $ \ T^* T \ $ 
From here we can find the matrix $ \ \ P \ $ i.e.,  $ \sqrt{T^* T} \ $
But my question is how to find the Isometry $ \ S \ $  so that $ \ T=S \sqrt{T^* T  } \ $
Help me out only to find $ \ S \ $ 

You want $T\vec{a_k}=S\sqrt{T^*T}\,\vec{a_k}$ for $k=1,2$. This does not uniquely determine $S$ because $\sqrt{T^*T}\,\vec{a_1}=0$ and $T\vec{a_1}=0$. The condition $S\sqrt{T^*T}\,\vec{a_2}=T\vec{a_2}$ determines $S$ on $\vec{a_2}$. Setting $S\,\vec{a_1}=e^{i\theta}\vec{b_1}$ defines an isometry $S$ if $\vec{b_1}$ is orthogonal to $T\vec{a_2}$; $S$ is not unique because $\theta$ can be any real number, which is to be expected in every case where $T$ is not invertible.
To be specific, $\sqrt{T^*T}\, \vec{a_2} = \frac{1}{5}\vec{a_2}$ where $\vec{a_2}$ is a unit vector
$$
          \vec{a_2} = \frac{1}{\sqrt{5}}\left[\begin{array}{c}2 \\ -1\end{array}\right].
$$
The vector $\vec{a_1}$ is orthogonal to $\vec{a_2}$, which gives
$$
     \vec{a_1} = \frac{1}{\sqrt{5}}\left[\begin{array}{c}1 \\ 2\end{array}\right]
$$
It's not hard to verify that
$$
           T\vec{a_1}=\vec{0},\;\; T^*T\vec{a_2}=\frac{1}{\sqrt{5}}\left[\begin{array}{c}50 \\ -25\end{array}\right] = 25\vec{a_2}
$$
The matrix $S$ must be defined so that $S\sqrt{T^*T}\,\vec{a_2}=T\vec{a_2}$ or $S(5\vec{a_2})=T\vec{a_2}$, and so that $S\vec{a_1}$ is a unit vector orthogonal to $T\vec{a_2}$. In summary,
$$
         S\vec{a_2} = \frac{1}{5}T\vec{a_2} \implies S\left[\begin{array}{c}2 \\ -1\end{array}\right] = \frac{1}{5}\left[\begin{array}{c}5\\10\end{array}\right] = \left[\begin{array}{c}1 \\ 2 \end{array}\right] \\
         S\vec{a_1}\perp T\vec{a_2} \implies S\left[\begin{array}{c}1\\2\end{array}\right] = e^{i\theta}\left[\begin{array}{c}2 \\ -1\end{array}\right] \\
      S\left[\begin{array}{cc}
             1 & 2  \\ 2 & -1
             \end{array}\right]
      =\left[\begin{array}{cc}
             2e^{i\theta} & 1 \\
             -e^{i\theta} & 2
             \end{array}\right] = \left[\begin{array}{cc} 2 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc} e^{i\theta} & 0 \\ 0 & 1\end{array}\right]
   \\
  S =\frac{1}{5}\left[\begin{array}{cc} 2 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc} e^{i\theta} & 0 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc} 1 & 2 \\ 2 & -1\end{array}\right]
$$
