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 \ $ 

\begin{align}
T &= S\sqrt{T^\ast T} \\
\sqrt{T^\ast T}S^\ast &= T^\ast \\
\end{align}
Now you can use Gaussian elimination to find $S^\ast$ by reducing:
$$[\sqrt{T^\ast T}\ |\ T^\ast].$$
If $\sqrt{T^\ast T}$ is invertible then it will reduce to
$$[I\ |\ (\sqrt{T^\ast T})^{-1}T^*].$$
Otherwise you will have multiple solutions for $S^\ast$ and you will have to invoke the constraint that $S^\ast$ has orthonormal columns or has orthonormal rows (whichever makes the algebra easier).
