Polar decomposition of a general matrix
So first let $A = \begin{pmatrix}a&-b\\b&a\end{pmatrix}=QS$.
Then, we have $A^{T}A=\begin{pmatrix}a&b\\-b&a\end{pmatrix}\begin{pmatrix}a&-b\\b&a\end{pmatrix}=\begin{pmatrix}a^2+b^2&0\\0&a^2+b^2\end{pmatrix}$.
The Eigenvalue(s) of this matrix is $a^2+b^2$, sometimes they're different but not too hard to handle.
The Eigenvectors are $\begin{pmatrix}1\\0 \end{pmatrix}$ and $\begin{pmatrix}0\\1 \end{pmatrix}$.
So then we can write $A^TA=\begin{pmatrix}1&0\\0&1\end{pmatrix}\begin{pmatrix}a^2+b^2&0\\0&a^2+b^2\end{pmatrix}\begin{pmatrix}1&0\\0&1\end{pmatrix}^{-1}$.
The first factor is just the Eigenvectors, the second is a diagonal matrix consisting of each of the corresponding eigenvalues, and the third is the inverse of the first.
We now define a second matrix $S=\begin{pmatrix}1&0\\0&1\end{pmatrix}\begin{pmatrix}\sqrt{a^2+b^2}&0\\0&\sqrt{a^2+b^2}\end{pmatrix}\begin{pmatrix}1&0\\0&1\end{pmatrix}^{-1}$.
The only difference here is that $S$ is constructed using the singular values, which are basically the square roots of the eigenvalues.
We have that $S=\begin{pmatrix}\sqrt{a^2+b^2}&0\\0&\sqrt{a^2+b^2}\end{pmatrix}$.
We also have that $Q=AS^{-1}= \begin{pmatrix}a&-b\\b&a\end{pmatrix}\begin{pmatrix}\sqrt{a^2+b^2}&0\\0&\sqrt{a^2+b^2}\end{pmatrix}^{-1}=\begin{pmatrix}\frac{a}{\sqrt{a^2+b^2}}&-\frac{b}{\sqrt{a^2+b^2}}\\\frac{b}{\sqrt{a^2+b^2}}&\frac{a}{\sqrt{a^2+b^2}}\end{pmatrix}$.
So finally, we have the polar decomposition: $\begin{pmatrix}a&-b\\b&a\end{pmatrix}=\begin{pmatrix}\frac{a}{\sqrt{a^2+b^2}}&-\frac{b}{\sqrt{a^2+b^2}}\\\frac{b}{\sqrt{a^2+b^2}}&\frac{a}{\sqrt{a^2+b^2}}\end{pmatrix}\begin{pmatrix}\sqrt{a^2+b^2}&0\\0&\sqrt{a^2+b^2}\end{pmatrix}$
Presumably, $A$ is invertible (otherwise, it would be the zero matrix). I'll also assume that $A$ has real entries (if not, change all occurrences of $A^T$ to $A^*$). I'm not going to discuss the proof of the polar decomposition, but consulting the proof will tell you that the matrices I give are the correct ones, and that they have the desired properties (you can check this without too much work).
We want to find the factorization $A=UP,$ where $U$ is orthogonal (unitary) and $P$ is positive semi-definite and symmetric (Hermitian). Compute $(A^TA)^{1/2},$ and set this to be $P$. Then, simply define $U$ by $U=AP^{-1}.$ Let's talk about finding the square root. Since $A$ is invertible, $A^TA$ is positive definite, so we can define the square root (all of the eigenvalues are positive; note that only positive semi-definiteness is needed). So, we can diagonalize $A^TA,$ getting the eigendecomposition $A^TA=S\Lambda S^{-1}.$ Now, the square root is defined as $(A^TA)^{1/2}=S\Lambda^{1/2}S^{-1}$, where $\Lambda^{1/2}=\text{diag}\left(\sqrt{\lambda_j}\right).$
So, here are the steps:
- Get an eigendecomposition of $A^TA$.
- Find the square root of $A^TA$, and define this to be $P$.
- Set $U=AP^{-1}.$
This gives us our decomposition $A=UP.$ We can do this for a matrix of your form, and I encourage you to follow these steps and see what you get. We don't need "numbers," as it's simple enough for a $2\times 2$.