The mystery of normal co-ordinates

I must begin with a confession, namely that I have never understood the point of normal co-ordinates mostly because we can only arrive at them after we have completely solved the problem and found all the solutions. The point of them to decouple our differential equations, but they decouple our differential equation after we have found all the solutions. Nevertheless, they are quite interesting and understanding them will give you an easier time understanding basis vectors and operators in quantum mechanics.
Now if we have two vectors that span \(\mathbf{R}^2\) then we can write any vector \(\vec{x}\) in \(\mathbf{R}^2\) as a linear combination of them i.e \(\vec{x} = \sum_{i=1}^N a_i\mathbf{e}_i\) for constants \(a_i\) and basis vectors \(\mathbf{e_i}\).  But supposing we needed to describe this vector in another set of co-ordinates because the vector looked simpler, then we need a transformation operation to move from on set of co-ordinate basis to another. This transformation operation is usually called a standard matrix in linear algebra. So quick example before we move forward.
Let's begin with the standard Cartesian co-ordinates \(\hat{x}\) = (1, 0) and \(\hat{y}\) = (0, 1). Supposing we needed to rotate this co-ordinates vectors  \(45^{\circ}\) counterclockwise. Then that means that \(\hat{x}\) moves to \(\hat{x}^{'}\) = (1,1) and \(\hat{y}\) moves to \(\hat{y}^{'}\) = (-1,1). Thus, our standard matrix does this: \(T(\hat{x})\) = (1, 1) and \(T(\hat{y})\) = (-1,1).We then make the standard matrix, T, from how it operates on our original basis vectors in the following manner:
\begin{equation}
  \begin{pmatrix}
    T(\hat{x}) & T(\hat{y}) \\
     1              &     -1\\
     1              &       1
  \end{pmatrix}
\end{equation}

So, now our matrix equation to move from one basis co-ordinate to another is the following
\begin{align}
 \begin{pmatrix}
  \hat{x}\\ \hat{y}
 \end{pmatrix}
 =   \begin{pmatrix}
     1              &     -1\\
     1              &       1
  \end{pmatrix}
 \begin{pmatrix}
  \hat{x}^{'}\\
 \hat{y}^{'}
\end{pmatrix}
\hspace{20mm} Eq.1
\end{align}
Now  going back to our problem we found our solution in matrix notation to be the following:

\begin{align}
\begin{pmatrix}x_a(t) \\x_b(t)\end{pmatrix}= \begin{pmatrix}1 & 1 \\ 1 & - 1\end{pmatrix}\begin{pmatrix} \cos \omega_1 t \\ \cos \omega_2 t \end{pmatrix}
\hspace{20mm} Eq.2
\end{align}
Notice the similarity, we can think of the solution we found as living naturally in another co-ordinate space that has been rotated \(45^{\circ}\) clockwise. In order to reconstruct our solution the matrix in eq.2 will have to be an identity matrix on the right hand side of the equation(because then the first solution is simply in terms of \(\cos \omega_1 t\) and the second solutions is simply in terms of \(\cos \omega_2 t\)). And how in the world do we do this ? We simply multiply both sides by from the left by the inverse of the matrix. The inverse matrix acting on \(x_a(t)\) and \(x_b(t\) gives us how to construct our normal co-ordinates. Our inverse matrix is $$ \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2}\end{pmatrix}$$  If we multiply this by the column vector [\(x_a (t) , x_b (t)\)] we arrive at our normal co-ordinates. This then is the systematic way of getting at normal co-ordinates.  I have eschewed from discussing similar matrices because that is always done and rarely is any connection made with this problem. Similar matrices will have to wait when I write a discussion on the matrix formulation of quantum mechanics.