$$\begin{aligned}
v_\beta(x) :=&& (-1)^{\theta-\nu+1}\left(\frac{\pi}{L}\right)^{-2(\theta-\nu)-1} \frac{(2\theta+2\nu-2)!}{(4\theta-1)!}\psi^{2(\theta-\nu)+1}_{2\nu-1}(x) \nonumber \\
&&= \cos (\tilde{x}) \sum_{n=\nu}^{\theta}  a_{2\nu-1,2n-1} \sin^{2n-1} (\tilde{x}),
\label{eq:expansion2}\end{aligned}$$ where $$a_{2\nu-1,2n-1} = \prod_{k=n}^{\theta-1}\frac{k(2k+1)-(2\nu-1)(\nu-1)}{2(k - \theta)(k+\theta)},$$ with $a_{2\nu-1,2\theta-1}:= 1$.

Since $\psi^{N-2\nu}_{2\nu}(x)$ ($\psi^{N-2\nu+1}_{2\nu-1}(x)$) and $u_{2\nu}(x)$ ($v_{2\nu-1}(x)$) are the same up to an overall constant, the only thing remaining to prove is to show that there is a linear transformation between the spaces of $\{ u_\beta(x), v_\beta(x)\}$ and $\{ \puni_n (x) \}^{N}_{n=0}$ such that it is invertible. From Eqs. (\[eq:expansion1\]) and (\[eq:expansion2\]), one finds the following relations: $$\begin{aligned}
\begin{pmatrix}
u_0(x)                \\
u_2(x)                \\
 \vdots            \\
u_N(x)                
\end{pmatrix}
&=&
\begin{pmatrix}
a_{0, 0} &  a_{0,2}&  \cdots & a_{0,N}  \\
        &  a_{2,2}&   \cdots      & a_{2,N}  \\
        &        &   \ddots&   \vdots   \\
        &        &         & a_{N,N}  
\end{pmatrix}
\begin{pmatrix}
  1            \\
 \sin^2(\tilde{x})  \\
 \vdots            \\
 \sin^N(\tilde{x})  
\end{pmatrix},
\label{vectorspaceu}\\
\begin{pmatrix}
v_1(x)                \\
v_3(x)                \\
 \vdots            \\
v_{N-1}(x)                
\end{pmatrix}
&=&
\begin{pmatrix}
a_{1, 1} &  a_{1,3}&  \cdots & a_{1,N-1}  \\
        &  a_{3,3}&    \cdots     & a_{3,N-1}  \\
        &        &   \ddots&   \vdots   \\
        &        &         & a_{N-1,N-1}  
\end{pmatrix}
\begin{pmatrix}
 \cos(\tilde{x}) \sin(\tilde{x})                \\
 \cos(\tilde{x}) \sin^3(\tilde{x})  \\
 \vdots            \\
 \cos(\tilde{x}) \sin^{N-1}(\tilde{x})  
\end{pmatrix},\end{aligned}$$ where matrix elements which are zero are left empty. These upper triangular matrices are non-singular and invertible since the diagonal element $a_{\beta, \beta}$ is nonzero for all 