Let $n$ be even and consider the discrete sine transform of type 5 which is the matrix
$$S=\left(\sin(k+1)(l+1)\frac{\pi}{n+\frac12}\right)_{k,l=0}^{n-1}$$
Let us denote by $s_{-,l}$ the $l^{\text{th}}$-column of $S$. It can be considered as an $n$-tuple in $\mathbb{R}^n$.
Q. I am looking for $n$-tuples $v=(v_0,\dotsc,v_{n-1})$ in $\mathbb{R}^n$ satisfying the following conditions:
$v_j=\sin2(j+1)\frac{\pi}{n+\frac12}$ if $j$ is even.
The following are valid concerning inner products: $$\langle v , s_{-,l} \rangle=\begin{cases} 1 & l=0 \\ 0 & l\neq 0~,~ \text{$l$ is even.} \end{cases} $$
P.S. Let us consider only entries of $S$ which are located on the even columns and odd rows and say it $\tilde{S}$. It is a submatrix of $S$. How can we prove that $\tilde{S}$ is invertible? If we could prove this point, the above mentioned problem would be solved.
\left\{\begin{array}…\end{array}\right.idiom is better rendered as\begin{cases}…\end{cases}. Also,\operatornameshould not be used for plain text in math mode; that's what\textis for. For example, when one adds sentence-ending punctuation, one gets $l \operatorname{is even.}$l \operatorname{is even.}vs. $\text{$l$ is even.}$\text{$l$ is even.}(which some people prefer to set asl\text{ is even.}). I have edited accordingly $\endgroup$