General matrix solution for a variant orthogonality condition An $n\times n$ complex matrix $X$ satisfies a constraint looking like a rank deficient version of the orthogonality condition
$$ X^TX = \text{diag}\left(1,\dots,1,0\right), $$
where $X^T$ is the transpose of $X$. 
How can I describe the most general form of $X$? Can it be factorized into some specific parts, or expressed as some constrained SVD?

I use the @ user161825 's notation.
Proposition 1. $X_n=0$ and $\{X_1,\cdots,X_{n-1}\}$ is a free system.
Proof. Clearly $rank(X)=n-1$ and it remains to show that $X_n=0$. Note that $\{X_1,\cdots,X_n\}$ is not free.
Case 1. $X_n=\sum_{i<n}a_iX_i$; then, for every $j<n$, $0=X_n^TX_j=\sum_{i<n}a_iX_i^TX_j=a_j$; we deduce that $X_n=0$.
Case 2. For example, $X_1=\sum_{1<i<n}a_iX_i$ and $1=X_1^TX_1=\sum_{1<i<n}a_iX_i^TX_1=0$, that is a contradiction. $\square$
The $n(n-1)$ complex parameters $X_1,\cdots,X_{n-1}$ are linked by the algebraic relations $X_i^TX_j=\delta_{ij}$ for $1\leq i\leq j\leq n-1$, that is, $(n-1)(n-2)/2+(n-1)=(n^2-n)/2$ relations. The previous relations are algebraically independent (cf. below); then our considered set $Z$ of complex matrices $X$ is a complex algebraic set of dimension $n(n-1)-(n^2-n)/2=(n^2-n)/2$.
Proposition 2. The dimension of the complex algebraic set $Z$ is $(n^2-n)/2$.
Proof. Let $Y=[Y_1,\cdots,Y_{n-1}]$. Since $Y^TY=I_{n-1}$, the tangent space of $Z$ is the set of $H=[H_1,\cdots,H_{n-1}]$ s.t. $H^TY+Y^TH=0$, that is s.t. $Y^TH$ is skew-symmetric ($Y^TH\in SK_{n-1}$). Note that $SK_{n-1}$ is a complex vector space of dimension $(n-1)(n-2)/2$. We may assume that $Y=[I_{n-1},0_n]^T$ and $Y^TH$ is formed by the first $n-1$ rows of $H$ (there is no conditions about the last row of $H$). Finally, the required dimension is $(n-1)(n-2)/2+(n-1)=(n^2-n)/2$ and we are done.
