Finding a matrix using constraints imposed on it I was implementing a method from a white paper, but got stuck on finding a certain matrix $\mathbf{D}$. The paper does not explicitly say how to get that Matrix, but only imposes constraints on it. These constraints are:

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*The matrix $\mathbf{D} ∈ \mathbb{R}^{(L−1)×L}$ contains an orthonormal basis for the nullspace of $\mathbf{y}∈ \mathbb{R}^{L}$ in its rows such that $\mathbf{Dy}$ = $\mathbf{0}$ and $\mathbf{DD}^H = \mathbf{I}$.

*$\mathbf{B = D}^H\mathbf{D = I} − \frac{1}{q_N}\mathbf{y y}^H $, where $\mathbf{B}∈ \mathbb{R}^{L×L}$ and positive semi-definite and $q_N=\frac{L}{4\pi}= ||\mathbf{y}||^2_2$.

Both $\mathbf{y}$ and $\mathbf{B}$ are known, but I could not find a way to get $\mathbf{D}$ with these constraints. For the first I've looked at many problems in the form of $\mathbf{Ax=0}$, but it is always the matrix which is known. For the second constraint I tried to decompose the matrix $\mathbf{B}$ via Cholesky and singular value decomposition, but that only yielded square matrices of dimension $\mathbb{R}^{L×L} \neq \mathbb{R}^{(L−1)×L}$. 
How can I go about finding $\mathbf{D}$?
 A: You don't say so, but I presume that $q_N$ is the squared norm (length) of the vector $y$, i.e. $q_N = y^Hy$. I also presume that the "nullspace of $y$" refers to its orthogonal complement.
The fact that $D^HD = I - \frac 1{q_N}yy^T$ is a consequence of the fact that the rows $D$ form an orthonormal basis for the orthogonal complement of $y$.
Indeed, the fact that the rows of $D$ form an orthonormal basis for the orthogonal complement of $y$ means that $DD^H$ will be the orthogonal projection matrix that projects onto this orthogonal complement. In general, for a matrix $M$ with orthonormal rows, $MM^H$ orthogonally projects onto the span of these rows.
On the other hand, from the fact that $\frac 1{q_{N}} yy^T$ is the orthogonal projection onto the span of $y$, it follows that $I - \frac 1{q_{N}} yy^T$ is the orthogonal projection onto the orthogonal complement of the span of $y$.
Since $DD^H$ and $I - \frac 1{q_n}yy^T$ are orthogonal projection matrix that project onto the same subspace, they must be the same matrix.
A: To find a matrix $\mathbf{D}$ which satisfies $\mathbf{Dy = 0}$, you take the outer product of $\mathbf{y}$ with itself to create a rank one matrix
$\mathbf{Y}=\mathbf{y}\mathbf{y}^T$. 
Taking the singular value decomposition of $\mathbf{Y}$ yields $\mathbf{Y}=\mathbf{U}\mathbf{\Sigma}\mathbf{V}^T$. The first row of $\mathbf{V}^T$ contains one orthonormal basis for the row space of $\mathbf{Y}$ (because $\mathbf{Y}$ is rank 1). The $L-1$ remaining rows contain orthonormal bases for the nullspace of $\mathbf{Y}$.
To get to a suitable matrix $\mathbf{D}$ you simply discard the first row of $\mathbf{V}^T$ and the resulting $\mathbb{R}^{(L-1)\times L}$ matrix satisfies $\mathbf{Dy = 0}$.
