Can we say something about the rows of the null space matrix

Question

I have $x_1=Mx_2$ where $x_1\in\mathbb{C}^{q\times1}$, $M\in\mathbb{C}^{q\times p}$ and $x\in\mathbb{C}^{p\times1}$.

Then we have $M:\mathbb{C}^{p}\longrightarrow \rm{Range}(M)$ and we know that $\mathbb{C}^p = \rm{Range}(M^T) \oplus \rm{Ker}(M)$ where $\rm{dim}(Ker(M))=s$. Let $X=Null(M)$ be the $p\times s$ matrix that contains the $s$ vectors that span the subspace $\rm{Ker}(M)$.

In matlab I have computed $x_{\rm{Range}(M^T)}$ so I can write $x_2 = x_{\rm{Range}(M^T)} + x_{\rm{ker}(M)}$. That is, $x_2 = x_{\rm{Range}(M^T)} + Xa$ where $a$ is a vector that contains the coefficients of the linear combination in the subspace $\rm{Ker}(M)$. Here I am trying to find the vector $a$ and I have noticed that there are exactly $s$ rows in $X$ where $x_2 = 0$.

Which means if I use a projection matrix $P\in\mathbb{C}^{p\times p}$ that has $s$ $1$'s in the diagonal and zeros elsewhere I can find $Px_2 = Px_{\rm{Range}(M^T)} + PXa = 0$, thus I can compute $a$.

Moreover, I have noticed that these $s$ rows are not unique there are many of them.

So my question is can we determine the subspace of these different sets of $s$ rows of $X$ so that I can find at least one projection matrix $P$ among many hence find the vector $a$ or at least can I find a statistical property between these $s$ rows which will help to reduce my search.

Any response would be really appreciated.

Answer 1

Notice that $Xa = x_2 - x_{\operatorname{Range}(M^T)}$. Since the columns of $X$ form a basis, there is exactly one $a$ to satisfy this. Solve it like a system of linear equations that it is. If you remove linearly dependent rows, it'll be a regular system, but you can also use QR factorization.

Then, as I wrote in my comment on your previous question, your matrix is $P = \operatorname{diag}(p_1,p_2,\dots,p_n)$, where $p_k = 0$ if $(x_2)_k \ne 0$. You can choose the values of $p_k$ for $(x_2)_k = 0$ as you see fit.

Now, I don't see why would $x_2$ have any zeros, but you claim that they do, so I'm guessing that it's not a problem.