Let $Ax = 0$ be a linear equation system. Assume that we already have a non-trivial solution $\tilde{x}$ which solves the linear equation system $A^t x = 0$ where $A^t$ is the transposed matrix of $A$.
Question Is there a way of obtaining a non-trivial solution of $Ax=0$ by using $\tilde{x}$ explicitly?
I do not only want the existence of such a non-trivial solution (which is fairly easy to show by using the determinant). I would like to compute a solution directly from $\tilde{x}$. Is there a way to do so?
Thanks in advance!
There can be no such method. In fact, we can construct an $A$ that has solutions $Ax = 0$ and $A^Ty = 0$ for any pair of vectors $x,y$.
For instance, we can use the following construction (motivated by singular value decomposition). Suppose $A$ is $m \times n$, and let $d = \min\{m,n\}$. Let $u_1,\dots,u_{d-1}$ be a collection of orthonormal unit vectors perpendicular to $x$, and let $v_1,\dots,v_{d-1}$ be a collection of orthonormal unit vectors perpendicular to $y$. Let $U$ be the matrix with columns $u_1,\dots,u_{d-1}$, and let $V$ be the matrix with columns $v_1,\dots,v_{d-1}$. Then the matrix $A = VDU^T$ will satisfy $Ax = 0$ and $A^Ty = 0$ whenever $D$ is a $(d-1) \times (d-1)$ diagonal matrix.