I have a squared real matrix $\mathbf{A} = A_{ij}$ and I define the vector $\mathbf{v}$ as a solution of the homogeneous system $\mathbf{A}\cdot \mathbf{v} = 0$. I assume that $\mathbf{v}$ is not the null vector, and thus $\mathbf{A}$ is singular. The vector $\mathbf{v}$ can be thought of as a function of the variables $A_{ij}$. So, for each choice of the parameters $A_{ij}$, I have a different $\mathbf{v}$ solution of $\mathbf{A}\cdot \mathbf{v} = 0$.
My question is: is there a "general" way of calculating the derivative $\frac{\partial \mathbf{v}}{\partial A_{ij}}$ ? That is, how the solution of the homogenous system changes for small variations of each of the components of the matrix $\mathbf{A}$.
I naively tried to calculate it by doing
$$ \frac{\partial}{\partial A_{ij}} \left(\mathbf{A}\cdot \mathbf{v} \right) = 0 \rightarrow I^{(ij)} \mathbf{v} + \mathbf{A}\cdot \frac{\partial \mathbf{v}}{\partial A_{ij}} = 0 \rightarrow \frac{\partial \mathbf{v}}{\partial A_{ij}} = \mathbf{A}^{-1} I^{(ij)} \mathbf{v}$$ (where I defined the matrix $I^{(ij)}$ as a matrix with all entries equal to zero, except for the entry $(ij)$ which is equal to 1). However, this is not very useful because $\mathbf{A}$ is singular and thus $\mathbf{A}^{-1}$ does not exist.
If $A$ is singular, the solution of $A v = 0$ is non-unique: you can always replace $v$ by $tv$ for any scalar $t$.
Also, in general a small change to $A_{ij}$ is likely to make your matrix nonsingular, in which case $v$ immediately becomes $0$.
In fact, the only case where changing $A_{ij}$ slightly does not make the matrix nonsingular is when the minor $M_{ij}$, the determinant of the matrix obtained from $A$ by removing row $i$ and column $j$, is $0$.
Here's an example of what you can do. For convenience I'll take $i=j=n$. Let your matrix be expressed in block form as $$ A = \pmatrix{B & b\cr c & x}$$ where $B$ is $(n-1) \times (n-1)$ and is singular. To address the first problem, I'll require $v$ to have its last entry $1$, thus $v = \pmatrix{u\cr 1}$. Now we want $$ A v = \pmatrix{B & b\cr c & x\cr} \pmatrix{u\cr 1\cr} = \pmatrix{B u + b\cr c u + x} = \pmatrix{0\cr 0\cr}$$ Since $B$ is singular, the solution of $B u + b = 0$ will be nonunique (if it exists). I'll suppose the null space of $B$ has dimension $1$, so we can write $u = u_p + t u_h$ where $u_p$ is a particular solution and $u_h$ spans the null space. In order for $v$ to be a solution of $Av = 0$ we then need $c (u_p + t u_h) + x = 0$, i.e. assuming $c u_h \ne 0$, $$t = \frac{-c u_p - x}{c u_h}$$ .
So in this case $$ \dfrac{\partial v}{\partial x} = \frac{-1}{c u_h}\pmatrix{u_h\cr 0} $$