I just want some help with the theory behind the possible set of solutions.
So this is in linear algebra so we assume A is an m x n matrix, x is is a vector $K^{n,1}$ and b is a vector in $K^{m,1}$. Presume K is infinite and that T is the linear map with T(u) = v $\leftrightarrow$ Ax = b
So in the homogeneous case, I guess {0} must always be a solution but are there any unique (aside from 0) non-zero solutions? My intuition says no in that if x is a non zero solution in the null space of A, then the kernel must have dimension greater than 0 but this leads to 'infinite' solutions?
For the general case where b is not 0, can someone check my reasoning in the 3 cases?
Case 1: No solutions. So in the scenario where b was 0, the 'image' of T(u) (the corresponding linear map) would be the kernel thus there would always be at least one solution (0). Whilst if b is not 0, then T(u) = v, so if $v \notin Im(T)$ there is no solutions.
Case 2: Suppose we have found a solution x of the equation. Note that if $y \in null space(A)$ then x+y is also a solution since x solves it and y gives an extra 0. Thus there is only a unique solution if nullspace = {0}.
Case 3: As above, but if nullspace has dimension greater than 0, then there are infinite 'y' that can be added to x to form solutions.
Thanks for the help
In the case $Ax=0$, $x=0$ is always a solution. If there is some nonzero $x_0$ with $Ax_0=0$, then the same is true of $cx_0$ for any scalar $c$. Thus if the base field is infinite then there are infinitely many solutions.
In the case $Ax=b$, the solutions are given by any solution $x_p$ that you choose, plus any $x_n$ with $Ax_n=0$. Thus either there are no solutions at all, or the space of solutions forms an affine space (it's not linear, since it doesn't contain $0$) with the same dimension as the kernel of $A$. Again if the base field is infinite and the kernel of $A$ has positive dimension, then this affine space is also infinite.
I think everything here agrees with what you've said except perhaps for the issue of whether $Ax=b$ has any solutions.