If T is linear transformation from $U \to V$ such that dim(U)=n, dim(V)=m, $T(x) = b$, and if T is onto then there is always a solution but may or may not be unique.
Can you tell me how to prove this?
$Tx = b$ means b is in range of T i.e., b is linearly dependent on linearly independent columns of T
i.e., always rank of T = rank of augmented matrix $[T|b]$
rank of T = n = number of variables = dimension of domain = dim(U), then unique solution for x; or T is 1-1 (from rank nullity theorem)
If rank of T < n, then multiple solutions
It means $dim (R(T)) < n \implies Nullity(T) = dim (U) - dim(R(T)) \geq 1$
Now able to proceed after that.
Since $T$ is onto, $m\le n$. Hence the linear system $T(x)=b$ has a solution for any $b\in V$. Such solution is unique if and only if $m=n$. You may see this by the rank-nullity theorem.
Let $v\in \text{Ker}T$, and let $b\in V$. Since $T$ is onto there exists $x\in U$ such that $T(x)=b$; on the other hand also $T(x+v)=T(x)+T(v)=b+0=b$; hence also $x+v$ is a solution. With this reasoning $x$ is unique if and only if the kernel of $T$ is trivial, hence $\dim U=\dim V$.