What is the connection between the invertibility of a matrix the kernel of the matrix If the kernel of a matrix A only has the zero vector (ker(A) = {0}) 
Then why is the matrix A considered invertible because of that?
I know that if ker(A) = 0, then there is only one unique solution of the matrix( assuming that the matrix is made up of a system of linear equations) but what does this tell me about whether or not the matrix is singular?
Can someone explain the logic behind this?
 A: A matrix is invertible if there is a unique solution to the problem $Ax=b$. If there is another vector $x$ besides zero where $Ax=0$ then there is not a unique solution to that problem.
A: The word "invertible" and its synonym "non-singular" are reserved for linear systems that have unique solutions. Say that the system $Ax = b$ has a solution. Denote such a solution by $x_0.$ Let $x_h$ (the $h$ stands for "homogenous") denote a nonzero solution to the system $Ax =0$ (called the "homogenous system"). It is true that $x_0$ and $x_0 + x_h$ are two separate solutions to the system $Ax=b.$ So, if there is any nonzero vector in $\ker A,$ then $Ax=b$ has two separate solutions, meaning that it cannot be invertible.
A: In general, for any function $f:V\to W$ that takes vectors to vectors to have an inverse, for every vector $b\in W$, there must be exactly one element $x\in V$ such that $f(x)=b$. Applying this to the function $f: V\to V$ given by $f: x\mapsto Ax$, this means that for $f$ to have an inverse the system $Ax=b$ must have a unique solution for every $b\in W$. This is precisely the case when the null space of $A$ is trivial.
