Very confused about linear Algebra My confusion is a result from the methodology used to solve some problems. 
As an example, in order to find the kernel I set $Ax=0$ and the way I find a basis I also set $Ax=0$ and to find something that is linearly independent I also set $Ax=0$, I also set $Ax=0$ to find the range, to find if a transformation is one to one. For any linear algebra problem I have better chances of solving it correctly if I set $Ax=0$. 
I believe that this is related to the fact that I am trying to show that the vectors are linearly independent or dependent. But I don't understand why there are so many different names for essentially the same thing. 
I am not looking at it from a specific definition point of view but from a practical stand-point. I am sure there are differences in the definitions of basis and kernel, but for both we try to find and show the same thing which make them as concepts equivalent. 
Please help me understand. 
 A: All these things are very closely related (except for one which you got a bit wrong - see below) and that is why they are all solved by looking at $A{\bf x}={\bf0}$.  But they all put the emphasis on differing aspects of the subject.  One of the best ways to get a grip on linear algebra is to work on understanding why these things are all so similar.  IMHO your question is an excellent sign that you are trying to do this (so +1).
Let's look at some specifics.  You find the kernel of $A$ by solving $A{\bf x}={\bf0}$, because this is what the kernel means:
$${\rm kernel}(A)=\{\,{\bf x}\,\mid\,A{\bf x}={\bf0}\,\}\ .$$
You can show that the columns of $A$ are linearly independent by looking at $A{\bf x}={\bf0}$.  There is a slight difference here though: in finding the kernel you need to determine the solutions of the system, whatever they are; in proving linear independence (or not) you need to show that the only solution is the zero vector (or not).  If we call the columns ${\bf a}_1,\ldots,{\bf a}_n$, saying that they are independent means that
$$\hbox{the system $x_1{\bf a}_1+\cdots+x_n{\bf a}_n={\bf0}$ has a unique solution}.$$
So for this problem you don't have to actually find the values of $x_1,\ldots,x_n$, you just have to go far enough to see whether or not there is a unique solution.
There is an important theorem which says that
$$\hbox{the transformation $T({\bf x})=A{\bf x}$ is one-to-one if and only if ${\rm kernel}(A)=\{\,{\bf0}\,\}$}$$
and this is why one-to-one is related to the kernel.  It is not quite true to say that you will always "find a basis" by looking at $A{\bf x}={\bf0}$: this will enable you to find a basis of the kernel of $A$, and again, that should not be surprising since this is what the kernel means.
You cannot find the range of $A$ by solving $A{\bf x}={\bf0}$.  To do this you will look at $A{\bf x}={\bf b}$, and find conditions on the components of ${\bf b}$ such that the system has a solution.  Another way is to reduce $A$ to echelon form and then look at the leading columns, giving a basis for the range; but this is only looking at the matrix $A$ itself, not the equation $A{\bf x}={\bf0}$.  Also, if you look into why this method works, you will find that although we are not writing down the vector ${\bf b}$ on the right hand side, there is an "implied" ("suppressed" maybe?) right hand side.
Hope this helps.  If it still seems that all these things are much the same, well, that's a fair enough point of view.  As I said at the top, a lot of this concerns changes of emphasis rather than anything else.
A: When talking about square matrices, the following three statements are equivalent:


*

*The kernel of a matrix is trivial

*The transformation represented by the matrix is one to one

*The columns of the matrix are linearly independent

*The rows of the matrix are linearly independent.


Even for non square matrices, the kernel of a matrix is trivial iff the columns of the matrix are linearly independent.
So, when you are saying that whatever you are doing, setting $Ax=0$ is the way to go, you really stumbled on this hidden fact: all three things you were doing were, at least to some extent, equivalent.
