I've been looking around for a method to find the basis of the kernel. What I've seen to be the simple method is:
- Do REF to the matrix A.
- Take the non-pivot columns we will call them k.
- Put them in [ k | 0 ] then solve.
Now the thing is that I have trouble in figuring out how this method would work with a dimension greater than 1. How do you get 2 vectors from this method if dim ker A = 2?
I now that by doing [k | 0 ] you will get either {0} or a general solution with free variables. You then need to set the free variable that we will call r to an arbitrary value to get the vector in your kernel basis.
So, when you have dim ker A = 2 and you get a general solution with a free variable r. Do you arbitrarily set r to any value? One time to get a first particular solution that represents your first vector of your basis and a second time to get a second particular solution for you second vector of your basis.
This way you get your two vectors needed to form a base of dimension 2. This is the problem, I am not sure if this is correct.
