Here , that means you are looking for <a, b, c, d> such that
That gives you the equations -a+ d= 0, a- b= 0, -a+ c= 0, b- d= 0, and -b+ c= 0. Those do not have a single solution (if they did it would have to be a= b= c= d= 0) but can be solved for b, c, and d in terms of a. That gives a single vector basis for the kernel.
Again, use the definition of "Image". The image of linear transformation, A, is the set of all vectors, v, such that Au= v for [b]some[\b] vector u. In other words, that would be the set of all v such that Au= v has a solution. If we let then we are looking at the equationand the Image of
You can "solve" that by row reduction of the augmented matrix
If there are any rows in which the first 6 terms are all 0, then the last term, which must be in terms of a, b, c, and d, must also be 0. That gives the equations you need to find the image. If there are no rows in which the first 6 numbers are 0, the kernel is all of .
I need it to calculate the homology, ie. using Ker i / Im j