Try this with A =
0 1 0 0
Note that A^2 = 0.
In general, ker(B) is contained in ker(AB) and im(BA) is contained in im(B). Try these out with B = A^k.
im(B) is contained in ker(A) if and only if AB = 0.
The intersection of two subspaces is always a subspace -- go back to the definition and check that: (0) 0 is in both (1) if v, w are in both the v+w is in both (2) if v is in both and c is a scalar then cv is in both.
Think about the union of the coordinate axes in the plane. It is not a subspace of the plane -- closure under + fails.
The columns of a SQUARE MATRIX are independent if and only if the rref is the identity. In general, when the columns are independent you get what looks like an identity at the top and then some rows of zeros. The columns cannot be independent unless the number of rows is at least as large as the number of columns.
1 2 3 6
the columns are dependent. The rref is
1 2 0 0
which has the same relation on its columns: the second is twice the first in both cases. Passing to the rref typically changes the column space but not the relations that hold among the columns.
Finding a basis for a subspace amounts to finding the smallest number of vectors that span (they are then indep. or you can drop at least one). It is also amounts to finding the largest number in the subspace that are indep. (These must span: a vector outside the span but in the subspace will enable you to enlarge the indep. set.) The smallest number that span and the largest number that are indep. both give the DIMENSION of the subspace.